how to print just the 'use%' column with all use % values by 'df -k' command

Question

I want to print Use% column with all % value in 'df -k' command. I want to compare all those value to a variable value.

Answer 1

 df -kP | awk 'NR>1 && $5+0 >= 90'

Would list the entries (in the portable output format of df (with -P) that also guarantees all entries are on a single line) for which the use% is greater or equal to 90. The +0 is to force numeric comparison (otherwise 100% would be considered less than 90).

Answer 2

Is this what you want?

df -k|sed 's/  */\t/g;s/%//'|cut -f 5

You can then eliminate the first line and compare the values, but you have no more device relation at this point.

[EDIT:] So I was right and you need the device. So you want something like this:

for item in `df -k|sed 's/  */_/g;s/%//'|cut -d_ -f 1,5`; do
  if (( ${item#*_} > 90 )); then
    echo ${item%_*} has ${item#*_}%
  fi
done

This is bash syntax. Adapt it for your needs.

Answer 3

With GNU df - Untested on aix

You can have the Use% values without parsing df output:

df --output=pcent