bash 如果逻辑布尔字符串

bash if logical boolean string

提问by user30772

I am notlooking for a different way to accomplish the apparent intention. I'm looking to understandwhy this exactsyntax is not working.

我不是在寻找不同的方式来实现明显的意图。我想了解为什么这种确切的语法不起作用。

[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

Here's a reference that's identical to so many others i found. I understandwhat it's doing, it's using the non logical's for AND and OR when everything I've read said that it should be using logical bools.

这是一个与我发现的许多其他参考相同的参考。我明白它在做什么，当我读过的所有内容都说它应该使用逻辑布尔值时，它使用非逻辑的 AND 和 OR。

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

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

Ok so here it is, with Nahuel's suggestion behaving how I had originally expected it to:

好的，就是这样，Nahuel 的建议表现出我最初的预期：

[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

采纳答案by Nahuel Fouilleul

The problem is that : [ "$ans" != "" -o "$ans" != "y" ] is always true because of the or and the negation. $ans cannot be equal to "" and to "y".

问题在于： [ "$ans" != "" -o "$ans" != "y" ] 由于 or 和否定，总是为真。$ans 不能等于“”和“y”。

Try replace these lines

尝试替换这些行

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

by these

通过这些

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

or these

或者这些

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

The easier is to do is a case:

更容易做的是一个案例：

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

also breakmust be used only in a foror whileloop, or in a selectbut it is missing in your post .

也break必须只在 afor或while循环中使用，或者在 aselect但它在您的帖子中丢失。

回答by V H

also logically its a flawed way of doing things. firstly using case would be best in this scenario, secondly you are looking for == n then stating if it is blank or not equal to yes - so although no is caught out in first if statement in theory it would still meet second criteria

从逻辑上讲，它也是一种有缺陷的做事方式。首先在这种情况下使用 case 是最好的，其次你正在寻找 == n 然后说明它是空白还是不等于是 - 所以尽管理论上第一个 if 语句中没有发现它仍然符合第二个条件

surely the most logical way to ensure input is 100% would be

确保输入 100% 的最合乎逻辑的方法肯定是

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

        else

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

回答by Rastislav Hasicek

I don't really understand, why do you use -o in the elif. I would use "||" or "OR" operator. When you use two conditions in if, you should use double [[ and ]]. So if you use:

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

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

it works fine.

它工作正常。