As the output of the free -m command, I get the following:
total used free shared buffers cached
Mem: 2496 2260 236 0 5 438
-/+ buffers/cache: 1816 680
Swap: 1949 68 1881
I want to get only used memory, like 2260, as output. I tried the following command:
free -m | grep Mem | cut -f1 -d " "
Help me to improve my command.
How can I get it as a percentage, like 35%?
You can use awk without the need for a separate grep pipe for this:
awk '/^Mem/ {print $3}' <(free -m)Where records/rows are filtered for those beginning with Mem and the third field/column ($3) is printed for the filtered record.
Or with sed:
free -m | sed -n 's/^Mem:\s\+[0-9]\+\s\+\([0-9]\+\)\s.\+/\1/p'
Another solution would be:
free -m | grep ^Mem | tr -s ' ' | cut -d ' ' -f 3
Credits for the second solution got to this post.
As for the added question of displaying as percentage (based on jasonwryan's answer):
awk '/^Mem/ {printf("%u%%", 100*$3/$2);}' <(free -m)
get percentage by diving 3rd field by 2nd and print as an integer (no rounding up!).
EDIT: added double '%' in printf (the first one escapes the literal character intended for printing).
with bash free and grep only
read junk total used free shared buffers cached junk < <(free -m | grep ^Mem)
echo $used
FNR =2 is 2nd row of output and {print $3} printing the 3rd column which is used
free -m | awk 'FNR == 2 {print $3}'
