参见英文答案 >
Simple logical operators in Bash5个
我不确定如何在shell中进行多次测试.我在编写这个脚本时遇到了麻烦：
我不确定如何在shell中进行多次测试.我在编写这个脚本时遇到了麻烦：
echo "You have provided the following arguments $arg1 $arg2 $arg3"
if [ "$arg1" = "$arg2" && "$arg1" != "$arg3" ]
then
echo "Two of the provided args are equal."
exit 3
elif [ $arg1 = $arg2 && $arg1 = $arg3 ]
then
echo "All of the specified args are equal"
exit 0
else
echo "All of the specified args are different"
exit 4
fi
问题是我每次都会收到此错误：
./compare.sh: [: missing `]’ command not found
sh正在解释&&作为shell运算符.将它更改为-a,这是[的连接操作符：
原文链接：https://www.f2er.com/bash/387155.html[ "$arg1" = "$arg2" -a "$arg1" != "$arg3" ]
此外,你应该总是引用变量,因为[当你放弃参数时会感到困惑.
