I need to get a PID of a process by name in bash script. I cannot use pidof as it is not available on my Linux distribution and I cannot install that.

$ ps aux | grep name_of_process 
user  1111  0.0  03:20   0:00 grep name_of_process 
user2 2222  2.4  03:11   0:13  -Dappsrv.root=/dir1/dir2/dir2/tomcat/name_of_process 

I tried the following solution but it gives my two pids each time and in random order thus it is no use:

OUTPUT=$( pgrep -f name_of_process )
VARS=( $OUTPUT )
echo $VARS  #1111 2222

I also tried:

ps -ef | awk '$8=="name_of_process" {print $2}'

But that does not return anything

I need to kill that process but each time from command

sudo -u root kill -9 `ps -ef | grep '[n]ame_of_process' | awk '{print $2}'`

170497
181212
kill 170497: Operation not permitted    
kill 181216: No such process

CodePudding user response：

Your pgrep approach looks basically reasonable for me, but why are you using -f?

VARS=( $(pgrep name_of_process) )

should be sufficient. But be careful with the process name: pgrep interprets the argument as extended regular expression, and when your process name contains i.e. . or or other regex characters, the outcome is not necessarily what you expect.

CodePudding user response：

If anyone will encounter the same problem: My script name contained name of the process therefore when I was trying to get the PID of the process by it's name it found two process: one for the actual process and second for my script name.

CodePudding user response：

It's simple:

1. Type ps aux - this command get you all process
2. Filter process by name with | grep PROCESS_NAME
3. Now from the output get the PID of the process from second column with |awk '{print $2}'

Like this:

ps aux| grep PROCESS_NAME|awk '{print $2}'