bash - get pid for a script using the script filename

Question

I have two scripts:

• running_script
• script_one

I need to get the PID for the/any instances of running_script running under a username, and then pkill to stop the running_script and daughter processes.

We expected something like:

ps -fu will | grep running_script 

to find the running_script process(es). However checking the PID against the ps command output show that the cmd as: "bin/bash" for the running_script process.

running_script runs as a detached process(& operator) which starts script_one. I print the PID-s at the start to compare with ps command's output.

running_script & echo $! -- $BASHPID 

In the real use-case, we won't have PIDs for some running_script processes running. Also, script_one may or may not be a detached process from the running_script parent.

For the purposes of the exercise, script_one just does loops.

while [ true ] do echo " $0 - 35sec ..." sleep 35 done 

However that's just the example. The requirement is to get PID for the parent, running_script process(es).

Is there an option on ps or another command that can give me the name of the script file and the PID? Or a method to set a name that can be searched for?

In the final use-case, there could be several instances of running_script so picking them out by name seems the best option to date.

example

I thought it might help to show what the ps command shows, since most responses appear to think that's going to work. I ran this example just a while ago.

$ ./running_script & $ echo $! - $BASHPID 9047 - 3261 $ ps -ef | grep will UID PID PPID C STIME TTY TIME CMD will 8862 2823 0 22:48 ? 00:00:01 gnome-terminal will 8868 8862 0 22:48 ? 00:00:00 gnome-pty-helper will 8869 8862 0 22:48 pts/4 00:00:00 bash * will 9047 3261 0 22:55 pts/2 00:00:00 /bin/bash will 9049 9047 0 22:55 pts/2 00:00:00 /bin/bash will 9059 2886 0 22:56 pts/0 00:00:00 man pgrep will 9070 9059 0 22:56 pts/0 00:00:00 pager -s will 10228 9049 0 23:31 pts/2 00:00:00 sleep 35 will 10232 8869 0 23:31 pts/4 00:00:00 ps -ef will 10233 8869 0 23:31 pts/4 00:00:00 grep --colour=auto william 

I have marked PID #9047, is simply shows: - will 9047 3261 0 22:55 pts/2 00:00:00 /bin/bash

Is there something like a a "jobname" attribute I could set on linux?

Answer 1

Try pgrep -f running_script -- the -f option uses the whole command line to match against, not just the process name

Answer 2

ps -o pid,args -C bash | awk '/running_script/ { print $1 }'

This uses ps to get the pid and args for all bash processes, then uses awk to print the pid (field 1) of the matching process.

BTW, ps -o pid,args -C bash gives you the pid and the name of the script file you asked for - the script's name is in the args of the bash command.

If you're going to use ps rather than pgrep at least use its full capabilities rather than the extraordinarily ugly ps ... | grep -v grep | grep | awk construct. The last grep isn't even needed as awk can perform the pattern match. In fact, neither of the greps are needed as awk can do it all: ps ax | awk '! /awk/ && /myprocessname/ { print $1}'

Answer 3

ps -ef | grep -v grep | grep <script_name> | awk '{print $2}' 

The above command will give the PID for the script_name. Also you can have the script write a temporary file with its running PID.

To save the current PID, add a line inside the script:

echo $$>/tmp/script_pid.tmp 

Answer 4

#!/bin/bash # pidofargs.sh # get process id by first argument of a process # using `ps -eo pid,args` and `read -a` # sample use: pidofargs.sh python /usr/bin/youtube-dl search_grep="$1" # first filter search_arg1="$2" # second filter # first filter ps -eo pid,args | grep "$search_grep" | while read pid args do read -a arg < <(echo "$args") # pid = process id # arg[0] = executable file of process # arg[1] = first argument of process # second filter if [[ "${arg[1]}" == "$search_arg1" ]] then # success echo $pid break fi done