How to test if variable is set and not empty, when using “set -o nounset”

Question

Yes there is a similar thread here: Test if a variable is set in bash when using "set -o nounset"

However there are so many different answers that it's not particularly clear.

Would the following be sufficient to test if a variable is set AND not empty?

#!/bin/bash 
set -o nounset

if [[ ! -z "${EXAMPLE-}" ]]; then
    echo "Variable is defined and is not empty..."
fi

This is covered by [BashFAQ #112](http://mywiki.wooledge.org/BashFAQ/112). — Charles Duffy, Nov 22 '17 at 14:47
BashFAQ #112 could use an update; it doesn't mention the `-v` operator. — chepner, Nov 22 '17 at 15:01

Charles Duffy · Accepted Answer · 2017-11-22T14:52:24.950

5

Yes, [[ ! -z "${EXAMPLE-}" ]] safely determines whether the variable named EXAMPLE has a non-empty value assigned, even with set -u active.

Personally, I would write [[ -n "${EXAMPLE-}" ]] or even [[ ${EXAMPLE-} ]] -- taking advantage of additional terseness made safe by [[ ]] and not trustworthy with [ ] -- but all these are correct.

edited Nov 22 '17 at 14:52

answered Nov 22 '17 at 14:48

Charles Duffy

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score 2 · Answer 2 · answered Nov 22 '17 at 15:02

2

You can use the -v operator to check if a name has been set to a value:

if [[ -v EXAMPLE ]]; then
    echo "Safe to expand: $EXAMPLE"
fi

answered Nov 22 '17 at 15:02

chepner

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Notice that this requires Bash 4.2 or newer. – Benjamin W. Nov 22 '17 at 15:10

How to test if variable is set and not empty, when using “set -o nounset”

2 Answers2