Running function or command concurrently in a bash script


 
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# 8  
Old 02-05-2009
Quote:
Originally Posted by dj_bridges
I never realised that by defining a variable in /var/run/ it would be assigned a pid.
That's not exactly what's happening, my friend. The $$ expands to the process ID of the currently running script. So your "keepalive" script echo's that out to the file.

Quote:
While this is a bit off-topic - are the other .pid files in /var/run/ dynamic i.e. if one were to kill the pid in that .pid file that .pid file would disappear?
Generally no. Those PID files are created by the "init" scripts. You can see these in the scripts in /etc/rc${RUNLEVEL}.d/ where RUNLEVEL is the number you see with /sbin/runlevel. These scripts use the PID files so that they can easily stop and restart the correct service. If you remove the pid file, these scripts can be confused and cause programs not to shutdown properly. If you have a mysql database running, for instance, and you shutdown, if the pid file is no longer there, mysql might be shutdown abruptly before it has a chance to flush its buffers, resulting in data corruption.
# 9  
Old 02-05-2009
Aahhh that makes a lot more sense! I was wondering what the $$ meant - not the easiest thing to search on google by the way to try and find out. Will leave the /var/run/ .pid files well alone!

Thanks for taking the time to explain. I have got a lot more learning to do......
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