I am trying to check if a process (assume it is called some_process) is running on a server. If it is, then echo 1, otherwise echo 0.
This is the command that I am using but it only works partially (more info below). Note that I need to write the script in one line.
ps aux | grep some_proces[s] > /tmp/test.txt && if [ $? -eq 0 ]; then echo 1; else echo 0; fi
Note: The [s] in some_proces[s] is to prevent grep from returning itself.
If some_process is running, then "1" gets echoed, which is fine. However, if some_process is not running, nothing gets echoed.
This question is related to
bash
pgrep -q some_process && echo 1 || echo 0
more oneliners here
&& means "and if successful"; by placing your if statement on the right-hand side of it, you ensure that it will only run if grep returns 0. To fix it, use ; instead:
ps aux | grep some_proces[s] > /tmp/test.txt ; if [ $? -eq 0 ]; then echo 1; else echo 0; fi
(or just use a line-break).
Use grep -vc to ignore grep in the ps output and count the lines simultaneously.
if [[ $(ps aux | grep process | grep -vc grep) > 0 ]] ; then echo 1; else echo 0 ; fi
You can make full use of the && and || operators like this:
ps aux | grep some_proces[s] > /tmp/test.txt && echo 1 || echo 0
For excluding grep itself, you could also do something like:
ps aux | grep some_proces | grep -vw grep > /tmp/test.txt && echo 1 || echo 0
Source: Stackoverflow.com