bash脚本中的参数数量

I want to test the number of arguments passed to a Linux shell script. If the number of arguments is not 2 or 4, it should print something. Unfortunately it does not work. Can anyone explain what I am doing wrong?

我想测试传递给Linux shell脚本的参数数量。如果参数的数量不是2或4，它应该打印一些东西。不幸的是，它不起作用。有人能解释一下我做错了什么吗?

#!/bin/bash
if [[ $# -ne 2 ]] || [[ $# -ne 4 ]];
then
    echo "here";
fi

2 个解决方案

#1

You should replace logical OR by logical AND, so :

你应该用逻辑或逻辑来代替逻辑，所以:

#!/bin/bash

if [[ $# -ne 2 && $# -ne 4 ]]; then
   echo "here"
fi

In arithmetic form:

在算术形式:

#!/bin/bash

if (($# != 2 && $# != 4)); then
   echo "here"
fi

As you can see, no need to use 2 [[ ]]

如您所见，不需要使用2 []

#2

Logic.

逻辑。

if [[ $# -ne 2 ]] && [[ $# -ne 4 ]]; then
  echo "here"
fi

#1