Code:
#!/bin/bash
EXISTS=""
echo ${#EXISTS} #print size of string
if [ -n $EXISTS ]; then
echo "It exists!"
else
echo "It is empty."
fi
if [ -z $EXISTS ]; then
echo "It is empty."
else
echo "It exists!"
fi
Output:
0
It exists!
It is empty.
From man bash:
-z string
True if the length of string is zero.
-n string
True if the length of string is non-zero.
Could someone explain this behavior of -n to me, and why it doesn't agree with -z? Thanks!
Quote the variable or better use [[...]] in BASH:
if [[ -n $EXISTS ]]; then echo "It exists!"; else echo "It is empty."; fi
It will print:
It is empty.
Similarly:
if [ -n "$EXISTS" ]; then echo "It exists!"; else echo "It is empty."; fi
It is empty.
$EXISTS is unquoted and the shell expands the variable in [ -n $EXISTS ] it becomes [ -n ] which is interpreted by [/test as [ -n -n ] (see the man page) whereas [[ doesn't have that empty variable problem and the quotes make it see [ -n '' ].
Commented
Jun 18, 2015 at 18:23
EXISTS=""
if [ -n $EXISTS ]; then
...
fi
The behavior of [ ... ] depends on the number of arguments before the closing ].
If there's exactly one argument, it tests whether that argument is non-empty. Since the string "-n" is non-empty, [ -n ] is true. -n is interpreted as an operator only if it's followed by another argument.
To fix this, quote the argument so it's interpreted as a single empty argument, rather than as no argument:
if [ -n "$EXISTS" ]; then
...
fi
[and always usetest(or even better, use[[ ... ]]when available and POSIX compatibility isn't a concern). You're less likely to be surprised by how the arguments totestare handled (because it actually looks like a command), and you don't need the ridiculous final]argument whose only purpose is to further the illusion that[is some sort of syntax.