Get separate used memory info from free -m command

Question

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%?

Solution 1

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).

Solution 2

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.

Solution 3

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.

Solution 4

with bash free and grep only

read junk total used free shared buffers cached junk < <(free -m  | grep ^Mem)
echo $used

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Comments

[Brian D]

note that -m part is optional and simply converts the output number from kB to mB.

[peterph]

@user34571 thanks for pointing out the double % issue. You may also want to put these things into comment next time (to get some credits :)).