BASH Total CPU Percentage

[eklikeroomys]

Hi,

I've been searching for a neat way of finding the total instantaneous CPU usage as a %, and found all sorts of tedious solutions.

I thought I'd share a simple one-liner that accomplishes the task really simply:


top -bn 1 | awk '{print $9}' | tail -n +8 | awk '{s+=$1} END {print s}'

Basically, it grabs the CPU column from top and sums all the values together. Simple as that.

Enjoy!

[Habitual]

very slick. :)

[Vaphell]

this will do the same

top -bn 1 | awk 'NR>7{s+=$9} END {print s}'

imho that code shows wrong number, you have to divide by number of cores.
in my case system monitor shows approx: 20/20/10/10
code: 60

top -bn 1 | awk 'NR>7{s+=$9} END {print s/4}'

why not print the 3rd line where 'ingredients' of CPU load are listed?

top -bn 1 | sed -n 3p
or assuming cpu load = 100% - idle%

top -bn 1 | awk 'BEGIN{FS="[ \t%]+"} NR==3{ print 100-$8 }'

[eklikeroomys]

I agree that dividing by the number of CPUs will give the actual percentage of total CPU usage if you have more than one core. Thanks!