Question

I am facing with the following bash script:

#! /bin/bash
processname=$1
x=1
while [ $x -eq 1 ] ; do
    ps -el | grep $processname | awk ' {if($15!="grep") print "The memory consumed by the process " $15 " is = "$9}  ' >> out.log
    sleep 30
done

and I am running this with :

$ ./myscript.sh Firefox

but when i see the output in the file, apart from the firefox process i am also getting information for /bin/bash process

The memory consumed by the process /Applications/Firefox.app/Contents/MacOS/firefox is = 911328  
The memory consumed by the process /bin/bash is = 768

Can some one help me with this so that I only want to get information related to Firefox process and nothing else(/bin.bash etc)

Answer 1

This is normal and because the $processname is Firefox. Since your command also monitors it, there is a process that uses it.

For example, try ps -el | grep Firefox and you will get two process lines matching (if you have one instance of Firefox running), one is Firefox, the other is the grep command looking for Firefox.

Piping your output in grep -v /bin/bash' should solve this. Eg:

ps -el | grep $processname | awk ...

becomes:

ps -el | grep $processname | grep -v 'grep' | awk ...