Extract leading numbers from a string in batch script

Question

I am new to batch scripting. although i have shell and python solutions for this problem but got stuck in batch script.

I have string like 123_happy,

Answer 1

What about this:

@echo off
setlocal EnableExtensions DisableDelayedExpansion

rem // Define the string here:
set "STR=123_happy"

setlocal EnableDelayedExpansion
rem // Get first character behind the numeric part:
for /F delims^=0123456789^ tokens^=*^ eol^= %%F in ("!STR!") do (
    endlocal
    set "SEP=%%F"
    setlocal EnableDelayedExpansion
    if defined SEP set "SEP=!SEP:~,1!"
)

rem // Split the string at the character behind the numeric part:
if not defined SEP goto :SKIP
for /F eol^=^%SEP%^ delims^=^%SEP% %%N in ("0!STR!") do (
    endlocal
    set "STR=%%N"
    setlocal EnableDelayedExpansion
    set "STR=!STR:~1!"
)
:SKIP

rem // Return the numeric part:
echo(!STR!
endlocal

endlocal
exit /B

The basic idea is to get the first character after the numeric part, which is the used as the delimiter for a for /F loop parsing the input string. This has got the great advantage that the limit for signed 32-bit integers does not apply, opposed to the approaches using set /A or for /L. In addition, leading zeros do not lead to unintentional interpretation as octal numbers, since this script treats the numeric part as string.