针对-n选项测试中的bash脚本总是返回true [英] Testing against -n option in BASH scripts always returns true

问题描述

我写一个bash脚本，其中我想检查是否有提供了具体的参数。我注意到 [-n参数] 测试的一个奇怪的（至少对我来说）的行为。对于下面的脚本：

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

我得到的结果如下：

I am getting results as follows:

1. 不带参数：

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

与参数：

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

我已经发现了封闭 $ 1 在双引号给我的结果不如预期，但我仍然不知道的为什么当报价均为测试返回true不使用和脚本被称为不带参数吗似乎 $ 1 为空的话，那么 [-n $ 1] 应该返回假的，不应该吗？

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?

推荐答案

引用它。

if [ -n "$1" ]; then 

如果没有引号，如果 $ 1 为空，则执行 [-n] ，这是真的*如果 $ 1 不为空，那么它显然如此。

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

*如果你给 [一个参数（不包括] ），它始终是真实的。 （顺便说一句，这是一个陷阱，许多新的用户分为时候，他们期望 [0] 是假的）。在这种情况下，单一的字符串是 -n 。

* 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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