bash: Run multiple command in background, store output in variable

Question

How do I run a sequence of commands in parallel, and store their output in a variable?

I've tried:

output=`(echo -n "started "; sleep 2; echo "stopped") &`

echo "output before=$output"
wait
echo "output after=$output"

And got a pause of two seconds, followed by:

output before=started stopped
output after=started stopped

I expected:

output before=
<2 seconds pause>
output after=started stopped

How do I run a sequence of commands in the background, and store their output in a variable?

All block within () is executed on background, so you get its output once everything is finished. Hence, the sleep is "hidden" to you. — fedorqui
– fedorqui, Commented Oct 14, 2013 at 11:16
What are you really trying to do? This example looks constructed to illustrate an issue that may not be the best way to solve the problem you really have. — chepner
– chepner, Commented Oct 14, 2013 at 12:08
Run ~10 commands in parallel, wait for termination, and process their standard output separately. — Adam Matan
– Adam Matan, Commented Oct 14, 2013 at 12:22

devnull · Accepted Answer · 2013-10-14 11:33:35Z

6

From the manual:

If a command is terminated by the control operator ‘&’, the shell executes the command asynchronously in a subshell. This is known as executing the command in the background. The shell does not wait for the command to finish, and the return status is 0 (true).

A workaround would be to store the output in a file and read from it. The following might work for you:

tempout=$(mktemp)
( echo -n "started "; sleep 2; echo "stopped" ) > "${tempout}" &
echo "output before=$(<$tempout)"
wait
echo "output after=$(<$tempout)"

answered Oct 14, 2013 at 11:33

devnull

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1 Comment

chepner Over a year ago

There might be buffering issues involved here, where nothing is available yet to read from $tempout when the first echo is called. Depending on processor load, the background task may not even be started before the first echo.

Greg · Accepted Answer · 2015-06-26 01:22:49Z

0

I think the reason is the evaluation of this line: output=`(echo -n "started "; sleep 2; echo "stopped") &` Even though you want to put the whole command in the background (and this results in return code 0, as devnull mentioned) - but the result is not known. And to assign value to $output - it has to wait for the original command to exit.

You did not put var assignment in the background.

But still @devnull's workaround applies. Instead of files you may use named pipes.

answered Jun 26, 2015 at 1:22

Greg

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