原始数据与期望结果
有表tb, 如下:
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id value
----------- --------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
方法一
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go SELECT *FROM tb SELECT A.id, B.value
FROM(
SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb
)A
OUTER APPLY(
SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v)
)B DROP TABLE tb
方法二
USE tempdb
GO
--1. 要增加一个分割表值函数
IF OBJECT_ID('[dbo].[Fun_String2ToStringArray]') IS NOT NULL
DROP FUNCTION [dbo].[Fun_String2ToStringArray]
GO
CREATE FUNCTION [dbo].[Fun_String2ToStringArray](@str NVARCHAR(MAX), @split NVARCHAR(10))
RETURNS @table TABLE ([item] NVARCHAR(max))
AS
BEGIN
IF LEN(@split) = 0
BEGIN
SET @split = N','
END DECLARE @xml XML;
SET @xml = CONVERT(XML, '<x><![CDATA[' + replace(CONVERT(VARCHAR(MAX), @str), @split, ']]></x><x><![CDATA[') + ']]></x>') INSERT INTO @table
SELECT item
FROM (SELECT c.value('text()[1]', 'nvarchar(4000)') [item]
FROM @xml.nodes('/x') t(c)) t
WHERE item IS NOT NULL RETURN
END
GO
--测试表及测试数据
IF OBJECT_ID('t') IS NOT NULL DROP TABLE t
GO
CREATE TABLE t(
result VARCHAR(20),
d DATETIME
)
GO
INSERT INTO t VALUES('23@34','2018-04-01 08:00')
INSERT INTO t VALUES('','2018-04-01 09:00')
INSERT INTO t VALUES('','2018-04-02 09:00')
INSERT INTO t VALUES('3@4.6@5.7@3.2@4','2018-04-03 08:00')
INSERT INTO t VALUES('','2018-04-04 09:00')
INSERT INTO t VALUES('','2018-04-05 09:00')
INSERT INTO t VALUES('6@4.6@4.7@3.2@3','2018-04-06 09:00')
INSERT INTO t VALUES('','2018-04-07 09:00')
INSERT INTO t VALUES('','2018-04-08 09:00') --实际的查询语句
;WITH cte AS (
SELECT f.item AS result,t.d FROM t
CROSS APPLY dbo.[Fun_String2ToStringArray](t.result,'@') AS f
)
,cte2 AS(
SELECT TOP 125 * FROM cte ORDER BY d
),cte3 AS(
SELECT TOP 125 * FROM cte ORDER BY d DESC
)
SELECT * FROM cte2
UNION
SELECT * FROM cte3
摘录于此贴