Question

我不正在寻找一种不同的方式来实现明显的意图。我希望理解为什么完全语法不起作用。

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" == "n" ];then
>                 echo
>                 echo "bye"
>                 exit
>         elif [ "$ans" != "" -o "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)? **"Should have continued"**

Invalid entry...
Would you like the script to check the second box ([y]n)? **"Should have continued"**
y
Invalid entry...
Would you like the script to check the second box ([y]n)? **"Correct behavior"**
alskjfasldasdjf
Invalid entry...
Would you like the script to check the second box ([y]n)? **"Correct behavior"**
n

bye

这是一个与我发现的其他许多人相同的参考文献。我了解它正在做什么，当我读过的所有内容都说它应该使用逻辑bool时，它使用非逻辑的AND和OR。

http://www.groupsrv.com/linux/about140851.html

好的，就是这样，Nahuel的建议表明我最初的预期：

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
asdfad
Invalid entry...
Would you like the script to check the second box ([y]n)?

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
y
[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
n

logout

Answer 1

问题是：[“$ ans”！=“” - o“$ ans”！=“y”]因为或否定而总是正确的。 $ ans不能等于“”和“y”。

尝试替换这些行

if [ "$ans" == "n" ];then 
elif [ "$ans" != "" -o "$ans" != "y" ];then

通过这些

if [ "$ans" = "n" ];then 
elif [ "$ans" != "" -a "$ans" != "y" ];then

或这些

if [[ $ans == n ]];then 
elif [[ $ans != "" && $ans != y ]];then

更容易做的是一个案例：

case $ans in
  y) echo "yes"
  ;;
  n) echo "no"
  ;;
  *)
  ;;
 esac

break也必须仅在for或while循环中使用，或在select中使用，但您的帖子中缺少它。

Answer 2

我真的不明白，你为什么在elif中使用-o。我会用“||”或“或”运算符。在if中使用两个条件时，应使用double [[and]]。所以如果你使用：

    elif [[ "$ans" != "" ||  "$ans" != "y" ]];then

它工作正常。

Answer 3

从逻辑上讲，它也是一种有缺陷的做事方式。首先使用案例在这种情况下是最好的，其次你正在寻找== n然后说明它是否为空或不等于是 - 所以尽管在理论上第一个if语句中没有被捕获它仍然符合第二个标准

确保输入的最合理方式是100％

 if [ "$ans" == "n" ];then
                 echo
                 echo "bye"
                 exit
         elif [ "$ans" == "y" ];then
                 echo Yes
                        break;

        else

                 echo "Invalid entry... >$ans<"
         fi

bash如果是逻辑布尔字符串

3 个答案: