Question

I’m struggling to parse the output of the time command in bash – and even to stop it from printing out its output when I call it. This is my test code:

#!/bin/bash
TIME=`time ls -lh > /dev/null`
echo "Testing..."
echo $TIME

This currently prints out:

{blank-line}
real    0m0.064s
user    0m0.002s
sys     0m0.005s
Testing
{blank-line}

So, it seems like the value assigned to $TIME is the blank line at the start of the time print-out. I need to get at the seconds value of the sys line – that is, the “0.005”. I am guaranteed that I will only ever have seconds, so I do not need anything before the “m” – however, the seconds part may be in the form of xx.xxx if it goes >= 10 seconds. I currently have no idea how to suppress the ‘time’ output, capture it all instead of the blank line, nor parse it to get the values I need.

Any help would be much appreciated…

Answer 1

If you use the Bash builtin time, you can control its output by setting the TIMEFORMAT variable:

TIMEFORMAT=%R

and you won’t have to do any parsing since that will cause time to only output the number of seconds.

and use this:

echo "Testing..."
TIME=$( { time ls -lh > /dev/null; } 2>&1 )
echo $TIME

or one of the other techniques from BashFAQ/032.