Alternatively you can monitor the process using pattern (no pidfile needed):
check process myprocess matching "/usr/bin/myprocess -a -b -c"
Regards,
Martin
On 07 May 2014, at 13:34, Anthony Griffiths <[email protected]> wrote:
> I think I realise now what the (simple) problem is.
> monit does not line '/bin/sh /root/stream-start.sh', I need to use something
> else but I don't know what.
>
>
> On Wed, May 7, 2014 at 7:09 AM, Weedy <[email protected]> wrote:
> On 06/05/14 06:03 PM, Anthony Griffiths wrote:
> > I wrote a script called stream-start.sh that follows your instructions
> > but even though ffmpeg starts fine it does not create a pid file. I
> > created this:
> > *long-ffmpeg-command &*
> > *pid=$!*
> > *echo $pid > /var/run/stream.pid*
> cat > /root/start-stream.sh
> #!/bin/sh
> if [ ! -d /var/run/ ]; then mkdir -p /var/run/; fi
> if [ -f /var/run/stream.pid ]; then rm /var/run/stream.pid; fi
>
> /usr/bin/ffmpeg -a -lot -of -options &
> sleep 3 # make sure it's given enough time to start/die??
> echo $(pidof --single-shot ffmpeg) > /var/run/stream.pid
>
> <hit ctrl+d after at least one blank line>
> chmod +x /root/start-stream.sh # not really needed
> > but this didn't work and and after trawling google and trying a few
> > variations the script still does not create a pid file.
> >
> > in monitrc I have:
> > *check process stream with pidfile /var/run/stream.pid*
> > * start = "/usr/bin /root/start-stream.sh"*
> > and monit starts without any complaints
>
> I... what? /usr/bin is a directory. This is probably why $! failed for
> you, your not running the file with a specific interpreter. Hard code
> the interpreter so this never happens again.
>
> start = "/bin/sh /root/start-stream.sh"
>
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