I have set up start-stop-daemon to start my script automatically
case "$1" in
start)
log_begin_msg "starting foo"
start-stop-daemon --start --chuid nobody --user nobody --pidfile \
/tmp/foo.pid --startas /usr/local/bin/foo.sh &
log_end_msg $?
the problem is, it always returns 0 (success),even if the process was not started.
How can I capture the return code of start-stop-daemon properly ?
You are not capturing return code of start-stop-daemon.
Your problem is that you are launching it in the background and it is started properly. I mean that you are capturing return code of starting something in background that wants to start something in background.
Try this:
rm /tmp/not_existent_file &
echo $?
This always prints 0.
In order to get the return code of a backgrounded process, you must wait for it to exit with wait. Here is an example:
rm /tmp/not_existent_file &
wait $!
echo $?
If you want to start process that is not forking on its own, try to use --background switch and remove & from the end of start-stop-daemon line.
wait command.
wait $! as suggested by you, then the start-stop-daemon never finishes. i.e. when I do service foo start it waits indefinitely.
wait command is there only for illustration how to get return code of backgrounded process. Use --background switch instead.
start-stop-deamonreturns0if the requested action was performed (note that it's not on if the process started). You can add some logic and usepidofto matchpid.