否定Bash中的多个条件 [英] Negating multiple conditions in Bash

问题描述

我意识到这是一个简单的问题，但是由于bash中严格的语法要求，我很难找到答案.我有以下脚本:

I realize this is a simple question, but I am finding a hard time getting an answer due to the finnicky syntax requirements in bash. I have the following script:

if ! [ [ -z "$1" ] || [ -z "$2" ] ]; then
  echo "both arguments are set!"
fi

运行时不带参数的输出如下:

I get the following output when I run it without arguments:

./test: line 3: [: -z: binary operator expected 
both arguments are set!

我不希望有任何输出-均未设置任何参数.我在做什么错了?

I am not expecting any output - neither argument is set. What am I doing wrong?

推荐答案

内置于[的test shell支持参数-a和-o，它们分别是逻辑AND和OR.

The test shell builtin [ supports the arguments -a and -o which are logical AND and OR respectively.

#!/bin/sh 

if [ -n "$1" -a -n "$2" ]
then echo "both arguments are set!"
fi

在这里，我使用-n来检查字符串的长度是否为非零而不是-z，这表明字符串 的长度为零，因此必须用!取反.

Here I use -n to check that the strings are non-zero length instead of -z which shows that they are zero length and therefore had to be negated with a !.

"-n string如果字符串的长度非零，则为true."

" -n string True if the length of string is nonzero."

-z字符串，如果字符串的长度为零，则为true."

"-z string True if the length of string is zero."

如果使用的是bash，它还支持更强大的测试类型[[]]，可以使用布尔运算符||和&&:

If you are using bash it also supports a more powerful type of test [[]] which can use the boolean operators || and &&:

#!/bin/bash 

if [[ -n "$1" && -n "$2" ]]
then echo "both arguments are set!"
fi

在评论中，解决了这些示例均未显示如何在shell中取反多个测试的问题，下面是一个示例:

In comments it was addressed that none of these samples show how to negate multiple tests in shell, here is an example which does:

#!/bin/sh 

if [ ! -z "$1" -a ! -z "$2" ]
then echo "both arguments are set!"
fi

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