使用Windows shell或Perl脚本搜索和替换字符串 [英] Search and replace a string using Windows shell or Perl script

问题描述

如何搜索 // packages / thirdparty / release / main.txt＃10
并替换＃10 使用批处理命令以及忽略空格与＃15 （设置VARIABLE =＃15）。 ＃10 可以是任何东西，让假设 #XXX 。

How can i search for //packages/thirdparty/release/main.txt #10 and replace #10 with #15 (set VARIABLE="#15") using batch commands and also ignoring spaces. #10 can be anything, lets assume #XXX.

testlog.txt

   //packages/thirdparty/release/config.txt                    #8
   //packages/thirdparty/release/config2.txt   #3
   //packages/thirdparty/release/config1.txt    #4
   //packages/thirdparty/release/main.txt            #10

这是我的尝试

@ECHO OFF
set $FindStr=//packages/thirdparty/release/main.txt#10
set $ReplString=//packages/thirdparty/release/main.txt#15

setlocal enabledelayedexpansion
for /f "delims=" %%a in ('type testlog.txt') do (
   set $Ver=%%a
   set $Ver=!$Ver: =!
   If /i !$Ver!==%$FindStr% set $Ver=%$ReplString%
   echo !$Ver! ) >> testlog.txt

推荐答案

您可以使用一行Perl：

You could use a one line Perl:

perl -pe"s!(//packages/thirdparty/release/main.txt *)#\d+!$1#15!;" testlog.txt 

会替换＃10 code>＃15

replaces #10 by #15

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