How to store output from a bash command in a variable

Question

Hello I am trying to store an OS X command in a variable and I am having problems doing so. Here is my code:

#! /bin/bash
Output=$(dscl . -read /Users/root AuthenticationAuthority)
Check="No such key: AuthenticationAuthority"  

if [ "$Output" = "$Check" ]
    then
       echo "OK"
      else
    echo "FALSE"
fi

I have done this before with commands such as "defaults read...." and it works fine but the dscl . -read will not store the output in the variable. Any ideas?

When you run that script do you see the output on your screen? (Is it returning that error to standard error instead of standard output?) — Etan Reisner
– Etan Reisner, Commented Mar 12, 2015 at 17:19

kojiro · Accepted Answer · 2015-03-12 17:23:38Z

7

On failure, the dscl command, as well as all standard shell commands, outputs the error message on stderr, whereas $(...) only captures stdout.

You have to merge the two streams first:

Output=$(dscl . -read /Users/root AuthenticationAuthority 2>&1)

edited Mar 12, 2015 at 17:23

kojiro

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answered Mar 12, 2015 at 17:21

0xFC

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Comments

Brian Campbell · Accepted Answer · 2015-03-12 17:23:02Z

0

When the dscl command succeeds, its output goes to stdout, which is captured by the command substitution.

When there is an error, the message is printed to stderr instead.

To capture either stdout or stderr, you can redirect stderr in the command to go to stdout:

Output=$(dscl . -read /Users/root AuthenticationAuthority 2>&1)

answered Mar 12, 2015 at 17:23

Brian Campbell

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