rroReturn类型方法丢失，YES_NO_OPTION无法解决一些错误

Question

我做了一个程序来识别正方形，平方根，数字立方体和字符串的反转我有一些错误我没有一个想法来解决这个问题。

这里是我的下面

import javax.swing.JOptionPane; 
    import java.util.Scanner; 

    public class FLABWORK3_ABUEL 
    { 
     static Scanner Scan new = Scanner (System.in); 
     public static void main(String[] args) 
     { 
     String choice; 
     String num1; 
     String string; 
     String Inverse; 

     int choicee, num2, response, length; 
     double squareroot; 



     do { 
      choice = JOptionPane.showInputDialog("Main Menu" + 
     "\n 1. Square of a number" + 
     "\n 2. Square root a number" + 
     "\n 3. Cube of a number" + 
     "\n 4. Length of number" + 
     "\n 5. Inverse of a string"); 

      choicee = Integer.parseInt(choice); 

      while (choicee > 5) 
      { 
       choice = JOptionPane.showInputDialog("Enter only 1-5!"); 
      } 

      switch (choicee) 
      { 
      case 1: 
      num1 = JOptionPane.showInputDialog("Enter an number."); 
      num2 = Integer.parseInt(num1); 
      num2 = num2*num2 ; 

      JOptionPane.showMessageDialog(null, "The square of the number: " + num2); 
      break; 

      case 2: 
       num1 = JOptionPane.showInputDialog("Enter a number."); 
       squareroot = Integer.parseInt(num1); 
      squareroot = Math.sqrt(squareroot); 

      JOptionPane.showMessageDialog(null, "Square root is: " + squareroot); 
      break; 

      case 3: 
       num1 = JOptionPane.showInputDialog("Enter a number."); 
       num2 = Integer.parseInt(num1); 
       num2 = num2*(num2*num2); 

       JOptionPane.showMessageDialog(null, "The cube is: " + num2); 
       break; 

      case 4: 
       string = JOptionPane.showInputDialog("Enter a sentence or a word."); 
       length = string.length(); 
       JOptionPane.showMessageDialog(null, "The length : " + "\n" + length + "\n\n" + 
       "is:" + string); 
       break; 

      case 5: 
       string = JOptionPane.showInputDialog("Enter a word."); 
       length = string.length(); 
       for (int i = length - 1; i >= 0; i--) 
        Inverse = Inverse + string.charAt(i); 

       JOptionPane.showInputDialog(null, "Would you like to ry again?" 
         JOptionPane.YES_NO_OPTION, 
         JOptionPane.Question_Message, null, options, options [0]); 

      } 
     } 
     while (response == JOptionPane.YES_OPTION); 
    } 
} 

代码是更好地做切换的情况下，或者如果else语句？ 错误是返回类型方法是缺少静态扫描仪和YES_NO_OPTION不能解决

Answer 1

这可能是你错过了一个逗号后，你的“你想再次吗？”字符串在下面：

JOptionPane.showInputDialog(null, "Would you like to ry again?" 
       JOptionPane.YES_NO_OPTION, 
       JOptionPane.Question_Message, null, options, options [0]); 

它应该是：

JOptionPane.showInputDialog(null, "Would you like to ry again?", 
       JOptionPane.YES_NO_OPTION, 
       JOptionPane.Question_Message, null, options, options [0]); 

而且你定义扫描仪如下：

static Scanner Scan new = Scanner (System.in); 

应

static Scanner Scan = new Scanner (System.in); 

另一件事是我没有看到你定义的选项anywh ERE。所以，你可能需要声明和初始化它使用在以下行之前：

JOptionPane 
        .showInputDialog(null, "Would you like to ry again?", 
          JOptionPane.YES_NO_OPTION, 
          JOptionPane.QUESTION_MESSAGE, null, options, 
          options[0]); 

注：Java是大小写敏感的，所以QUESTION_MESSAGE不一样QUESTION_MESSAGE。