[SOLVED] bash: propogate exit status

[zmunkz]

Hello again,

I had a second question, but I figured it should be a separate post. I am running a command pretty similar to the one below:

sudo -u <user> sh -c "export status=0; source my_job.sh |tee my_log.txt 2>&1; exit \$status"

In order to get the SUDO to exit the same as my_job.sh, I need to capture the exit status of that part of the command. What I am trying to do is set a variable that can be modified within my_job.sh, and used to exit. However, no matter what I set "status" to in my_job.sh, it exits with "0".

What can I do differently? Any help would be appreciated!

[nmset]

Quoting bash man pages concerning source command:

The return status is the status of the last command exited within the script (0 if no commands are executed), and false if filename is not found or cannot be read.

I have not tested through ssh but just in a local shell. If the return status of the last command is assigned to $status, its value can be echoed simply, whether 0 or any other value.

It seems you should assign the return status of every command in your script to $status and exit after the first failure (This seems quite huge a task if it's a big script.)

[lomcevak]

Problem is you're using the exit status of the tee command and not your script. When a pipe is used enviroment variable PIPESTATUS is set. PIPESTATUS is an array. The exit status of the 1st command is in element zero so try:

exit \$PIPESTATUS[0]

[zmunkz]

Exiting with the pipestatus worked! I just had to add braces:

exit \${PIPESTATUS[0]}

Thanks for the help!