I tried this:
top -bn1 | grep "Cpu(s)" | \
> sed "s/.*, *\([0-9.]*\)%* id.*/\1/" | \
> awk '{print 100 - $1"%"}'
2.1%
what's the meaning ?
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Andreas Louv
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LeoChu
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I think it gets the CPU usage of the system idle process, then inverts it. – Jun 23 '16 at 10:50
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On my Debian, it seems to extract the idle %, then subtract that from 100, which should get all the "non-idle" fields in one number. – tripleee Jun 23 '16 at 10:55
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Refactoring the [useless use of `grep | sed | awk`](http://www.iki.fi/era/unix/award.html#grep) should reduce the load, albeit only slightly. – tripleee Jun 23 '16 at 10:56
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@tripleee Dead link – 123 Jun 23 '16 at 10:58
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@123 Sorry, should be back up later today. – tripleee Jun 23 '16 at 11:02