Benoît Xylo Benoît Xylo - 4 months ago 10
Bash Question

Testing against -n option in BASH scripts always returns true

I am writing a bash script, in which I am trying to check if there are particular parameters provided. I've noticed a strange (at least for me) behavior of

[ -n arg ]
test. For the following script:

#!/bin/bash

if [ -n $1 ]; then
echo "The 1st argument is of NON ZERO length"
fi

if [ -z $1 ]; then
echo "The 1st argument is of ZERO length"
fi


I am getting results as follows:


  1. with no parameters:

    xylodev@ubuntu:~$ ./my-bash-script.sh
    The 1st argument is of NON ZERO length
    The 1st argument is of ZERO length

  2. with parameters:

    xylodev@ubuntu:~$ ./my-bash-script.sh foobar
    The 1st argument is of NON ZERO length



I've already found out that enclosing
$1
in double quotes gives me the results as expected, but I still wonder why both tests return true when quotes are not used and the script is called with no parameters? It seems that
$1
is null then, so
[ -n $1 ]
should return false, shouldn't it?

Answer

Quote it.

if [ -n "$1" ]; then 

Without the quotes, if $1 is empty, you execute [ -n ], which is true*, and if $1 is not empty, then it's obviously true.

* If you give [ a single argument (excluding ]), it is always true. (Incidentally, this is a pitfall that many new users fall into when they expect [ 0 ] to be false). In this case, the single string is -n.

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