Why is this command in the bash script not running in the background?

I have a bash script that includes the following lines:

    .
    .
    if [ -z "$(pgrep mplayer)" ]; then
      /usr/bin/mplayer -slave -input file=/home/administrator/files/mplayer-control.pipe http:/www.musicserveraddress.com/ &
    fi
    .
    .
    Other things to execute
    .
    exit

what happens is that mplayer connects to the streamingserver and s开发者_运维知识库tart playing the stream. However, the script never moves on. I added an ampersand to move this process to the background so that the script should continue to run and then exit itself (keeping the audio stream playing).

How should I do to achieve that? Thanks in advance/J

Edit: It runs as planned when I run the script from the command line, but it is intended to be run as a cron job (and the pgrep is intended to start mplayer only if it has crashed since last cron job). When run as a cron job, nothing happens...

---

Try with a nohup at the beginning of the command :

nohup /usr/bin/mplayer -slave -input file=/home/administrator/files/mplayer-control.pipe http:/www.musicserveraddress.com/ &

---

The command you've used works perfectly fine for me. The only thing that can confuse people is that you don't get the bash prompt when the script finishes, but it's actually there, try pressing Enter.

As you want the player to be controlled via a pipe and be in the background, I'd recommend to redirect standard streams from the console as well and send all the output to a logfile:

(
  exec </dev/null
  exec >/dev/null
  exec 2>/dev/null

  umask 0
  cd /

  exec setsid mplayer -slave -idle -input /home/user/control.pipe http://server.com > /var/log/mplayer.log 2>&1
) &

---

You may want to use setsid as well. That worked for me.