Output of a command not passing to assigned variable

Question

I was trying to test nginx config in a script:

reply="$(nginx -t)"
echo $reply

But $reply is not containing the returned text. If i put first line in a terminal, nginx texts are popping in the terminal, but if i put second line in terminal, nothing outputs. My interest is to use that nginx test's output in a if statement

reply="$(nginx -t)"
    if [[ ${reply} == *"nginx: configuration file /etc/nginx/nginx.conf test failed"* ]]
    then
            echo "It's there!";
    fi

Is there any way to get the output returned by the command nginx -t in the variable ?

What does `$reply` contain? Make sure to use `echo "$reply"` and not `echo $reply` to check — that other guy, May 26 '16 at 20:41

score 3 · Accepted Answer · answered May 26 '16 at 20:44

3

Judging by your conditional you're testing for an error message, which means that nginx most likely printed that to stderr, whereas $(...) only captures stdout output by default.

In order to capture both streams, use:

reply="$(nginx -t 2>&1)"

2>&1 redirects stderr to stdout, so that stderr is also captured by $(...), the command substitution.

answered May 26 '16 at 20:44

mklement0

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A very straightforward answer that works perfectly. So are those stdout and stderr are some kind of streams ? – cowboysaif May 26 '16 at 20:55
Yes, stdin, stdout, and stderr are standard streams in Unix-like operating systems - see https://en.wikipedia.org/wiki/Standard_streams – mklement0 May 26 '16 at 20:57

Output of a command not passing to assigned variable

1 Answers1