Capturing bash time output in script variable

Question

I am trying to do this:

TIMEFORMAT=%R; foo=$(time wget http://www.mysite.com) echo $foo 

and when I execute, I see the number I want in the output, but not in the variable foo (echo $ foo does not print anything).

Why is this?

Answer 1

You don't write anything to foo because time sends its output to stderr . The problem is that the wget command also sends most of its output to stderr . To separate the two streams (and throw the output from wget ), you will need to use subshell :

 TIMEFORMAT=%R; foo=$( time ( wget http://www.example.com 2>/dev/null 1>&2 ) 2>&1 ) echo $foo 

Here is an explanation of what is happening ...

The inside of this command is:

 ( wget http://www.example.com 2>/dev/null 1>&2 ) 

Sends both stderr and stdout to /dev/null , essentially discarding them.

Outside:

 foo=$( time ( ... ) 2>&1 ) 

Sends stderr from the time command to the same place where stdout sent, so that it can be captured using the replacement ( $() ).

Update:

If you want to become really smart, you can pass the output of wget to stderr by manipulating file descriptors as follows:

 foo=$( time ( wget http://www.example.com 2>&1 ) 3>&1 1>&2 2>&3 )