How to set a timeout for a process under Windows 7?

Question

I would like to start a program with a windows batch file. But the program should stop after a certain timeout value. For example: Run the program 60 seconds and stop it after 60 seconds.

Under Linux, there is this nice timeout command to do what I want. Windows has also a timeout command, but its just to pause a command, to delay its execution. Is there something else under Windows to do so ?

Setup: Windows 7, 64 Bit, Professional

Answer 1

start yourprogram.exe timeout /t 60 taskkill /im yourprogram.exe /f 

Answer 2

Bali C gave a concise and to the point answer. I needed something a little more featureful and reusable. Based on Bali C's Example. I came up with this. If anyone should need the same as me.

your.bat

REM...  CALL STARTwaitKILL..bat /relative/path/your_program.exe  REM... 

STARTwaitKILL.BAT

@ECHO OFF  IF[%1]==[] GOTO EOF IF NOT EXIST %1 GOTO EOF  REM SET PRIORITY=/NORMAL REM ADJUST PRIORITY, BELOWNORMAL LETS BATCH FILE RUN MORE SMOOTHLY REM WITH A PROGRAM THAT CONSUMES MORE CPU. SEE ABOUT MAXWAIT BELLOW SET PRIORITY=/BELOWNORMAL REM SET PRIORITY=/LOW REM 0 NORMAL WINDOW :: 1 NO WINDOW :: 2 MINIMIZED WINDOW SET /A HIDDEN=1 REM MAXWAIT HERE IS MORE LIKE MINIMUM WAIT IN WINDOWS. SET MAXWAIT=10 SET WAITCOUNT=0  SET ARGS=/I %PRIORITY% IF %HIDDEN% EQU 1 SET ARGS=%ARGS% /B IF %HIDDEN% EQU 2 SET ARGS=%ARGS% /MIN  START %ARGS% %1  :WAIT IF %WAITCOUNT% GEQ %MAXWAIT% GOTO KILL_IT  TIMEOUT /T 1 > NUL SET /A WAITCOUNT+=1 FOR /F "delims=" %%a IN ('TASKLIST ^| FIND /C "%~nx1"') DO IF %%a EQU 0 GOTO RUN_DONE GOTO WAIT  :KILL_IT TASKKILL /IM %~nx1 /F > NUL :RUN_DONE 

Could be fleshed out ore to take more arguments for priority and such, but I don't have the need for it. Shouldn't be hard to add.