How to get position of the sub string in string in batch

@echo OFFSETLOCALSet "str1=This is Test string"Set "sstr=Test"SET stemp=%str1%&SET pos=0:loopSET /a pos+=1echo %stemp%|FINDSTR /b /c:"%sstr%" >NULIF ERRORLEVEL 1 (SET stemp=%stemp:~1%IF DEFINED stemp GOTO loopSET pos=0)ECHO Pos of "%sstr%" IN "%str1%" = %pos%

(This returns "9", counting the first position as "1". The definition is in the mind of the user...)

And an additional alternative:

@echo off &setlocal enabledelayedexpansionSet "str1=This is Test string"Set "sstr=Test"set /a position=0Set "sst0=!str1:*%sstr%=!"if "%sst0%"=="%str1%" echo "%sstr%" not found in "%str1%"&goto :eofSet "sst1=!str1:%sstr%%sst0%=!"if "%sst1%" neq "" for /l %%i in (0,1,8189) do if "!sst1:~%%i,1!" neq "" set /a position+=1echo.Position of %sstr% is %position%endlocal

If %str1% contains more than one %sstr%, the code will find the first position.

Try this:

@echo off &setlocal enabledelayedexpansionSet "str1=This is Test string"Set "sstr=Test"call :strlen str1 len1call :strlen sstr len2set /a stop=len1-len2if %stop% gtr 0 for /l %%i in (0,1,%stop%) do if "!str1:~%%i,%len2%!"=="%sstr%" set /a position=%%iif defined position (echo.Position of %sstr% is %position%) else echo."%sstr%" not found in "%str1%"goto :eof:strlen:: list string length up to 8189 (and reports 8189 for any string longer than 8189:: function from http://ss64.org/viewtopic.php?pid=6478#p6478(   setlocal enabledelayedexpansion & set /a "}=0"    if "%~1" neq "" if defined %~1 (        for %%# in (4096 2048 1024 512 256 128 64 32 16) do (            if "!%~1:~%%#,1!" neq "" set "%~1=!%~1:~%%#!" & set /a "}+=%%#"        )        set "%~1=!%~1!0FEDCBA9876543211" & set /a "}+=0x!%~1:~32,1!!%~1:~16,1!"    ))endlocal & set /a "%~2=%}%" & exit /bendlocal

If %str1% contains more than one %sstr%, the code will find the last position.