Grep time command output

Using time ls, I have the following output:

$ time ls -l 
total 2
-rwx------+ 1 FRIENDS None 97 Jun 23 08:59 location.txt
-rw-r--r--+ 1 FRIENDS None 10 Jun 23 09:06 welcome
real    0m0.040s
user    0m0.000s    
sys     0m0.031s

Now, when I try to grep only the real value line, the actual result is:

$ time ls -l | grep real
real    0m0.040s
user    0m0.000s
sys     0m0.031s

My question is, how to get only the real value as output? In this case, 0m0.040s.

Answer

time writes its output to stderr, so you need to pipe stderr instead of stdout. But it’s also important to remember that time is part of the syntax of bash, and it times an entire pipeline. Consequently, you need to wrap the pipeline in braces, or run it in a subshell:

 $ { time ls -l >/dev/null; } 2>&1 | grep real
 real   0m0.005s

With Bash v4.0 (probably universal on Linux distributions but still not standard on Mac OS X), you can use |& to pipe both stdout and stderr:

{ time ls -l >/dev/null; } |& grep real

Alternatively, you can use the time utility, which allows control of the output format. On my system, that utility is found in /usr/bin/time:

/usr/bin/time -f%e ls -l >/dev/null 

man time for more details on the time utility.