bash脚本代码运行不正常

以下代码工作不正常，总是从第一个

elif

打印

echo

。这是在bbb上运行的。有人能找到错误吗？我知道这很愚蠢，但我找不到

#!/bin/bash

#bashUSR.sh


path=/sys/class/leds/beaglebone:green:usr

echo "starting the LED bash script."

if [ "$#" == 0 ]; then
        echo "to few arguments. enter either 0 or 1 followed by on or off."
        exit 2

elif [ "$2" != "on" ] || [ "$2" != "off" ]; then
        echo "invalid second argument. please input either on or off."

elif [ "$1" == "0" ] && [ "$2" == "on" ]; then
        echo "turning on usr0 LED."
        export path0/brightness 1

elif [ "$1" == "0" ] && [ "$2" == "off" ]; then
        echo "turning off usr0 LED."
        export path0/brightness 0

elif [ "$1" == "1" ] && [ "$2" == "on" ]; then
        echo "turning on usr1 LED."
        export path1/brightness 1

elif [ "$1" == "1" ] && [ "$2" == "off" ]; then
        echo "turning off usr1 LED."
        export path1/brightness 0

else
        echo "invalid user number. please input a number between 0 and 1."
fi

替换：

elif [ "$2" != "on" ] || [ "$2" != "off" ]; then
        echo "invalid second argument. please input either on or off."

与：

讨论 此逻辑测试始终为真：

[ "$2" != "on" ] || [ "$2" != "off" ]

无论

$2

的值是多少，上述两个测试中的一个将为真。因为这两者都是用逻辑or连接的，所以整个语句都是正确的

我怀疑你想要的是：

[ "$2" != "on" ] && [ "$2" != "off" ]

例子 让我们测试一下

$2的三个值：

for x in on off other
do
    set -- 1 "$x"
    if [ "$2" != "on" ] && [ "$2" != "off" ]
    then
        echo "$2 is bad"
    else
        echo "$2 is OK"
    fi
done

上述代码生成：

on is OK
off is OK
other is bad

顺便说一句——当使用
[]
时，它最好使用的唯一字符串比较运算符是POSIX指定的，
=
（而不是
=
）；谢谢，我没看到
on is OK
off is OK
other is bad