Save in a variable the number of seconds a process took to run

Question

I want to run a process in bash and save in an env variable the number of seconds it took to run. How would I do such a thing?

Answer 1

Using GNU time,

\time -p -o time.log $COMMAND

and then read time.log.

(Use either \time or command time, otherwise you'll be using Bash's time built-in, which doesn't support these options.)

This will work even when $COMMAND prints to stderr (which would confuse Oli's answer), and keeps stdout/stderr (which Farzy's answer doesn't).

-o ... tells time to send its output to a file rather than to stderr (as is the default), and -p generates the traditional

real 0.00
user 0.00
sys 0.00

rather than GNU time's default of

0.00user 0.00system 0:00.01elapsed 8%CPU (0avgtext+0avgdata 0maxresident)k
80inputs+0outputs (1major+188minor)pagefaults 0swaps