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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 ?

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1
  • it seems start-stop-deamon returns 0 if the requested action was performed (note that it's not on if the process started). You can add some logic and use pidof to match pid.
    – Simply_Me
    Commented Aug 8, 2014 at 22:45

1 Answer 1

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3

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.

See start-stop-deamon manpage

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  • Thanks @jordanm for completing my answer. I forgot on the wait command.
    – Chemik
    Commented Aug 9, 2014 at 0:06
  • when I add wait $! as suggested by you, then the start-stop-daemon never finishes. i.e. when I do service foo start it waits indefinitely.
    – Martin Vegter
    Commented Aug 9, 2014 at 9:26
  • Read the answer properly. The wait command is there only for illustration how to get return code of backgrounded process. Use --background switch instead.
    – Chemik
    Commented Aug 9, 2014 at 14:20

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