linux - 为什么 [ -n $var ] 在空字符串上为真？

Question

代码:

#!/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

输出:

0
It exists!
It is empty.

来自 man bash:

-z string

True if the length of string is zero.

-n 字符串

True if the length of string is non-zero.

有人可以向我解释 -n 的这种行为，以及为什么它与 -z 不一致吗？谢谢!

Answer 1

在 BASH 中引用变量或更好地使用 [[...]]:

if [[ -n $EXISTS ]]; then echo "It exists!"; else echo "It is empty."; fi

它将打印:

It is empty.

类似地:

if [ -n "$EXISTS" ]; then echo "It exists!"; else echo "It is empty."; fi
It is empty.