script sourcing issue.

[rakeshou]

I have a script say abc.sh which needs to be executed by another script as

. ../abc.sh

but the control never returns to the calling script. The control does come back if I use
../abc.sh

but then abc.sh is run as child script.

Please tell me how do I bring the control back to the calling script so that the calling script can read the exported variables from the called script.


abc.sh (just for demonstration) is as under:

#/bin/ksh -x

DEFAULT_DB=tektips
export DEFAULT_DB

if [ $DEFAULT_DB = "tektips" ]; then
exit 0
else
exit 1
fi

 

[SamBones]

To start, your first line is wrong. You're missing an exclamation point ("!"). Also, this line isn't executed when the file is sourced, only when the script is run as it's own process.

You should also change the "exit"s to "return"s.

#!/bin/ksh -x

DEFAULT_DB=tektips
export DEFAULT_DB

if [ $DEFAULT_DB = "tektips" ]; then
   return 0
else
   return 1
fi

Still, you can only set a variable in a script if it's being sourced.

 

[SamBones]

The reason it doesn't return when you are sourcing it is because of the "exit"s. Since you are sourcing it, the "exit" is run by the calling script, not abc.sh, and causes the calling script to exit. So the "exit" kills the parent and just appears to not return.