Get separate used memory info from free -m comand

Question

As a output of free -m command I got 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 following command.

free -m | grep Mem | cut -f1 -d " " 

Help me to improve my command.

How can I get it in percentage like 35% ?

Answer 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).

Answer 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.

Answer 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.