How to exit a batch program upon error?

Question

I've got a batch file that does several things. If one of them fails, I want to exit the whole program. For example:

@echo off
type foo.txt 2>> error.txt >> success.txt
mkdir bob

If the file foo.txt isn't found then I want the stderr message appended to the error.txt file, else the contents of foo.txt is appended to success.txt. Basically, if the type command returns a stderr then I want the batch file to exit and not create a new directory. How can you tell if an error occurred and decide if you need to continue to the next command or not?

I added the code IF NOT ERRORLEVEL 0 EXIT /B echo %errorlevel% before the mkdir bob command, but regardless of the value of ERRORLEVEL (i.e. 0 or 1) the directory is still created. So basically, ERRORLEVEL is being set with a different value whether the type command finds the file or not, but the program is not exiting. Thoughts? — Notorious2tall
– Notorious2tall, Commented Jul 21, 2010 at 21:07

danio · Accepted Answer · 2014-04-02 16:26:41Z

10

use ERRORLEVEL to check the exit code of the previous command:

 if ERRORLEVEL 1 exit /b

EDIT: documentation says "condition is true if the exit code of the last command is EQUAL or GREATER than X" (you can check this with if /?). aside from this, you could also check if the file exists with

 if exist foo.txt echo yada yada

to execute multple commands if the condition is true:

 if ERRORLEVEL 1 ( echo error in previous command & exit /b )

or

 if ERRORLEVEL 1 (
    echo error in previous command
    exit /b
 )

edited Apr 2, 2014 at 16:26

danio

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answered Jul 21, 2010 at 20:39

akira

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