Question

以下代码无法正常运行，并始终从第一个echo打印elif。这是在一个bbb上运行。任何人都能找到错误我知道它的东西是愚蠢但我找不到它。

#!/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

Answer 1

替换：

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

使用：

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

讨论

此逻辑测试始终为true：

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

无论$2的价值如何，上述两项测试中的一项都是正确的。因为这两者是逻辑关联的，所以整个陈述都是正确的。

我怀疑你想要的是：

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

实施例

让我们对$2的三个值进行测试：

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

上面的代码产生：

on is OK
off is OK
other is bad

bash脚本代码运行不正常

1 个答案:

讨论

实施例