Unix Scripting - Not returning to calling script 1

[wdellin]

I have a unix korn script that executes another script. If the called script detects an error it does an exit or exit 1. This does not return to the calling script. How do I code the script to return regardless of what happens in the calling script.

Script A

. ${path}/script_B

Script_B
echo 'This is executed from script_A'
if [ -e ${path}/goodfile ]
then
echo 'File exists'
else
echo 'when I do this exit it does not go back to script_A'
exit 1
fi
if [ -e ${path}/badfile ]
then
echo 'when I do this exit it does not go back to script_A'
exit 2
fi
echo 'If it comes here it will return to script_A'

 

[PHV]

This is just because scriptA is sourcing scriptB.
Replace this line:

. ${path}/script_B

by this:

${path}/script_B

And you'll be able to test the exit code ($?) of scriptB.



Hope This Help, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884

 

[wdellin]

Thanks for the answer and the quick reply.

 

[SamBones]

It may need to be sourced if the called script is defining environment variables that the calling script needs. In this case, just replace the "[tt]exit[/tt]" with "[tt]return[/tt]". This will return to the called script passing the return value to it.

Hope this helps.