Process grep with pipe returns itself. How do I exclude it?

Question

root@test:~# ps x | grep 'vsftpd'
  568 ?        Ss     0:00 /usr/sbin/vsftpd
28694 pts/0    S+     0:00 grep --color=auto vsftpd

How can I exclude the self grep process itself? Also, how do I fetch the process Id (pid) given a part of the name of the process?

I am looking for something along the lines that it will give me pid given the name and excludes the self grep process.

More thorough answers [here](https://serverfault.com/questions/367921/how-to-prevent-ps-reporting-its-own-process) — Unknown, Sep 26 '22 at 05:50

score 10 · Answer 1 · answered Jun 12 '14 at 17:14

10

The usual trick is

ps x | grep '[v]sftpd'

answered Jun 12 '14 at 17:14

choroba

score 8 · Accepted Answer · answered Jun 12 '14 at 17:13

The traditional way would be:

ps x | grep 'vsftpd'| grep -v grep

in which grep -v expr returns everything not matching expr

You can then use awk to extract the relevant field (the pid in your case)

ps x | grep 'vsftpd'| grep -v grep | awk '{ print $2 }'

(the $2 corresponds to the relevant field/column)

2 Answers2