How to get the pid of the last executed command in shell script?

Question

I want to have a shell script like this:

my-app &
echo $my-app-pid

But I do not know how the get the pid of the just executed command.

I know I can just use the jobs -p my-app command to grep the pid. But if I want to execute the shell multiple times, this method will not work. Because the jobspec is ambiguous.

Solution 1

The PID of the last executed command is in the $! shell variable:

my-app &
echo $!

Solution 2

Get PID:

#!/bin/bash
my-app &
echo $!

Save PID in variable:

#!/bin/bash
my-app &
export APP_PID=$!

Save all instances PID in text file:

#!/bin/bash
my-app &
echo $! >>/tmp/my-app.pid

Save output, errors and PID in separated files:

#!/bin/bash
my-app >/tmp/my-app.log 2>/tmp/my-app.error.log &
echo $! >>/tmp/my-app.pid

echo "my-app PID's: $(cat /tmp/my-app.pid)"

Solution 3

Try something like

pidof my_app >> /tmp/my_app.pid

Comments

[Arsen Karapetjan]

did you try echo $!? right after the command, you ran?

[Mat]

That's very fragile at best and doesn't work if my-app is executed more than once - davidshen84 specifically worries about that cas.

[user3153014]

It is printing pid as for eg. [1] 893 . I want only number.

[ramrunner]

It should be noted that this stands for programs started in the background. If no background processes have been started the parameter is not set.

[Willem]

If you even go this route, you should use pgrep instead.

[imz -- Ivan Zakharyaschev]

Another worthy solution is suggested in (a comment to an answer to) How to get pid of just started process: oh, and the "oneliner": /bin/sh -c 'echo $$>/tmp/my.pid && exec program args' & – sysfault Nov 24 '10 at 14:28

[S.D]

@user3153014 String s like "[2] 2625" are printed by shell after starting background task. This is not related to output of echo $!

[Eric Renouf]

The question didn't ask about redirection, then your answer is the mostly the same as the accepted one except that if an instance of my-app finishes, and perhaps even worse its PID gets reused later, you'll have bad information in your file. I don't think this answer adds much of value to the existing accepted answer really

[Eduardo Cuomo]

@EricRenouf post updated!

[mlathe]

another derivative that might be useful. This gets the PID and at the same time treats it as a (mostly) foreground process: sleep 4 | grep INVALID & export MY_PID=$!; fg; echo $MY_PID returned with $? `

[MrMesees]

isnt APP=main & a way to grab PID?

[papo]

external command, and slow, if system can give PID directly why searching through all processes to find just that? and also, can't be certain you'll get correct value

[Vrata Blazek]

I usually use such code in a script when I need to wait until the process ends. my-app & myVar=$! ; fg. fg brings the process to foreground again. I can then print echo $myVar later and I'm pretty sure that the my-app has already finished.

[Michele Piccolini]

If you are using that inside su {user} -c {command} as command, be sure to escape the $! (use \$!). See unix.stackexchange.com/questions/147916/…

[Michael]

@VrataBlazek fg: no job control

[Michael]

@imz--IvanZakharyaschev Why is it necessary to pipe the PID to a file? If I try to wrap the shell call with $() it hangs at that point... why can't I get it into a shell variable directly? When I try to pipe it to a file I have another problem - now the PID of my command isn't the same as reported, whereas if I don't pipe it's the same!

[imz -- Ivan Zakharyaschev]

@Michael Hiw do you want to get it into a shell variable directly? If you suggest a method, I could try to say whether it would work or not. As for $() wrapping, I can't try to explain the behavior without seeing the exact code.