Hi all,

I am calling script second.sh from script one.sh.
Though I made the last line of second.sh as "exit" , all the other lines down the second.sh in first.sh is not executed.

How can I exit from second.sh and execute the remaining lines in first.sh


Cheers

Think we'd need to see the code.

1. the exit in the called (2nd) script is redundant (you don't need it)
2. sounds like you might be including/sourcing the 2nd script instead of calling it eg

. 2nd.sh

or

source 2nd.sh

either of those syntaxes make 2nd script part of 1st script, instead of calling it externally, so the exit in the 2nd script will exit both scripts.

How are you calling second.sh in first.sh? If you are using one of these:
... that would explain the behavior. sourcing or executing with the dot notation means "read the contents of this file and execute it in the current shell". Thus, when "exit" is called, the first (only) shell will exit.

If you're using exec, then that will also exhibit the same behaviour, because the exec command means "replace the current process with this command", and at that point, your origial script goes away, never to be returned to.

What you probably want to do is execute second.sh in a sub-shell. If you have set the permissions so the file is executable, you can just use the path to the file like this:
The caveat is that a sub-process cannot modify the environment of the parent. If you need to do this you can pass output on stdout, or can source a file which defines functions, which you can then execute in the "parent" script.

Matthew,Chrism,


Sorry to all

Please Ignore this Post. I made a mistake in my script itself .

Thanks for your efforts