How to prevent zombie process?

Black Chaos · 11-16-2006, 09:39 AM

shell_script.sh

Code:

#!/bin/bash
cd /var/www/html/
php somescript.php &
php otherscript.php &
exit 0

When I run that with just a ./shell_script.sh it runs just fine. But when I run it from cron I get a zombie process of sh. Would anyone know a fix for this?

Thanks.

matthewg42 · 11-16-2006, 10:39 AM

zombie processes are processes that have terminated, but have not had the wait() system call made for them (to process their error level and finalise removal from the process table).

As a programmer, you are expected to make sure that if your programs spawn children, that the wait() system call is made for them, just like you should always make sure you call close() on file handles.

The shell has a built-in command called "wait", which you could call like this:

Code:

#!/bin/bash
cd /var/www/html/
php somescript.php &
php otherscript.php &
wait  
# $? is the return status of the last process "wait"ed for
exit $?

This would make the shell wait until all child processes have finished and will call the wait system call for them.

If you want the script itself to run in the background (so you can close the terminal which you run it from), call it like this:

Code:

nohup shell_script.sh &

sundialsvcs · 11-17-2006, 09:22 AM

Yes. When you created your process-tree, the one spawned two children and then died. Those orphans became children of cron, which does not know to wait for them.

When a process dies, it becomes a "zombie" specifically so that its parent can collect ("reap") its ending-status. (Processes that have no live parent become orphans of "process #1 (init)", which immediately reaps them.)

With the change as-proposed, the first process launches two children, then waits for them to complete. It probably doesn't care about their ending status, but by waiting, it causes them to be reaped.