Testing against -n option in BASH scripts always returns true

问题

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?

回答1:

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.

来源：https://stackoverflow.com/questions/20437067/testing-against-n-option-in-bash-scripts-always-returns-true