java - 制作菜单时在Java中创建while循环

Question

这个问题已经有答案了:

How do I compare strings in Java? (23 个回答)

已关闭 6 年前。

我一直在尝试使用 Java 中的“do”功能创建一个菜单。我首先为用户提供一系列选项，然后请求用户通过键盘输入他们的选项，然后创建字符串变量“选择”。但是，当我在循环表达式中引用“选择”变量时，会出现错误指示“选择无法解析为变量”。为什么会这样？我该如何纠正这个问题？

请引用下面的代码。

do{
       //output presenting the user with options
        System.out.print("A simple Menu\n");
        System.out.print("Y - Yolk\n");
        System.out.print("N - Naan\n");
        System.out.print("X - Exit\n\n");

        //request user input using the import.util.*; feature
        System.out.print("Enter your selection: ");
        String selection=keyboard.nextLine();
        // So here the 'selection' string is made

       }
       //Now here, whatever the user has selected as an option will be 
       // passed through and more options will appear.
       while(selection!="X");

       //however I get prompted with the red squiggly line underneath the 
       //'selection' variable that seems to not recognize 'selection' as a string

Answer 1

当循环尝试再次检查它时，您的selection变量超出了范围(或者您可以认为它已经被破坏)。所以在循环之外定义它。

String selection ="";
        do{
           //output presenting the user with options
            System.out.print("A simple Menu\n");
            System.out.print("Y - Yolk\n");
            System.out.print("N - Naan\n");
            System.out.print("X - Exit\n\n");

            //request user input using the import.util.*; feature
            System.out.print("Enter your selection: ");
            selection=keyboard.nextLine();
            // So here the 'selection' string is made

           }
           //Now here, whatever the user has selected as an option will be 
           // passed through and more options will appear.
           while(!selection.equals("X"));

编辑

我完全同意@Tim Biegeleisen 的观点。但他的错误消息不是关于字符串与逻辑运算符的比较。不管怎样，我编辑了我的答案。非常感谢您指出这一点。