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Old 08-23-2010, 07:27 AM   #1
arashi256
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Registered: Jan 2008
Location: Brighton, UK
Distribution: Ubuntu 12.04 / CentOS 6.5
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Bash script to get PID of a specific process.


I'm trying to write a script that involves storing a specific PID in a variable. I've got as far as: -

Code:
$PS_OUTPUT=ps -elf | grep <process name> | awk '{print $4}'
This stored the PID of the grep itself and the actual process I want. The output of $PS_OUTPUT is: -

Code:
3548
10747
I only want the second number as the first is the PID of the grep. I either want to suppress the PID of the grep from the output or just grab the second PID. Since the output is on two lines, I thought there must be a newline (\n) character in there somewhere, but if I use: -

Code:
ps -elf | grep <process name> | awk '{print $4}' | awk -F"\n" '{print $2}'
...I get nothing and {print $1} replaced in the above gives me both PIDs again on two lines. Which character is outputting the "ps" command on two lines that isn't a "\n"? Can I somehow write a regex that will exclude the PID of the grep itself?

Thanks for any help.
 
Old 08-23-2010, 07:34 AM   #2
vinaytp
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Is this what you are looking for ?

Code:
$PS_OUTPUT=`ps -elf | grep <process name> | grep -v grep | awk '{print $4}`
 
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Old 08-23-2010, 07:36 AM   #3
arashi256
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Perfect, thank you! I didn't know about the -v (invert) switch for grep. That will be very useful in future.
 
Old 08-23-2010, 07:37 AM   #4
bathory
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Or, you can use
Code:
pidof <process name>
that will give just the PID
 
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Old 08-23-2010, 07:38 AM   #5
arashi256
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Ah, that's good also Didn't know about that command.
 
Old 08-23-2010, 07:40 AM   #6
rgdacosta
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Why don't you use pgrep?


Code:
thanatos:~ # pgrep cron
3396
 
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Old 08-23-2010, 07:45 AM   #7
arashi256
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Cool - didn't know that one either Does the same job as pidof by the looks of it.
 
  


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