[SOLVED] Bash script to get PID of a specific process.

arashi256 · 08-23-2010, 07:27 AM

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.

vinaytp · 08-23-2010, 07:34 AM

Is this what you are looking for ?

Code:

$PS_OUTPUT=`ps -elf | grep <process name> | grep -v grep | awk '{print $4}`

arashi256 · 08-23-2010, 07:36 AM

Perfect, thank you! I didn't know about the -v (invert) switch for grep. That will be very useful in future.

bathory · 08-23-2010, 07:37 AM

Or, you can use

Code:

pidof <process name>

that will give just the PID

arashi256 · 08-23-2010, 07:38 AM

Ah, that's good also

Didn't know about that command.

rgdacosta · 08-23-2010, 07:40 AM

Why don't you use pgrep?

Code:

thanatos:~ # pgrep cron
3396

arashi256 · 08-23-2010, 07:45 AM

Cool - didn't know that one either

Does the same job as pidof by the looks of it.