Bash: Suppress standard output, capture error output

Question

So I'm running commands in a bash shell script, and I want to ONLY capture error output in a variable. Any standard output I want suppressed.

So far what I have definitely isn't working. So I'm trying to come up with a better idea, and struggling to find an answer.

Here's a code sample:

ERROR=$(svn switch "$NEW_URL" --accept postpone 1>/dev/null 2>&1) &

Looks like everything is getting suppressed. Any help would be appreciated. Thanks.

anubhava · Accepted Answer · 2016-04-07 20:50:56Z

2

It should be like that:

error=$({ echo "stdout"; echo "stderr" >&2; } 2>&1 >/dev/null)
echo "$error"
stderr

i.e. redirect stderr->stdout first and then redirect stdout to /dev/null to suppress stdout.

For your command it should be:

error=$(svn switch "$NEW_URL" --accept postpone 2>&1 >/dev/null)

You also need to remove & to avoid pushing your command in background.

answered Apr 7, 2016 at 20:50

anubhava

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