List process by name excluding grep

Question

I am trying to list all the ffmpeg processes that are currently running on Debian machine (Ubuntu 15).

I use the following command:

ps aux | grep 'ffmpeg'

If only one ffmpeg process is running, I still get two results. One for the actual process, and one for grep that is looking for ffmpeg in the process list.

max      21599 13.2  3.0 503848 92288 ?        Rl   01:39   1:18 ffmpeg -f video4linux2 -i /dev/video0 -f mpeg1video -b:v 800k -r 30 http://127.0.0.1:8082/oops/1024/640/ -nostdin -nostats -loglevel fatal
max      23789  0.0  0.0  13688  2172 pts/3    S+   01:49   0:00 grep --color=auto ffmpeg

How can I modify my request so that the grep result which is actually my request is omitted from the output?

Solution

You can use a trick :

ps aux | grep '[f]fmpeg'

Here [] is a format used for indicating range of characters to match using grep, we are using [f] to match only f. As the [f] is converted into f already, the literal ps aux | grep 'ffmpeg' won't show up in ps output.

Also note that appearance of ps aux | grep 'ffmpeg' on the output of ps aux | grep 'ffmpeg' mainly depends on the time it takes for the process table to be shown.

Note that there is a program named pgrep to search for processes, you can us that if ps aux is not absolutely needed.