How to extract a substring on batch file

Question

Now I need get the substring form a string on batch File

I have this

set path="c:\test\branches\9.1\_build"

I need get the first value after the branches value : 9.1

But this value can be in other position like

c:\xxx\yyyy\zzzz\branches\otherbranch\zzz\uuuu\iii

In this case, I need get this: otherbranch I need one generic solution, Thank you guys..

Recomendation unrelated to your question: Don't ever use PATH as your own variable. It is reserved to hold the folders that are to be searched when looking for a command. Changing the value can cause problems later down the line. — dbenham, Sep 10 '13 at 19:54
Thanks man.. this is only a example, but good recommendation.. — davdomin, Sep 10 '13 at 19:57

Endoro · Accepted Answer · 2013-09-10T20:08:11.783

17

set "mypath=c:\test\branches\9.1\_build"
set "value=%mypath:*\branches\=%"
if "%value%"=="%mypath%" echo "\branches\" not found &goto :eof
for /f "delims=\" %%a in ("%value%") do set "value=%%~a"
echo %value%

edited Sep 10 '13 at 20:08

answered Sep 10 '13 at 19:57

Endoro

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2

+1, but you probably should add one more \: `set "mypath=%mypath:*\branches\=%"` – dbenham Sep 10 '13 at 20:04
4

You also might want to put in a check to verify that \branches\ exists by checking the first substitution result does not match the original string. – dbenham Sep 10 '13 at 20:06

npocmaka · Answer 2 · 2013-09-10T20:03:40.277

It uses :split subroutine that can be found here http://ss64.org/viewtopic.php?id=1687

@echo off
setlocal
rem not good idea to overwrite path
set "_path="c:\test\branches\9.1\_build""
call :split %_path% branches 2 part
call :split %part% \ 2 part2

echo %part2%
endlocal
goto :eof
:split [%1 - string to be splitted;%2 - split by;%3 - possition to get; %4 - if defined will store the result in variable with same name]

setlocal EnableDelayedExpansion

set "string=%~1"
set "splitter=%~2"
set /a position=%~3-1

set LF=^


rem ** Two empty lines are required
echo off
for %%L in ("!LF!") DO (
    for /f "delims=" %%R in ("!splitter!") do ( 
        set "var=!string:%%R=%%L!"
        if "!var!" EQU "!string!" (
            echo "!string!" does not contain "!splitter!" >2
            exit /B 1
        )
    )
)

if !position! equ 0 ( set "_skip=" ) else (set  "_skip=skip=%position%")

for /f  "%_skip% delims=" %%P in (""!var!"") DO ( 

    set "part=%%~P"
    goto :end_for
)
:end_for
if "!part!" equ ""  echo Index Out Of Bound >2 &exit /B 2
endlocal & if "%~4" NEQ "" (set %~4=%part%) else echo %part%
exit /b 0

just give me 5 minutes to check it :) – npocmaka Sep 10 '13 at 19:48 — npocmaka, Sep 10 '13 at 19:48

score 3 · Answer 3 · edited May 23 '17 at 10:27

Here is a simple solution that uses a handy hybrid JScript/batch utility called REPL.BAT that performs regex search and replace on stdin and writes the result to stdout. It is pure script that runs on any modern Windows machine from XP onward - no 3rd party executeable required.

It has many options, including an A option that only lists substitued lines that match, and S that reads from an environment variable instead of stdin. Complete documentation is embedded within the script.

Assuming REPL.BAT is in your current directory, or better yet, somewhere within your path, then the following will set value appropriately. It will be undefined if \branches\ could not be found, or if nothing follows \branches\:

set "mypath=c:\test\branches\9.1\_build"
set "value="
for /f "eol=\ delims=" %%A in (
  'repl.bat ".*\\branches\\([^\\]*).*" "$1" AS mypath'
) do set "value=%%A"
echo %value%

How to extract a substring on batch file

3 Answers3

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