Bash脚本正则表达式……如何查找和替换所有匹配项?

I am writing a bash script that reads a file line by line.

我正在编写一个逐行读取文件的bash脚本。

The file is a .csv file which contains many dates in the format DD/MM/YYYY but I would like to change them to YYYY-MM-DD.

该文件是一个.csv文件，其中包含许多日期，格式为DD/MM/ yyyyy，但我想将它们更改为yyyyy -MM-DD。

I would to match the data using a regular expression, and replace it such that all of the dates in the file are correctly formatted as YYYY-MM-DD.

我将使用正则表达式匹配数据，并替换它，这样文件中的所有日期都正确格式化为YYYY-MM-DD。

I believe this regular expression would match the dates:

我相信这个正则表达式会匹配日期:

([0-9][0-9]?)/([0-9][0-9]?)/([0-9][0-9][0-9][0-9])

But I do not know how to find regex matches and replace them with the new format, or if this is even possible in a bash script. Please help!

但是我不知道如何找到regex匹配项并将它们替换为新的格式，或者在bash脚本中是否可能这样做。请帮助!

3 个解决方案

#1

Try this using sed:

使用sed试试这个:

line='Today is 10/12/2010 and yesterday was 9/11/2010'
echo "$line" | sed -r 's#([0-9]{1,2})/([0-9]{1,2})/([0-9]{4})#\3-\2-\1#g'

OUTPUT:
Today is 2010-12-10 and yesterday was 2010-11-9

PS: On mac use sed -E instead of sed -r

在mac上使用sed -E而不是sed -r

#2

Pure Bash.

纯粹的Bash。

infile='data.csv'

while read line ; do
  if [[ $line =~ ^(.*),([0-9]{1,2})/([0-9]{1,2})/([0-9]{4}),(.*)$ ]] ; then
    echo "${BASH_REMATCH[1]},${BASH_REMATCH[4]}-${BASH_REMATCH[3]}-${BASH_REMATCH[2]},${BASH_REMATCH[5]}"
  else
    echo "$line"
  fi
done < "$infile"

The input file

输入文件

xxxxxxxxx,11/03/2011,yyyyyyyyyyyyy          
xxxxxxxxx,10/04/2011,yyyyyyyyyyyyy          
xxxxxxxxx,10/05/2012,yyyyyyyyyyyyy          
xxxxxxxxx,10/06/2011,yyyyyyyyyyyyy

gives the following output:

给下面的输出:

xxxxxxxxx,2011-03-11,yyyyyyyyyyyyy
xxxxxxxxx,2011-04-10,yyyyyyyyyyyyy
xxxxxxxxx,2012-05-10,yyyyyyyyyyyyy
xxxxxxxxx,2011-06-10,yyyyyyyyyyyyy

#3

You can do it using sed

您可以使用sed完成

echo "11/12/2011" | sed -E 's/([0-9][0-9]?)\/([0-9][0-9]?)\/([0-9][0-9][0-9][0-9])/\3-\2-\1/'

#1