Redirect output of time command in unix into a variable in bash?

Question

I just want to capture the output of a time command i.e:

X=$(time ls)

or

$(time ls) | grep real

The time function spits it to the console though. How do I do this?

Answer 1

X=`(time ls) 2>&1 | grep real`

Answer 2

See BashFAQ/032.

$ # captures output of command and time
$ time=$( TIMEFORMAT="%R"; { time ls; } 2>&1 )    # note the curly braces

$ # captures the time only, passes stdout through
$ exec 3>&1 4>&2
$ time=$(TIMEFORMAT="%R"; { time ls 1>&3 2>&4; } 2>&1)
bar baz
$ exec 3>&- 4>&-

The time will look like "0.000" using TIMEFORMAT="%R" which will be the "real" time.

Answer 3

Time writes its output to STDERR rather than STDOUT. Making matters worse, by default 'time' is a shell builtin command, so if you attempt 'time ls 2>&1' the '2>&1' only applies to 'ls'.

The solution would probably be something like:

/usr/bin/time -f 'real %e' -o OUTPUT_FILE ls > /dev/null 2>&1<br>
REALTIME=$(cat OUTPUT_FILE | cut -f 2 -d ' ')

There are more fancy ways to do it, but that is the clear/simple way.