如何将逗号分隔的值拆分为列?

时间:2022-01-04 00:16:17

I have a table like this

我有一张这样的桌子。

Value   String
-------------------
1       Cleo, Smith

I want to separate the comma delimited string into two columns

我想把逗号分隔的字符串分隔成两列。

Value  Name Surname
-------------------
1      Cleo   Smith

I need only two fixed extra columns

我只需要两个固定的额外列。

30 个解决方案

#1


-2  

ALTER FUNCTION [dbo].[StringListTo] (@StringList Nvarchar(max),@Separators char(1),@start int, @index int )
RETURNS nvarchar(max)
AS
BEGIN
declare @out Nvarchar(max)
declare @i int
declare @start_old int
set @start=@start+1
set @i=1
while(@i<=@index)
begin
    set @start_old=@start
    set @start=CHARINDEX('.',@StringList,@start+1)
    if (@start>0)
    begin
        set @out=Substring(@StringList,@start_old+1,@start-@start_old-1)
    end
else
begin
    set @out=Substring(@StringList,@start_old+1,len(@StringList)-1)
end
set @i=@i+1
end
RETURN @out
END;

#2


100  

Your purpose can be solved using following query -

您的目的可以通过以下查询来解决。

Select Value  , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as  Surname
from Table1

There is no readymade Split function in sql server, so we need to create user defined function.

在sql server中没有现成的拆分函数,因此我们需要创建用户定义的函数。

CREATE FUNCTION Split (
      @InputString                  VARCHAR(8000),
      @Delimiter                    VARCHAR(50)
)

RETURNS @Items TABLE (
      Item                          VARCHAR(8000)
)

AS
BEGIN
      IF @Delimiter = ' '
      BEGIN
            SET @Delimiter = ','
            SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
      END

      IF (@Delimiter IS NULL OR @Delimiter = '')
            SET @Delimiter = ','

--INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic
--INSERT INTO @Items VALUES (@InputString) -- Diagnostic

      DECLARE @Item           VARCHAR(8000)
      DECLARE @ItemList       VARCHAR(8000)
      DECLARE @DelimIndex     INT

      SET @ItemList = @InputString
      SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      WHILE (@DelimIndex != 0)
      BEGIN
            SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
            INSERT INTO @Items VALUES (@Item)

            -- Set @ItemList = @ItemList minus one less item
            SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
            SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      END -- End WHILE

      IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString
      BEGIN
            SET @Item = @ItemList
            INSERT INTO @Items VALUES (@Item)
      END

      -- No delimiters were encountered in @InputString, so just return @InputString
      ELSE INSERT INTO @Items VALUES (@InputString)

      RETURN

END -- End Function
GO

---- Set Permissions
--GRANT SELECT ON Split TO UserRole1
--GRANT SELECT ON Split TO UserRole2
--GO

#3


32  

;WITH Split_Names (Value,Name, xmlname)
AS
(
    SELECT Value,
    Name,
    CONVERT(XML,'<Names><name>'  
    + REPLACE(Name,',', '</name><name>') + '</name></Names>') AS xmlname
      FROM tblnames
)

 SELECT Value,      
 xmlname.value('/Names[1]/name[1]','varchar(100)') AS Name,    
 xmlname.value('/Names[1]/name[2]','varchar(100)') AS Surname
 FROM Split_Names

and also check the link below for reference

并检查下面的链接以供参考。

http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html

http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html

#4


29  

xml base answer is simple and clean

xml基本答案简单明了。

refer this

请参考这个

DECLARE @S varchar(max),
        @Split char(1),
        @X xml

SELECT @S = 'ab,cd,ef,gh,ij',
       @Split = ','

SELECT @X = CONVERT(xml,' <root> <myvalue>' +
REPLACE(@S,@Split,'</myvalue> <myvalue>') + '</myvalue>   </root> ')

SELECT  T.c.value('.','varchar(20)'),--retrieve all values at once
  T.c.value('(/root/myvalue)[1]','VARCHAR(20)')  , --retrieve index 1 only, which is the 'ab'
  T.c.value('(/root/myvalue)[2]','VARCHAR(20)'),
  T.c.value('(/root/myvalue)[3]','VARCHAR(20)')
 FROM @X.nodes('/root/myvalue') T(c)

#5


24  

With CROSS APPLY

与交叉应用

select ParsedData.* 
from MyTable mt
cross apply ( select str = mt.String + ',,' ) f1
cross apply ( select p1 = charindex( ',', str ) ) ap1
cross apply ( select p2 = charindex( ',', str, p1 + 1 ) ) ap2
cross apply ( select Nmame = substring( str, 1, p1-1 )                   
                 , Surname = substring( str, p1+1, p2-p1-1 )
          ) ParsedData

#6


20  

I think this is cool

我觉得这很酷。

SELECT value,
    PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
    PARSENAME(REPLACE(String,',','.'),1) 'Sur Name'
FROM table WITH (NOLOCK)

#7


13  

Try this (change instances of ' ' to ',' or whatever delimiter you want to use)

试试这个(更改“to”的实例,或者你想使用的任何分隔符)

CREATE FUNCTION dbo.Wordparser
(
  @multiwordstring VARCHAR(255),
  @wordnumber      NUMERIC
)
returns VARCHAR(255)
AS
  BEGIN
      DECLARE @remainingstring VARCHAR(255)
      SET @remainingstring=@multiwordstring

      DECLARE @numberofwords NUMERIC
      SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1)

      DECLARE @word VARCHAR(50)
      DECLARE @parsedwords TABLE
      (
         line NUMERIC IDENTITY(1, 1),
         word VARCHAR(255)
      )

      WHILE @numberofwords > 1
        BEGIN
            SET @word=LEFT(@remainingstring, CHARINDEX(' ', @remainingstring) - 1)

            INSERT INTO @parsedwords(word)
            SELECT @word

            SET @remainingstring= REPLACE(@remainingstring, Concat(@word, ' '), '')
            SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1)

            IF @numberofwords = 1
              BREAK

            ELSE
              CONTINUE
        END

      IF @numberofwords = 1
        SELECT @word = @remainingstring
      INSERT INTO @parsedwords(word)
      SELECT @word

      RETURN
        (SELECT word
         FROM   @parsedwords
         WHERE  line = @wordnumber)

  END

Example usage:

使用示例:

SELECT dbo.Wordparser(COLUMN, 1),
       dbo.Wordparser(COLUMN, 2),
       dbo.Wordparser(COLUMN, 3)
FROM   TABLE

#8


11  

There are multiple ways to solve this and many different ways have been proposed already. Simplest would be to use LEFT / SUBSTRING and other string functions to achieve the desired result.

有多种方法可以解决这个问题,已经提出了许多不同的方法。最简单的方法是使用左/子字符串和其他字符串函数来实现预期的结果。

Sample Data

样本数据

DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))

INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');

Using String Functions like LEFT

使用像LEFT这样的字符串函数。

SELECT
    Value,
    LEFT(String,CHARINDEX(',',String)-1) as Fname,
    LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1

This approach fails if there are more 2 items in a String. In such a scenario, we can use a splitter and then use PIVOT or convert the string into an XML and use .nodes to get string items. XML based solution have been detailed out by aads and bvr in their solution.

如果字符串中有更多2项,则此方法失败。在这种情况下,我们可以使用splitter,然后使用PIVOT或将字符串转换为XML,并使用.nodes获取字符串项。基于XML的解决方案已经在其解决方案中被aads和bvr详细描述。

The answers for this question which use splitter, all use WHILE which is inefficient for splitting. Check this performance comparison. One of the best splitters around is DelimitedSplit8K, created by Jeff Moden. You can read more about it here

对于这个问题的答案,使用分配器,所有的使用,而这是低效的分裂。检查这个性能比较。其中最好的分割器是由Jeff Moden创建的DelimitedSplit8K。你可以在这里读到更多。

Splitter with PIVOT

分束器和主

DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))

INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');


SELECT t3.Value,[1] as Fname,[2] as Lname
FROM @tbl1 as t1
CROSS APPLY [dbo].[DelimitedSplit8K](String,',') as t2
PIVOT(MAX(Item) FOR ItemNumber IN ([1],[2])) as t3

Output

输出

Value   Fname   Lname
1   Cleo    Smith
2   John    Mathew

DelimitedSplit8K by Jeff Moden

由杰夫DelimitedSplit8K语音识别

CREATE FUNCTION [dbo].[DelimitedSplit8K]
/**********************************************************************************************************************
 Purpose:
 Split a given string at a given delimiter and return a list of the split elements (items).

 Notes:
 1.  Leading a trailing delimiters are treated as if an empty string element were present.
 2.  Consecutive delimiters are treated as if an empty string element were present between them.
 3.  Except when spaces are used as a delimiter, all spaces present in each element are preserved.

 Returns:
 iTVF containing the following:
 ItemNumber = Element position of Item as a BIGINT (not converted to INT to eliminate a CAST)
 Item       = Element value as a VARCHAR(8000)

 Statistics on this function may be found at the following URL:
 http://www.sqlservercentral.com/Forums/Topic1101315-203-4.aspx

 CROSS APPLY Usage Examples and Tests:
--=====================================================================================================================
-- TEST 1:
-- This tests for various possible conditions in a string using a comma as the delimiter.  The expected results are
-- laid out in the comments
--=====================================================================================================================
--===== Conditionally drop the test tables to make reruns easier for testing.
     -- (this is NOT a part of the solution)
     IF OBJECT_ID('tempdb..#JBMTest') IS NOT NULL DROP TABLE #JBMTest
;
--===== Create and populate a test table on the fly (this is NOT a part of the solution).
     -- In the following comments, "b" is a blank and "E" is an element in the left to right order.
     -- Double Quotes are used to encapsulate the output of "Item" so that you can see that all blanks
     -- are preserved no matter where they may appear.
 SELECT *
   INTO #JBMTest
   FROM (                                               --# & type of Return Row(s)
         SELECT  0, NULL                      UNION ALL --1 NULL
         SELECT  1, SPACE(0)                  UNION ALL --1 b (Empty String)
         SELECT  2, SPACE(1)                  UNION ALL --1 b (1 space)
         SELECT  3, SPACE(5)                  UNION ALL --1 b (5 spaces)
         SELECT  4, ','                       UNION ALL --2 b b (both are empty strings)
         SELECT  5, '55555'                   UNION ALL --1 E
         SELECT  6, ',55555'                  UNION ALL --2 b E
         SELECT  7, ',55555,'                 UNION ALL --3 b E b
         SELECT  8, '55555,'                  UNION ALL --2 b B
         SELECT  9, '55555,1'                 UNION ALL --2 E E
         SELECT 10, '1,55555'                 UNION ALL --2 E E
         SELECT 11, '55555,4444,333,22,1'     UNION ALL --5 E E E E E 
         SELECT 12, '55555,4444,,333,22,1'    UNION ALL --6 E E b E E E
         SELECT 13, ',55555,4444,,333,22,1,'  UNION ALL --8 b E E b E E E b
         SELECT 14, ',55555,4444,,,333,22,1,' UNION ALL --9 b E E b b E E E b
         SELECT 15, ' 4444,55555 '            UNION ALL --2 E (w/Leading Space) E (w/Trailing Space)
         SELECT 16, 'This,is,a,test.'                   --E E E E
        ) d (SomeID, SomeValue)
;
--===== Split the CSV column for the whole table using CROSS APPLY (this is the solution)
 SELECT test.SomeID, test.SomeValue, split.ItemNumber, Item = QUOTENAME(split.Item,'"')
   FROM #JBMTest test
  CROSS APPLY dbo.DelimitedSplit8K(test.SomeValue,',') split
;
--=====================================================================================================================
-- TEST 2:
-- This tests for various "alpha" splits and COLLATION using all ASCII characters from 0 to 255 as a delimiter against
-- a given string.  Note that not all of the delimiters will be visible and some will show up as tiny squares because
-- they are "control" characters.  More specifically, this test will show you what happens to various non-accented 
-- letters for your given collation depending on the delimiter you chose.
--=====================================================================================================================
WITH 
cteBuildAllCharacters (String,Delimiter) AS 
(
 SELECT TOP 256 
        'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789',
        CHAR(ROW_NUMBER() OVER (ORDER BY (SELECT NULL))-1)
   FROM master.sys.all_columns
)
 SELECT ASCII_Value = ASCII(c.Delimiter), c.Delimiter, split.ItemNumber, Item = QUOTENAME(split.Item,'"')
   FROM cteBuildAllCharacters c
  CROSS APPLY dbo.DelimitedSplit8K(c.String,c.Delimiter) split
  ORDER BY ASCII_Value, split.ItemNumber
;
-----------------------------------------------------------------------------------------------------------------------
 Other Notes:
 1. Optimized for VARCHAR(8000) or less.  No testing or error reporting for truncation at 8000 characters is done.
 2. Optimized for single character delimiter.  Multi-character delimiters should be resolvedexternally from this 
    function.
 3. Optimized for use with CROSS APPLY.
 4. Does not "trim" elements just in case leading or trailing blanks are intended.
 5. If you don't know how a Tally table can be used to replace loops, please see the following...
    http://www.sqlservercentral.com/articles/T-SQL/62867/
 6. Changing this function to use NVARCHAR(MAX) will cause it to run twice as slow.  It's just the nature of 
    VARCHAR(MAX) whether it fits in-row or not.
 7. Multi-machine testing for the method of using UNPIVOT instead of 10 SELECT/UNION ALLs shows that the UNPIVOT method
    is quite machine dependent and can slow things down quite a bit.
-----------------------------------------------------------------------------------------------------------------------
 Credits:
 This code is the product of many people's efforts including but not limited to the following:
 cteTally concept originally by Iztek Ben Gan and "decimalized" by Lynn Pettis (and others) for a bit of extra speed
 and finally redacted by Jeff Moden for a different slant on readability and compactness. Hat's off to Paul White for
 his simple explanations of CROSS APPLY and for his detailed testing efforts. Last but not least, thanks to
 Ron "BitBucket" McCullough and Wayne Sheffield for their extreme performance testing across multiple machines and
 versions of SQL Server.  The latest improvement brought an additional 15-20% improvement over Rev 05.  Special thanks
 to "Nadrek" and "peter-757102" (aka Peter de Heer) for bringing such improvements to light.  Nadrek's original
 improvement brought about a 10% performance gain and Peter followed that up with the content of Rev 07.  

 I also thank whoever wrote the first article I ever saw on "numbers tables" which is located at the following URL
 and to Adam Machanic for leading me to it many years ago.
 http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-numbers-table.html
-----------------------------------------------------------------------------------------------------------------------
 Revision History:
 Rev 00 - 20 Jan 2010 - Concept for inline cteTally: Lynn Pettis and others.
                        Redaction/Implementation: Jeff Moden 
        - Base 10 redaction and reduction for CTE.  (Total rewrite)

 Rev 01 - 13 Mar 2010 - Jeff Moden
        - Removed one additional concatenation and one subtraction from the SUBSTRING in the SELECT List for that tiny
          bit of extra speed.

 Rev 02 - 14 Apr 2010 - Jeff Moden
        - No code changes.  Added CROSS APPLY usage example to the header, some additional credits, and extra 
          documentation.

 Rev 03 - 18 Apr 2010 - Jeff Moden
        - No code changes.  Added notes 7, 8, and 9 about certain "optimizations" that don't actually work for this
          type of function.

 Rev 04 - 29 Jun 2010 - Jeff Moden
        - Added WITH SCHEMABINDING thanks to a note by Paul White.  This prevents an unnecessary "Table Spool" when the
          function is used in an UPDATE statement even though the function makes no external references.

 Rev 05 - 02 Apr 2011 - Jeff Moden
        - Rewritten for extreme performance improvement especially for larger strings approaching the 8K boundary and
          for strings that have wider elements.  The redaction of this code involved removing ALL concatenation of 
          delimiters, optimization of the maximum "N" value by using TOP instead of including it in the WHERE clause,
          and the reduction of all previous calculations (thanks to the switch to a "zero based" cteTally) to just one 
          instance of one add and one instance of a subtract. The length calculation for the final element (not 
          followed by a delimiter) in the string to be split has been greatly simplified by using the ISNULL/NULLIF 
          combination to determine when the CHARINDEX returned a 0 which indicates there are no more delimiters to be
          had or to start with. Depending on the width of the elements, this code is between 4 and 8 times faster on a
          single CPU box than the original code especially near the 8K boundary.
        - Modified comments to include more sanity checks on the usage example, etc.
        - Removed "other" notes 8 and 9 as they were no longer applicable.

 Rev 06 - 12 Apr 2011 - Jeff Moden
        - Based on a suggestion by Ron "Bitbucket" McCullough, additional test rows were added to the sample code and
          the code was changed to encapsulate the output in pipes so that spaces and empty strings could be perceived 
          in the output.  The first "Notes" section was added.  Finally, an extra test was added to the comments above.

 Rev 07 - 06 May 2011 - Peter de Heer, a further 15-20% performance enhancement has been discovered and incorporated 
          into this code which also eliminated the need for a "zero" position in the cteTally table. 
**********************************************************************************************************************/
--===== Define I/O parameters
        (@pString VARCHAR(8000), @pDelimiter CHAR(1))
RETURNS TABLE WITH SCHEMABINDING AS
 RETURN
--===== "Inline" CTE Driven "Tally Table" produces values from 0 up to 10,000...
     -- enough to cover NVARCHAR(4000)
  WITH E1(N) AS (
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
                ),                          --10E+1 or 10 rows
       E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
       E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
 cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
                     -- for both a performance gain and prevention of accidental "overruns"
                 SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
                ),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
                 SELECT 1 UNION ALL
                 SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
                ),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
                 SELECT s.N1,
                        ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
                   FROM cteStart s
                )
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
 SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
        Item       = SUBSTRING(@pString, l.N1, l.L1)
   FROM cteLen l
;

GO

#9


10  

With SQL Server 2016 we can use string_split to accomplish this:

使用SQL Server 2016,我们可以使用string_split来完成这个任务:

create table commasep (
 id int identity(1,1)
 ,string nvarchar(100) )

insert into commasep (string) values ('John, Adam'), ('test1,test2,test3')

select id, [value] as String from commasep 
 cross apply string_split(string,',')

#10


9  

I think PARSENAME is the neat function to use for this example, as described in this article: http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/

我认为PARSENAME是这个示例中使用的简洁函数,如本文所描述的:http://www.sqlshack.com/parsing/rotating-delimiteddata -sql-server-2012/。

The PARSENAME function is logically designed to parse four-part object names. The nice thing about PARSENAME is that it’s not limited to parsing just SQL Server four-part object names – it will parse any function or string data that is delimited by dots.

PARSENAME函数在逻辑上被设计用来解析四部分对象名称。PARSENAME的优点在于,它不仅限于解析SQL Server四部分对象名称——它将解析任何由点分隔的函数或字符串数据。

The first parameter is the object to parse, and the second is the integer value of the object piece to return. The article is discussing parsing and rotating delimited data - company phone numbers, but it can be used to parse name/surname data also.

第一个参数是要解析的对象,第二个参数是要返回的对象块的整数值。本文讨论了解析和旋转分隔数据—公司电话号码,但也可以用来解析姓名/姓氏数据。

Example:

例子:

USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';

The article also describes using a Common Table Expression (CTE) called ‘replaceChars’, to run PARSENAME against the delimiter-replaced values. A CTE is useful for returning a temporary view or result set.

本文还描述了使用一个名为“replaceChars”的公共表表达式来运行PARSENAME,以防止被分隔符替换的值。CTE用于返回临时视图或结果集。

After that, the UNPIVOT function has been used to convert some columns into rows; SUBSTRING and CHARINDEX functions have been used for cleaning up the inconsistencies in the data, and the LAG function (new for SQL Server 2012) has been used in the end, as it allows referencing of previous records.

在那之后,将一些列转换成行来使用UNPIVOT函数;SUBSTRING和CHARINDEX函数用于清理数据中的不一致性,最后使用滞后函数(SQL Server 2012的新功能),因为它允许引用以前的记录。

#11


8  

We can create a function as this

我们可以像这样创建一个函数。

CREATE Function [dbo].[fn_CSVToTable] 
(
    @CSVList Varchar(max)
)
RETURNS @Table TABLE (ColumnData VARCHAR(100))
AS
BEGIN
    IF RIGHT(@CSVList, 1) <> ','
    SELECT @CSVList = @CSVList + ','

    DECLARE @Pos    BIGINT,
            @OldPos BIGINT
    SELECT  @Pos    = 1,
            @OldPos = 1

    WHILE   @Pos < LEN(@CSVList)
        BEGIN
            SELECT  @Pos = CHARINDEX(',', @CSVList, @OldPos)
            INSERT INTO @Table
            SELECT  LTRIM(RTRIM(SUBSTRING(@CSVList, @OldPos, @Pos - @OldPos))) Col001

            SELECT  @OldPos = @Pos + 1
        END

    RETURN
END

We can then seperate the CSV values into our respective columns using a SELECT statement

然后,我们可以使用SELECT语句将CSV值分离到各自的列中。

#12


6  

select id,SUBSTRING(name,0,charindex(',',name))as firstname
,SUBSTRING(name,charindex(',',name),len(name)+1)as lastname from spilt

#13


5  

Using instring function :)

使用instring功能:)

select Value, 
       substring(String,1,instr(String," ") -1) Fname,  
       substring(String,instr(String,",") +1) Sname 
from tablename;

Used two functions,
1. substring(string, position, length) ==> returns string from positon to length
2. instr(string,pattern) ==> returns position of pattern.

使用两个函数,1。子字符串(string, position, length) ==>从positon返回字符串长度为2。instr(string,pattern) ==>返回模式的位置。

If we don’t provide length argument in substring it returns until end of string

如果在子字符串中不提供长度参数,则返回到字符串结束。

#14


4  

Use Parsename() function

使用Parsename()函数

with cte as(
    select 'Aria,Karimi' as FullName
    Union
    select 'Joe,Karimi' as FullName
    Union
    select 'Bab,Karimi' as FullName
)

SELECT PARSENAME(REPLACE(FullName,',','.'),2) as Name, 
       PARSENAME(REPLACE(FullName,',','.'),1) as Family
    FROM cte

Result

结果

Name    Family
-----   ------
Aria    Karimi
Bab     Karimi
Joe     Karimi

#15


4  

DECLARE @INPUT VARCHAR (MAX)='N,A,R,E,N,D,R,A'
DECLARE @ELIMINATE_CHAR CHAR (1)=','
DECLARE @L_START INT=1
DECLARE @L_END INT=(SELECT LEN (@INPUT))
DECLARE @OUTPUT CHAR (1)

WHILE @L_START <=@L_END
BEGIN
    SET @OUTPUT=(SUBSTRING (@INPUT,@L_START,1))
    IF @OUTPUT!=@ELIMINATE_CHAR
    BEGIN
        PRINT @OUTPUT
    END
    SET @L_START=@L_START+1
END

#16


3  

I encountered a similar problem but a complex one and since this is the first thread i found regarding that issue i decided to post my finding. i know it is complex solution to a simple problem but i hope that i could help other people who go to this thread looking for a more complex solution. i had to split a string containing 5 numbers (column name: levelsFeed) and to show each number in a separate column. for example: 8,1,2,2,2 should be shown as :

我遇到了一个类似的问题,但是一个复杂的问题,因为这是我发现的第一个关于这个问题的线程,我决定发布我的发现。我知道这是一个简单问题的复杂的解决方案,但我希望我可以帮助其他的人去寻找一个更复杂的解决方案。我必须拆分一个包含5个数字(列名称:levelsFeed)的字符串,并在单独的列中显示每个数字。例如:8、1、2、2、2应显示为:

1  2  3  4  5
-------------
8  1  2  2  2

Solution 1: using XML functions: this solution for the slowest solution by far

解决方案1:使用XML函数:到目前为止最慢的解决方案。

SELECT Distinct FeedbackID, 
, S.a.value('(/H/r)[1]', 'INT') AS level1
, S.a.value('(/H/r)[2]', 'INT') AS level2
, S.a.value('(/H/r)[3]', 'INT') AS level3
, S.a.value('(/H/r)[4]', 'INT') AS level4
, S.a.value('(/H/r)[5]', 'INT') AS level5
FROM (            
    SELECT *,CAST (N'<H><r>' + REPLACE(levelsFeed, ',', '</r><r>')  + '</r> </H>' AS XML) AS [vals]
    FROM Feedbacks 
)  as d
CROSS APPLY d.[vals].nodes('/H/r') S(a)

Solution 2: using Split function and pivot. (the split function split a string to rows with the column name Data)

解决方案2:使用Split函数和pivot。(split函数将一个字符串与列名称数据分隔开)

SELECT FeedbackID, [1],[2],[3],[4],[5]
FROM (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY feedbackID ORDER BY (SELECT  null)) as rn 
FROM (
    SELECT FeedbackID, levelsFeed
    FROM Feedbacks 
) as a
CROSS APPLY dbo.Split(levelsFeed, ',')
) as SourceTable
PIVOT
(
    MAX(data)
    FOR rn IN ([1],[2],[3],[4],[5])
)as pivotTable

Solution 3: using string manipulations functions - fastest by small margin over solution 2

解决方案3:使用字符串操作函数——在解决方案2上以小幅度的速度最快。

SELECT FeedbackID,
SUBSTRING(levelsFeed,0,CHARINDEX(',',levelsFeed)) AS level1,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),4) AS level2,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),3) AS level3,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),2) AS level4,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),1) AS level5
FROM Feedbacks

since the levelsFeed contains 5 string values i needed to use the substring function for the first string.

由于levelsFeed包含5个字符串值,所以我需要为第一个字符串使用子字符串函数。

i hope that my solution will help other that got to this thread looking for a more complex split to columns methods

我希望我的解决方案能够帮助其他到达这个线程的人寻找一个更复杂的列方法。

#17


3  

Try this:

试试这个:

declare @csv varchar(100) ='aaa,bb,csda,daass';
set @csv = @csv+',';

with cte as
(
    select SUBSTRING(@csv,1,charindex(',',@csv,1)-1) as val, SUBSTRING(@csv,charindex(',',@csv,1)+1,len(@csv)) as rem 
    UNION ALL
    select SUBSTRING(a.rem,1,charindex(',',a.rem,1)-1)as val, SUBSTRING(a.rem,charindex(',',a.rem,1)+1,len(A.rem)) 
    from cte a where LEN(a.rem)>=1
    ) select val from cte

#18


2  

mytable:

mytable:

Value  ColOne
--------------------
1      Cleo, Smith

The following should work if there aren't too many columns

如果没有太多的列,那么下面的内容应该是有效的。

ALTER TABLE mytable ADD ColTwo nvarchar(256);
UPDATE mytable SET ColTwo = LEFT(ColOne, Charindex(',', ColOne) - 1);
--'Cleo' = LEFT('Cleo, Smith', Charindex(',', 'Cleo, Smith') - 1)
UPDATE mytable SET ColTwo = REPLACE(ColOne, ColTwo + ',', '');
--' Smith' = REPLACE('Cleo, Smith', 'Cleo' + ',')
UPDATE mytable SET ColOne = REPLACE(ColOne, ',' + ColTwo, ''), ColTwo = LTRIM(ColTwo);
--'Cleo' = REPLACE('Cleo, Smith', ',' + ' Smith', '') 

Result:

结果:

Value  ColOne ColTwo
--------------------
1      Cleo   Smith

#19


2  

it is so easy, you can take it by below query:

它很简单,你可以在下面查询:

DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))

#20


2  

This worked for me

这为我工作

CREATE FUNCTION [dbo].[SplitString](
    @delimited NVARCHAR(MAX),
    @delimiter NVARCHAR(100)
) RETURNS @t TABLE ( val NVARCHAR(MAX))
AS
BEGIN
    DECLARE @xml XML
    SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'
    INSERT INTO @t(val)
    SELECT  r.value('.','varchar(MAX)') as item
    FROM  @xml.nodes('/t') as records(r)
    RETURN
END

#21


2  

I think following function will work for you:

我认为以下功能对你有用:

You have to create a function in SQL first. Like this

CREATE FUNCTION [dbo].[fn_split](
@str VARCHAR(MAX),
@delimiter CHAR(1)
)
RETURNS @returnTable TABLE (idx INT PRIMARY KEY IDENTITY, item VARCHAR(8000))
AS
BEGIN
DECLARE @pos INT
SELECT @str = @str + @delimiter
WHILE LEN(@str) > 0 
    BEGIN
        SELECT @pos = CHARINDEX(@delimiter,@str)
        IF @pos = 1
            INSERT @returnTable (item)
                VALUES (NULL)
        ELSE
            INSERT @returnTable (item)
                VALUES (SUBSTRING(@str, 1, @pos-1))
        SELECT @str = SUBSTRING(@str, @pos+1, LEN(@str)-@pos)       
    END
RETURN
END

You can call this function, like this:

你可以称这个函数为:

select * from fn_split('1,24,5',',')

Implementation:

Declare @test TABLE (
ID VARCHAR(200),
Data VARCHAR(200)
)

insert into @test 
(ID, Data)
Values
('1','Cleo,Smith')


insert into @test 
(ID, Data)
Values
('2','Paul,Grim')

select ID,
(select item from fn_split(Data,',') where idx in (1)) as Name ,
(select item from fn_split(Data,',') where idx in (2)) as Surname
 from @test

Result will like this:

如何将逗号分隔的值拆分为列?

#22


2  

You may find the solution in SQL User Defined Function to Parse a Delimited String helpful (from The Code Project).

您可能会在SQL用户定义函数中找到解决方案,以解析有限的字符串(从代码项目中)。

This is the code part from this page:

这是本页面的代码部分:

CREATE FUNCTION [fn_ParseText2Table]
  (@p_SourceText VARCHAR(MAX)
  ,@p_Delimeter VARCHAR(100)=',' --default to comma delimited.
  )
 RETURNS @retTable
  TABLE([Position] INT IDENTITY(1,1)
   ,[Int_Value] INT
   ,[Num_Value] NUMERIC(18,3)
   ,[Txt_Value] VARCHAR(MAX)
   ,[Date_value] DATETIME
   )
AS
/*
********************************************************************************
Purpose: Parse values from a delimited string
  & return the result as an indexed table
Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>)
Posted to the public domain Aug, 2004
2003-06-17 Rewritten as SQL 2000 function.
 Reworked to allow for delimiters > 1 character in length
 and to convert Text values to numbers
2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient.
********************************************************************************
*/


BEGIN
 DECLARE @w_xml xml;
 SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>';


 INSERT INTO @retTable
     ([Int_Value]
    , [Num_Value]
    , [Txt_Value]
    , [Date_value]
     )
     SELECT CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT)
      END AS [Int_Value]
    , CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3))
      END AS [Num_Value]
    , [i].value('.', 'VARCHAR(MAX)') AS [txt_Value]
    , CASE
       WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME)
      END AS [Num_Value]
     FROM @w_xml.nodes('//root/i') AS [Items]([i]);
 RETURN;
END;
GO

#23


2  

This function is most fast:

这个函数的速度最快:

CREATE FUNCTION dbo.F_ExtractSubString
(
  @String VARCHAR(MAX),
  @NroSubString INT,
  @Separator VARCHAR(5)
)
RETURNS VARCHAR(MAX) AS
BEGIN
    DECLARE @St INT = 0, @End INT = 0, @Ret VARCHAR(MAX)
    SET @String = @String + @Separator
    WHILE CHARINDEX(@Separator, @String, @End + 1) > 0 AND @NroSubString > 0
    BEGIN
        SET @St = @End + 1
        SET @End = CHARINDEX(@Separator, @String, @End + 1)
        SET @NroSubString = @NroSubString - 1
    END
    IF @NroSubString > 0
        SET @Ret = ''
    ELSE
        SET @Ret = SUBSTRING(@String, @St, @End - @St)
    RETURN @Ret
END
GO

Example usage:

使用示例:

SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
       dbo.F_ExtractSubString(COLUMN, 2, ', '),
       dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM   TABLE

#24


1  

I found that using PARSENAME as above caused any name with a period to get nulled.

我发现在上面使用PARSENAME时,会导致任何名称的周期为空。

So if there was an initial or a title in the name followed by a dot they return NULL.

如果名字中有一个初始的或一个标题后面有一个点,它们返回NULL。

I found this worked for me:

我发现这对我很有效:

SELECT 
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1

#25


1  

CREATE FUNCTION [dbo].[fnSplit](@sInputList VARCHAR(8000), @sDelimiter VARCHAR(8000) = ',')
RETURNS @List TABLE (item VARCHAR(8000))
BEGIN

    DECLARE @sItem VARCHAR(8000)
    WHILE CHARINDEX(@sDelimiter, @sInputList, 0) <> 0
    BEGIN

        SELECT @sItem = RTRIM(LTRIM(SUBSTRING(@sInputList, 1, CHARINDEX(@sDelimiter, @sInputList,0) - 1))),
               @sInputList = RTRIM(LTRIM(SUBSTRING(@sInputList, CHARINDEX(@sDelimiter, @sInputList, 0) + LEN(@sDelimiter),LEN(@sInputList))))

        -- Indexes to keep the position of searching
        IF LEN(@sItem) > 0

        INSERT INTO @List SELECT @sItem

    END

    IF LEN(@sInputList) > 0
    BEGIN

        INSERT INTO @List SELECT @sInputList -- Put the last item in

    END

    RETURN

END

#26


1  

You might try this query:

您可以试试这个查询:

 select substr('abc lmn pqr',1,instr('abc lmn pqr',' ',1)) as first, 
    substr('abc lmn pqr',instr('abc lmn pqr',' ',1,1),instr('abc lmn pqr',' ',-1,2)) as middle,
    substr('abc lmn pqr',instr('abc lmn pqr',' ',1,2)) as last
    from dual;

#27


1  

select distinct modelFileId,F4.*
from contract
cross apply (select XmlList=convert(xml, '<x>'+replace(modelFileId,';','</x><x>')+'</x>').query('.')) F2
cross apply (select mfid1=XmlNode.value('/x[1]','varchar(512)')
,mfid2=XmlNode.value('/x[2]','varchar(512)')
,mfid3=XmlNode.value('/x[3]','varchar(512)')
,mfid4=XmlNode.value('/x[4]','varchar(512)') from XmlList.nodes('x') F3(XmlNode)) F4
where modelFileId like '%;%'
order by modelFileId

#28


1  

You can use an inbuilt STRING_SPLIT function which is available only under compatibility level 130. If your database compatibility level is lower than 130, SQL Server will not be able to find and execute STRING_SPLIT function. You can change a compatibility level of database using the following command:

您可以使用仅在兼容性级别130以下的内置STRING_SPLIT函数。如果您的数据库兼容性级别低于130,SQL Server将无法找到并执行STRING_SPLIT函数。您可以使用以下命令更改数据库的兼容性级别:

ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130

Syntax

语法

STRING_SPLIT ( string , separator )

see documentation here

看到文档

#29


-1  

You can use split function.

你可以使用split函数。

SELECT 
(select top 1 item from dbo.Split(FullName,',') where id=1 ) as Name,
(select top 1 item from dbo.Split(FullName,',') where id=2 ) as Surname,
FROM MyTbl

#30


-1  

Select distinct PROJ_UID,PROJ_NAME,RES_UID from E2E_ProjectWiseTimesheetActuals
where   CHARINDEX(','+cast(PROJ_UID as varchar(8000))+',', @params) > 0 and  CHARINDEX(','+cast(RES_UID as varchar(8000))+',', @res) > 0

#1


-2  

ALTER FUNCTION [dbo].[StringListTo] (@StringList Nvarchar(max),@Separators char(1),@start int, @index int )
RETURNS nvarchar(max)
AS
BEGIN
declare @out Nvarchar(max)
declare @i int
declare @start_old int
set @start=@start+1
set @i=1
while(@i<=@index)
begin
    set @start_old=@start
    set @start=CHARINDEX('.',@StringList,@start+1)
    if (@start>0)
    begin
        set @out=Substring(@StringList,@start_old+1,@start-@start_old-1)
    end
else
begin
    set @out=Substring(@StringList,@start_old+1,len(@StringList)-1)
end
set @i=@i+1
end
RETURN @out
END;

#2


100  

Your purpose can be solved using following query -

您的目的可以通过以下查询来解决。

Select Value  , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as  Surname
from Table1

There is no readymade Split function in sql server, so we need to create user defined function.

在sql server中没有现成的拆分函数,因此我们需要创建用户定义的函数。

CREATE FUNCTION Split (
      @InputString                  VARCHAR(8000),
      @Delimiter                    VARCHAR(50)
)

RETURNS @Items TABLE (
      Item                          VARCHAR(8000)
)

AS
BEGIN
      IF @Delimiter = ' '
      BEGIN
            SET @Delimiter = ','
            SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
      END

      IF (@Delimiter IS NULL OR @Delimiter = '')
            SET @Delimiter = ','

--INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic
--INSERT INTO @Items VALUES (@InputString) -- Diagnostic

      DECLARE @Item           VARCHAR(8000)
      DECLARE @ItemList       VARCHAR(8000)
      DECLARE @DelimIndex     INT

      SET @ItemList = @InputString
      SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      WHILE (@DelimIndex != 0)
      BEGIN
            SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
            INSERT INTO @Items VALUES (@Item)

            -- Set @ItemList = @ItemList minus one less item
            SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
            SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      END -- End WHILE

      IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString
      BEGIN
            SET @Item = @ItemList
            INSERT INTO @Items VALUES (@Item)
      END

      -- No delimiters were encountered in @InputString, so just return @InputString
      ELSE INSERT INTO @Items VALUES (@InputString)

      RETURN

END -- End Function
GO

---- Set Permissions
--GRANT SELECT ON Split TO UserRole1
--GRANT SELECT ON Split TO UserRole2
--GO

#3


32  

;WITH Split_Names (Value,Name, xmlname)
AS
(
    SELECT Value,
    Name,
    CONVERT(XML,'<Names><name>'  
    + REPLACE(Name,',', '</name><name>') + '</name></Names>') AS xmlname
      FROM tblnames
)

 SELECT Value,      
 xmlname.value('/Names[1]/name[1]','varchar(100)') AS Name,    
 xmlname.value('/Names[1]/name[2]','varchar(100)') AS Surname
 FROM Split_Names

and also check the link below for reference

并检查下面的链接以供参考。

http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html

http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html

#4


29  

xml base answer is simple and clean

xml基本答案简单明了。

refer this

请参考这个

DECLARE @S varchar(max),
        @Split char(1),
        @X xml

SELECT @S = 'ab,cd,ef,gh,ij',
       @Split = ','

SELECT @X = CONVERT(xml,' <root> <myvalue>' +
REPLACE(@S,@Split,'</myvalue> <myvalue>') + '</myvalue>   </root> ')

SELECT  T.c.value('.','varchar(20)'),--retrieve all values at once
  T.c.value('(/root/myvalue)[1]','VARCHAR(20)')  , --retrieve index 1 only, which is the 'ab'
  T.c.value('(/root/myvalue)[2]','VARCHAR(20)'),
  T.c.value('(/root/myvalue)[3]','VARCHAR(20)')
 FROM @X.nodes('/root/myvalue') T(c)

#5


24  

With CROSS APPLY

与交叉应用

select ParsedData.* 
from MyTable mt
cross apply ( select str = mt.String + ',,' ) f1
cross apply ( select p1 = charindex( ',', str ) ) ap1
cross apply ( select p2 = charindex( ',', str, p1 + 1 ) ) ap2
cross apply ( select Nmame = substring( str, 1, p1-1 )                   
                 , Surname = substring( str, p1+1, p2-p1-1 )
          ) ParsedData

#6


20  

I think this is cool

我觉得这很酷。

SELECT value,
    PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
    PARSENAME(REPLACE(String,',','.'),1) 'Sur Name'
FROM table WITH (NOLOCK)

#7


13  

Try this (change instances of ' ' to ',' or whatever delimiter you want to use)

试试这个(更改“to”的实例,或者你想使用的任何分隔符)

CREATE FUNCTION dbo.Wordparser
(
  @multiwordstring VARCHAR(255),
  @wordnumber      NUMERIC
)
returns VARCHAR(255)
AS
  BEGIN
      DECLARE @remainingstring VARCHAR(255)
      SET @remainingstring=@multiwordstring

      DECLARE @numberofwords NUMERIC
      SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1)

      DECLARE @word VARCHAR(50)
      DECLARE @parsedwords TABLE
      (
         line NUMERIC IDENTITY(1, 1),
         word VARCHAR(255)
      )

      WHILE @numberofwords > 1
        BEGIN
            SET @word=LEFT(@remainingstring, CHARINDEX(' ', @remainingstring) - 1)

            INSERT INTO @parsedwords(word)
            SELECT @word

            SET @remainingstring= REPLACE(@remainingstring, Concat(@word, ' '), '')
            SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1)

            IF @numberofwords = 1
              BREAK

            ELSE
              CONTINUE
        END

      IF @numberofwords = 1
        SELECT @word = @remainingstring
      INSERT INTO @parsedwords(word)
      SELECT @word

      RETURN
        (SELECT word
         FROM   @parsedwords
         WHERE  line = @wordnumber)

  END

Example usage:

使用示例:

SELECT dbo.Wordparser(COLUMN, 1),
       dbo.Wordparser(COLUMN, 2),
       dbo.Wordparser(COLUMN, 3)
FROM   TABLE

#8


11  

There are multiple ways to solve this and many different ways have been proposed already. Simplest would be to use LEFT / SUBSTRING and other string functions to achieve the desired result.

有多种方法可以解决这个问题,已经提出了许多不同的方法。最简单的方法是使用左/子字符串和其他字符串函数来实现预期的结果。

Sample Data

样本数据

DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))

INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');

Using String Functions like LEFT

使用像LEFT这样的字符串函数。

SELECT
    Value,
    LEFT(String,CHARINDEX(',',String)-1) as Fname,
    LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1

This approach fails if there are more 2 items in a String. In such a scenario, we can use a splitter and then use PIVOT or convert the string into an XML and use .nodes to get string items. XML based solution have been detailed out by aads and bvr in their solution.

如果字符串中有更多2项,则此方法失败。在这种情况下,我们可以使用splitter,然后使用PIVOT或将字符串转换为XML,并使用.nodes获取字符串项。基于XML的解决方案已经在其解决方案中被aads和bvr详细描述。

The answers for this question which use splitter, all use WHILE which is inefficient for splitting. Check this performance comparison. One of the best splitters around is DelimitedSplit8K, created by Jeff Moden. You can read more about it here

对于这个问题的答案,使用分配器,所有的使用,而这是低效的分裂。检查这个性能比较。其中最好的分割器是由Jeff Moden创建的DelimitedSplit8K。你可以在这里读到更多。

Splitter with PIVOT

分束器和主

DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))

INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');


SELECT t3.Value,[1] as Fname,[2] as Lname
FROM @tbl1 as t1
CROSS APPLY [dbo].[DelimitedSplit8K](String,',') as t2
PIVOT(MAX(Item) FOR ItemNumber IN ([1],[2])) as t3

Output

输出

Value   Fname   Lname
1   Cleo    Smith
2   John    Mathew

DelimitedSplit8K by Jeff Moden

由杰夫DelimitedSplit8K语音识别

CREATE FUNCTION [dbo].[DelimitedSplit8K]
/**********************************************************************************************************************
 Purpose:
 Split a given string at a given delimiter and return a list of the split elements (items).

 Notes:
 1.  Leading a trailing delimiters are treated as if an empty string element were present.
 2.  Consecutive delimiters are treated as if an empty string element were present between them.
 3.  Except when spaces are used as a delimiter, all spaces present in each element are preserved.

 Returns:
 iTVF containing the following:
 ItemNumber = Element position of Item as a BIGINT (not converted to INT to eliminate a CAST)
 Item       = Element value as a VARCHAR(8000)

 Statistics on this function may be found at the following URL:
 http://www.sqlservercentral.com/Forums/Topic1101315-203-4.aspx

 CROSS APPLY Usage Examples and Tests:
--=====================================================================================================================
-- TEST 1:
-- This tests for various possible conditions in a string using a comma as the delimiter.  The expected results are
-- laid out in the comments
--=====================================================================================================================
--===== Conditionally drop the test tables to make reruns easier for testing.
     -- (this is NOT a part of the solution)
     IF OBJECT_ID('tempdb..#JBMTest') IS NOT NULL DROP TABLE #JBMTest
;
--===== Create and populate a test table on the fly (this is NOT a part of the solution).
     -- In the following comments, "b" is a blank and "E" is an element in the left to right order.
     -- Double Quotes are used to encapsulate the output of "Item" so that you can see that all blanks
     -- are preserved no matter where they may appear.
 SELECT *
   INTO #JBMTest
   FROM (                                               --# & type of Return Row(s)
         SELECT  0, NULL                      UNION ALL --1 NULL
         SELECT  1, SPACE(0)                  UNION ALL --1 b (Empty String)
         SELECT  2, SPACE(1)                  UNION ALL --1 b (1 space)
         SELECT  3, SPACE(5)                  UNION ALL --1 b (5 spaces)
         SELECT  4, ','                       UNION ALL --2 b b (both are empty strings)
         SELECT  5, '55555'                   UNION ALL --1 E
         SELECT  6, ',55555'                  UNION ALL --2 b E
         SELECT  7, ',55555,'                 UNION ALL --3 b E b
         SELECT  8, '55555,'                  UNION ALL --2 b B
         SELECT  9, '55555,1'                 UNION ALL --2 E E
         SELECT 10, '1,55555'                 UNION ALL --2 E E
         SELECT 11, '55555,4444,333,22,1'     UNION ALL --5 E E E E E 
         SELECT 12, '55555,4444,,333,22,1'    UNION ALL --6 E E b E E E
         SELECT 13, ',55555,4444,,333,22,1,'  UNION ALL --8 b E E b E E E b
         SELECT 14, ',55555,4444,,,333,22,1,' UNION ALL --9 b E E b b E E E b
         SELECT 15, ' 4444,55555 '            UNION ALL --2 E (w/Leading Space) E (w/Trailing Space)
         SELECT 16, 'This,is,a,test.'                   --E E E E
        ) d (SomeID, SomeValue)
;
--===== Split the CSV column for the whole table using CROSS APPLY (this is the solution)
 SELECT test.SomeID, test.SomeValue, split.ItemNumber, Item = QUOTENAME(split.Item,'"')
   FROM #JBMTest test
  CROSS APPLY dbo.DelimitedSplit8K(test.SomeValue,',') split
;
--=====================================================================================================================
-- TEST 2:
-- This tests for various "alpha" splits and COLLATION using all ASCII characters from 0 to 255 as a delimiter against
-- a given string.  Note that not all of the delimiters will be visible and some will show up as tiny squares because
-- they are "control" characters.  More specifically, this test will show you what happens to various non-accented 
-- letters for your given collation depending on the delimiter you chose.
--=====================================================================================================================
WITH 
cteBuildAllCharacters (String,Delimiter) AS 
(
 SELECT TOP 256 
        'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789',
        CHAR(ROW_NUMBER() OVER (ORDER BY (SELECT NULL))-1)
   FROM master.sys.all_columns
)
 SELECT ASCII_Value = ASCII(c.Delimiter), c.Delimiter, split.ItemNumber, Item = QUOTENAME(split.Item,'"')
   FROM cteBuildAllCharacters c
  CROSS APPLY dbo.DelimitedSplit8K(c.String,c.Delimiter) split
  ORDER BY ASCII_Value, split.ItemNumber
;
-----------------------------------------------------------------------------------------------------------------------
 Other Notes:
 1. Optimized for VARCHAR(8000) or less.  No testing or error reporting for truncation at 8000 characters is done.
 2. Optimized for single character delimiter.  Multi-character delimiters should be resolvedexternally from this 
    function.
 3. Optimized for use with CROSS APPLY.
 4. Does not "trim" elements just in case leading or trailing blanks are intended.
 5. If you don't know how a Tally table can be used to replace loops, please see the following...
    http://www.sqlservercentral.com/articles/T-SQL/62867/
 6. Changing this function to use NVARCHAR(MAX) will cause it to run twice as slow.  It's just the nature of 
    VARCHAR(MAX) whether it fits in-row or not.
 7. Multi-machine testing for the method of using UNPIVOT instead of 10 SELECT/UNION ALLs shows that the UNPIVOT method
    is quite machine dependent and can slow things down quite a bit.
-----------------------------------------------------------------------------------------------------------------------
 Credits:
 This code is the product of many people's efforts including but not limited to the following:
 cteTally concept originally by Iztek Ben Gan and "decimalized" by Lynn Pettis (and others) for a bit of extra speed
 and finally redacted by Jeff Moden for a different slant on readability and compactness. Hat's off to Paul White for
 his simple explanations of CROSS APPLY and for his detailed testing efforts. Last but not least, thanks to
 Ron "BitBucket" McCullough and Wayne Sheffield for their extreme performance testing across multiple machines and
 versions of SQL Server.  The latest improvement brought an additional 15-20% improvement over Rev 05.  Special thanks
 to "Nadrek" and "peter-757102" (aka Peter de Heer) for bringing such improvements to light.  Nadrek's original
 improvement brought about a 10% performance gain and Peter followed that up with the content of Rev 07.  

 I also thank whoever wrote the first article I ever saw on "numbers tables" which is located at the following URL
 and to Adam Machanic for leading me to it many years ago.
 http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-numbers-table.html
-----------------------------------------------------------------------------------------------------------------------
 Revision History:
 Rev 00 - 20 Jan 2010 - Concept for inline cteTally: Lynn Pettis and others.
                        Redaction/Implementation: Jeff Moden 
        - Base 10 redaction and reduction for CTE.  (Total rewrite)

 Rev 01 - 13 Mar 2010 - Jeff Moden
        - Removed one additional concatenation and one subtraction from the SUBSTRING in the SELECT List for that tiny
          bit of extra speed.

 Rev 02 - 14 Apr 2010 - Jeff Moden
        - No code changes.  Added CROSS APPLY usage example to the header, some additional credits, and extra 
          documentation.

 Rev 03 - 18 Apr 2010 - Jeff Moden
        - No code changes.  Added notes 7, 8, and 9 about certain "optimizations" that don't actually work for this
          type of function.

 Rev 04 - 29 Jun 2010 - Jeff Moden
        - Added WITH SCHEMABINDING thanks to a note by Paul White.  This prevents an unnecessary "Table Spool" when the
          function is used in an UPDATE statement even though the function makes no external references.

 Rev 05 - 02 Apr 2011 - Jeff Moden
        - Rewritten for extreme performance improvement especially for larger strings approaching the 8K boundary and
          for strings that have wider elements.  The redaction of this code involved removing ALL concatenation of 
          delimiters, optimization of the maximum "N" value by using TOP instead of including it in the WHERE clause,
          and the reduction of all previous calculations (thanks to the switch to a "zero based" cteTally) to just one 
          instance of one add and one instance of a subtract. The length calculation for the final element (not 
          followed by a delimiter) in the string to be split has been greatly simplified by using the ISNULL/NULLIF 
          combination to determine when the CHARINDEX returned a 0 which indicates there are no more delimiters to be
          had or to start with. Depending on the width of the elements, this code is between 4 and 8 times faster on a
          single CPU box than the original code especially near the 8K boundary.
        - Modified comments to include more sanity checks on the usage example, etc.
        - Removed "other" notes 8 and 9 as they were no longer applicable.

 Rev 06 - 12 Apr 2011 - Jeff Moden
        - Based on a suggestion by Ron "Bitbucket" McCullough, additional test rows were added to the sample code and
          the code was changed to encapsulate the output in pipes so that spaces and empty strings could be perceived 
          in the output.  The first "Notes" section was added.  Finally, an extra test was added to the comments above.

 Rev 07 - 06 May 2011 - Peter de Heer, a further 15-20% performance enhancement has been discovered and incorporated 
          into this code which also eliminated the need for a "zero" position in the cteTally table. 
**********************************************************************************************************************/
--===== Define I/O parameters
        (@pString VARCHAR(8000), @pDelimiter CHAR(1))
RETURNS TABLE WITH SCHEMABINDING AS
 RETURN
--===== "Inline" CTE Driven "Tally Table" produces values from 0 up to 10,000...
     -- enough to cover NVARCHAR(4000)
  WITH E1(N) AS (
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
                ),                          --10E+1 or 10 rows
       E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
       E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
 cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
                     -- for both a performance gain and prevention of accidental "overruns"
                 SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
                ),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
                 SELECT 1 UNION ALL
                 SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
                ),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
                 SELECT s.N1,
                        ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
                   FROM cteStart s
                )
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
 SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
        Item       = SUBSTRING(@pString, l.N1, l.L1)
   FROM cteLen l
;

GO

#9


10  

With SQL Server 2016 we can use string_split to accomplish this:

使用SQL Server 2016,我们可以使用string_split来完成这个任务:

create table commasep (
 id int identity(1,1)
 ,string nvarchar(100) )

insert into commasep (string) values ('John, Adam'), ('test1,test2,test3')

select id, [value] as String from commasep 
 cross apply string_split(string,',')

#10


9  

I think PARSENAME is the neat function to use for this example, as described in this article: http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/

我认为PARSENAME是这个示例中使用的简洁函数,如本文所描述的:http://www.sqlshack.com/parsing/rotating-delimiteddata -sql-server-2012/。

The PARSENAME function is logically designed to parse four-part object names. The nice thing about PARSENAME is that it’s not limited to parsing just SQL Server four-part object names – it will parse any function or string data that is delimited by dots.

PARSENAME函数在逻辑上被设计用来解析四部分对象名称。PARSENAME的优点在于,它不仅限于解析SQL Server四部分对象名称——它将解析任何由点分隔的函数或字符串数据。

The first parameter is the object to parse, and the second is the integer value of the object piece to return. The article is discussing parsing and rotating delimited data - company phone numbers, but it can be used to parse name/surname data also.

第一个参数是要解析的对象,第二个参数是要返回的对象块的整数值。本文讨论了解析和旋转分隔数据—公司电话号码,但也可以用来解析姓名/姓氏数据。

Example:

例子:

USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';

The article also describes using a Common Table Expression (CTE) called ‘replaceChars’, to run PARSENAME against the delimiter-replaced values. A CTE is useful for returning a temporary view or result set.

本文还描述了使用一个名为“replaceChars”的公共表表达式来运行PARSENAME,以防止被分隔符替换的值。CTE用于返回临时视图或结果集。

After that, the UNPIVOT function has been used to convert some columns into rows; SUBSTRING and CHARINDEX functions have been used for cleaning up the inconsistencies in the data, and the LAG function (new for SQL Server 2012) has been used in the end, as it allows referencing of previous records.

在那之后,将一些列转换成行来使用UNPIVOT函数;SUBSTRING和CHARINDEX函数用于清理数据中的不一致性,最后使用滞后函数(SQL Server 2012的新功能),因为它允许引用以前的记录。

#11


8  

We can create a function as this

我们可以像这样创建一个函数。

CREATE Function [dbo].[fn_CSVToTable] 
(
    @CSVList Varchar(max)
)
RETURNS @Table TABLE (ColumnData VARCHAR(100))
AS
BEGIN
    IF RIGHT(@CSVList, 1) <> ','
    SELECT @CSVList = @CSVList + ','

    DECLARE @Pos    BIGINT,
            @OldPos BIGINT
    SELECT  @Pos    = 1,
            @OldPos = 1

    WHILE   @Pos < LEN(@CSVList)
        BEGIN
            SELECT  @Pos = CHARINDEX(',', @CSVList, @OldPos)
            INSERT INTO @Table
            SELECT  LTRIM(RTRIM(SUBSTRING(@CSVList, @OldPos, @Pos - @OldPos))) Col001

            SELECT  @OldPos = @Pos + 1
        END

    RETURN
END

We can then seperate the CSV values into our respective columns using a SELECT statement

然后,我们可以使用SELECT语句将CSV值分离到各自的列中。

#12


6  

select id,SUBSTRING(name,0,charindex(',',name))as firstname
,SUBSTRING(name,charindex(',',name),len(name)+1)as lastname from spilt

#13


5  

Using instring function :)

使用instring功能:)

select Value, 
       substring(String,1,instr(String," ") -1) Fname,  
       substring(String,instr(String,",") +1) Sname 
from tablename;

Used two functions,
1. substring(string, position, length) ==> returns string from positon to length
2. instr(string,pattern) ==> returns position of pattern.

使用两个函数,1。子字符串(string, position, length) ==>从positon返回字符串长度为2。instr(string,pattern) ==>返回模式的位置。

If we don’t provide length argument in substring it returns until end of string

如果在子字符串中不提供长度参数,则返回到字符串结束。

#14


4  

Use Parsename() function

使用Parsename()函数

with cte as(
    select 'Aria,Karimi' as FullName
    Union
    select 'Joe,Karimi' as FullName
    Union
    select 'Bab,Karimi' as FullName
)

SELECT PARSENAME(REPLACE(FullName,',','.'),2) as Name, 
       PARSENAME(REPLACE(FullName,',','.'),1) as Family
    FROM cte

Result

结果

Name    Family
-----   ------
Aria    Karimi
Bab     Karimi
Joe     Karimi

#15


4  

DECLARE @INPUT VARCHAR (MAX)='N,A,R,E,N,D,R,A'
DECLARE @ELIMINATE_CHAR CHAR (1)=','
DECLARE @L_START INT=1
DECLARE @L_END INT=(SELECT LEN (@INPUT))
DECLARE @OUTPUT CHAR (1)

WHILE @L_START <=@L_END
BEGIN
    SET @OUTPUT=(SUBSTRING (@INPUT,@L_START,1))
    IF @OUTPUT!=@ELIMINATE_CHAR
    BEGIN
        PRINT @OUTPUT
    END
    SET @L_START=@L_START+1
END

#16


3  

I encountered a similar problem but a complex one and since this is the first thread i found regarding that issue i decided to post my finding. i know it is complex solution to a simple problem but i hope that i could help other people who go to this thread looking for a more complex solution. i had to split a string containing 5 numbers (column name: levelsFeed) and to show each number in a separate column. for example: 8,1,2,2,2 should be shown as :

我遇到了一个类似的问题,但是一个复杂的问题,因为这是我发现的第一个关于这个问题的线程,我决定发布我的发现。我知道这是一个简单问题的复杂的解决方案,但我希望我可以帮助其他的人去寻找一个更复杂的解决方案。我必须拆分一个包含5个数字(列名称:levelsFeed)的字符串,并在单独的列中显示每个数字。例如:8、1、2、2、2应显示为:

1  2  3  4  5
-------------
8  1  2  2  2

Solution 1: using XML functions: this solution for the slowest solution by far

解决方案1:使用XML函数:到目前为止最慢的解决方案。

SELECT Distinct FeedbackID, 
, S.a.value('(/H/r)[1]', 'INT') AS level1
, S.a.value('(/H/r)[2]', 'INT') AS level2
, S.a.value('(/H/r)[3]', 'INT') AS level3
, S.a.value('(/H/r)[4]', 'INT') AS level4
, S.a.value('(/H/r)[5]', 'INT') AS level5
FROM (            
    SELECT *,CAST (N'<H><r>' + REPLACE(levelsFeed, ',', '</r><r>')  + '</r> </H>' AS XML) AS [vals]
    FROM Feedbacks 
)  as d
CROSS APPLY d.[vals].nodes('/H/r') S(a)

Solution 2: using Split function and pivot. (the split function split a string to rows with the column name Data)

解决方案2:使用Split函数和pivot。(split函数将一个字符串与列名称数据分隔开)

SELECT FeedbackID, [1],[2],[3],[4],[5]
FROM (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY feedbackID ORDER BY (SELECT  null)) as rn 
FROM (
    SELECT FeedbackID, levelsFeed
    FROM Feedbacks 
) as a
CROSS APPLY dbo.Split(levelsFeed, ',')
) as SourceTable
PIVOT
(
    MAX(data)
    FOR rn IN ([1],[2],[3],[4],[5])
)as pivotTable

Solution 3: using string manipulations functions - fastest by small margin over solution 2

解决方案3:使用字符串操作函数——在解决方案2上以小幅度的速度最快。

SELECT FeedbackID,
SUBSTRING(levelsFeed,0,CHARINDEX(',',levelsFeed)) AS level1,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),4) AS level2,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),3) AS level3,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),2) AS level4,
PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),1) AS level5
FROM Feedbacks

since the levelsFeed contains 5 string values i needed to use the substring function for the first string.

由于levelsFeed包含5个字符串值,所以我需要为第一个字符串使用子字符串函数。

i hope that my solution will help other that got to this thread looking for a more complex split to columns methods

我希望我的解决方案能够帮助其他到达这个线程的人寻找一个更复杂的列方法。

#17


3  

Try this:

试试这个:

declare @csv varchar(100) ='aaa,bb,csda,daass';
set @csv = @csv+',';

with cte as
(
    select SUBSTRING(@csv,1,charindex(',',@csv,1)-1) as val, SUBSTRING(@csv,charindex(',',@csv,1)+1,len(@csv)) as rem 
    UNION ALL
    select SUBSTRING(a.rem,1,charindex(',',a.rem,1)-1)as val, SUBSTRING(a.rem,charindex(',',a.rem,1)+1,len(A.rem)) 
    from cte a where LEN(a.rem)>=1
    ) select val from cte

#18


2  

mytable:

mytable:

Value  ColOne
--------------------
1      Cleo, Smith

The following should work if there aren't too many columns

如果没有太多的列,那么下面的内容应该是有效的。

ALTER TABLE mytable ADD ColTwo nvarchar(256);
UPDATE mytable SET ColTwo = LEFT(ColOne, Charindex(',', ColOne) - 1);
--'Cleo' = LEFT('Cleo, Smith', Charindex(',', 'Cleo, Smith') - 1)
UPDATE mytable SET ColTwo = REPLACE(ColOne, ColTwo + ',', '');
--' Smith' = REPLACE('Cleo, Smith', 'Cleo' + ',')
UPDATE mytable SET ColOne = REPLACE(ColOne, ',' + ColTwo, ''), ColTwo = LTRIM(ColTwo);
--'Cleo' = REPLACE('Cleo, Smith', ',' + ' Smith', '') 

Result:

结果:

Value  ColOne ColTwo
--------------------
1      Cleo   Smith

#19


2  

it is so easy, you can take it by below query:

它很简单,你可以在下面查询:

DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))

#20


2  

This worked for me

这为我工作

CREATE FUNCTION [dbo].[SplitString](
    @delimited NVARCHAR(MAX),
    @delimiter NVARCHAR(100)
) RETURNS @t TABLE ( val NVARCHAR(MAX))
AS
BEGIN
    DECLARE @xml XML
    SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'
    INSERT INTO @t(val)
    SELECT  r.value('.','varchar(MAX)') as item
    FROM  @xml.nodes('/t') as records(r)
    RETURN
END

#21


2  

I think following function will work for you:

我认为以下功能对你有用:

You have to create a function in SQL first. Like this

CREATE FUNCTION [dbo].[fn_split](
@str VARCHAR(MAX),
@delimiter CHAR(1)
)
RETURNS @returnTable TABLE (idx INT PRIMARY KEY IDENTITY, item VARCHAR(8000))
AS
BEGIN
DECLARE @pos INT
SELECT @str = @str + @delimiter
WHILE LEN(@str) > 0 
    BEGIN
        SELECT @pos = CHARINDEX(@delimiter,@str)
        IF @pos = 1
            INSERT @returnTable (item)
                VALUES (NULL)
        ELSE
            INSERT @returnTable (item)
                VALUES (SUBSTRING(@str, 1, @pos-1))
        SELECT @str = SUBSTRING(@str, @pos+1, LEN(@str)-@pos)       
    END
RETURN
END

You can call this function, like this:

你可以称这个函数为:

select * from fn_split('1,24,5',',')

Implementation:

Declare @test TABLE (
ID VARCHAR(200),
Data VARCHAR(200)
)

insert into @test 
(ID, Data)
Values
('1','Cleo,Smith')


insert into @test 
(ID, Data)
Values
('2','Paul,Grim')

select ID,
(select item from fn_split(Data,',') where idx in (1)) as Name ,
(select item from fn_split(Data,',') where idx in (2)) as Surname
 from @test

Result will like this:

如何将逗号分隔的值拆分为列?

#22


2  

You may find the solution in SQL User Defined Function to Parse a Delimited String helpful (from The Code Project).

您可能会在SQL用户定义函数中找到解决方案,以解析有限的字符串(从代码项目中)。

This is the code part from this page:

这是本页面的代码部分:

CREATE FUNCTION [fn_ParseText2Table]
  (@p_SourceText VARCHAR(MAX)
  ,@p_Delimeter VARCHAR(100)=',' --default to comma delimited.
  )
 RETURNS @retTable
  TABLE([Position] INT IDENTITY(1,1)
   ,[Int_Value] INT
   ,[Num_Value] NUMERIC(18,3)
   ,[Txt_Value] VARCHAR(MAX)
   ,[Date_value] DATETIME
   )
AS
/*
********************************************************************************
Purpose: Parse values from a delimited string
  & return the result as an indexed table
Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>)
Posted to the public domain Aug, 2004
2003-06-17 Rewritten as SQL 2000 function.
 Reworked to allow for delimiters > 1 character in length
 and to convert Text values to numbers
2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient.
********************************************************************************
*/


BEGIN
 DECLARE @w_xml xml;
 SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>';


 INSERT INTO @retTable
     ([Int_Value]
    , [Num_Value]
    , [Txt_Value]
    , [Date_value]
     )
     SELECT CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT)
      END AS [Int_Value]
    , CASE
       WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3))
      END AS [Num_Value]
    , [i].value('.', 'VARCHAR(MAX)') AS [txt_Value]
    , CASE
       WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1
       THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME)
      END AS [Num_Value]
     FROM @w_xml.nodes('//root/i') AS [Items]([i]);
 RETURN;
END;
GO

#23


2  

This function is most fast:

这个函数的速度最快:

CREATE FUNCTION dbo.F_ExtractSubString
(
  @String VARCHAR(MAX),
  @NroSubString INT,
  @Separator VARCHAR(5)
)
RETURNS VARCHAR(MAX) AS
BEGIN
    DECLARE @St INT = 0, @End INT = 0, @Ret VARCHAR(MAX)
    SET @String = @String + @Separator
    WHILE CHARINDEX(@Separator, @String, @End + 1) > 0 AND @NroSubString > 0
    BEGIN
        SET @St = @End + 1
        SET @End = CHARINDEX(@Separator, @String, @End + 1)
        SET @NroSubString = @NroSubString - 1
    END
    IF @NroSubString > 0
        SET @Ret = ''
    ELSE
        SET @Ret = SUBSTRING(@String, @St, @End - @St)
    RETURN @Ret
END
GO

Example usage:

使用示例:

SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
       dbo.F_ExtractSubString(COLUMN, 2, ', '),
       dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM   TABLE

#24


1  

I found that using PARSENAME as above caused any name with a period to get nulled.

我发现在上面使用PARSENAME时,会导致任何名称的周期为空。

So if there was an initial or a title in the name followed by a dot they return NULL.

如果名字中有一个初始的或一个标题后面有一个点,它们返回NULL。

I found this worked for me:

我发现这对我很有效:

SELECT 
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1

#25


1  

CREATE FUNCTION [dbo].[fnSplit](@sInputList VARCHAR(8000), @sDelimiter VARCHAR(8000) = ',')
RETURNS @List TABLE (item VARCHAR(8000))
BEGIN

    DECLARE @sItem VARCHAR(8000)
    WHILE CHARINDEX(@sDelimiter, @sInputList, 0) <> 0
    BEGIN

        SELECT @sItem = RTRIM(LTRIM(SUBSTRING(@sInputList, 1, CHARINDEX(@sDelimiter, @sInputList,0) - 1))),
               @sInputList = RTRIM(LTRIM(SUBSTRING(@sInputList, CHARINDEX(@sDelimiter, @sInputList, 0) + LEN(@sDelimiter),LEN(@sInputList))))

        -- Indexes to keep the position of searching
        IF LEN(@sItem) > 0

        INSERT INTO @List SELECT @sItem

    END

    IF LEN(@sInputList) > 0
    BEGIN

        INSERT INTO @List SELECT @sInputList -- Put the last item in

    END

    RETURN

END

#26


1  

You might try this query:

您可以试试这个查询:

 select substr('abc lmn pqr',1,instr('abc lmn pqr',' ',1)) as first, 
    substr('abc lmn pqr',instr('abc lmn pqr',' ',1,1),instr('abc lmn pqr',' ',-1,2)) as middle,
    substr('abc lmn pqr',instr('abc lmn pqr',' ',1,2)) as last
    from dual;

#27


1  

select distinct modelFileId,F4.*
from contract
cross apply (select XmlList=convert(xml, '<x>'+replace(modelFileId,';','</x><x>')+'</x>').query('.')) F2
cross apply (select mfid1=XmlNode.value('/x[1]','varchar(512)')
,mfid2=XmlNode.value('/x[2]','varchar(512)')
,mfid3=XmlNode.value('/x[3]','varchar(512)')
,mfid4=XmlNode.value('/x[4]','varchar(512)') from XmlList.nodes('x') F3(XmlNode)) F4
where modelFileId like '%;%'
order by modelFileId

#28


1  

You can use an inbuilt STRING_SPLIT function which is available only under compatibility level 130. If your database compatibility level is lower than 130, SQL Server will not be able to find and execute STRING_SPLIT function. You can change a compatibility level of database using the following command:

您可以使用仅在兼容性级别130以下的内置STRING_SPLIT函数。如果您的数据库兼容性级别低于130,SQL Server将无法找到并执行STRING_SPLIT函数。您可以使用以下命令更改数据库的兼容性级别:

ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130

Syntax

语法

STRING_SPLIT ( string , separator )

see documentation here

看到文档

#29


-1  

You can use split function.

你可以使用split函数。

SELECT 
(select top 1 item from dbo.Split(FullName,',') where id=1 ) as Name,
(select top 1 item from dbo.Split(FullName,',') where id=2 ) as Surname,
FROM MyTbl

#30


-1  

Select distinct PROJ_UID,PROJ_NAME,RES_UID from E2E_ProjectWiseTimesheetActuals
where   CHARINDEX(','+cast(PROJ_UID as varchar(8000))+',', @params) > 0 and  CHARINDEX(','+cast(RES_UID as varchar(8000))+',', @res) > 0