JOptionPane 错误疑难解答

Question

我有一个问题，我有一个程序想测试用户记住随机颜色列表的能力。根据用户输入的正确或错误，它会询问下一种颜色。

所以我得到了所有的工作，直到用户输入第一个颜色。在用户输入第一种颜色之前。程序已经假设用户输入错误，即使它没有要求任何输入。

从以前的知识中我知道我可以刷新缓冲区，你能用 JOptionPane 做到这一点吗？

或者这是我没有看到的另一个问题？

import java.util.*;
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import javax.swing.JOptionPane;

public class Testing
{
   //Intialization of the whole program for everything to pass information
   public JFrame main;
   public JLabel lbInsturctions;
   public JLabel welcomeMessage;
   public JLabel homeScreen;
   public JLabel homeScreenColors;
   public JTextField txtInput;
   public int num = 1;
   public String colorList = "";
   public String[] color = {"red","green","blue","brown","yellow", "gold", "orange", "silver"};
   public String[] solution = new String[5];

   //sets up the window and random colors
   public Testing ()
   {
      Random r = new Random();

      for (int i = 0; i<solution.length; i++)
      {
         solution[i] = color[r.nextInt(7)];
         colorList = Arrays.toString(solution);
      }

      JOptionPane.showMessageDialog(null, "Lets test your memory. Memorize these colors: " + colorList);

      main = new JFrame ();
      main.setSize (500,300);
      main.setTitle ("Ultimate Colors");
      main.setDefaultCloseOperation(main.EXIT_ON_CLOSE);
      main.setLayout(new FlowLayout());

      intializeGame();
      main.setVisible(true); 
   }


   public void intializeGame () 
   {
      //All Intiazations
      lbInsturctions = new JLabel();
      homeScreen = new JLabel();
      txtInput= new JTextField(null, 15);  


       //Need to delete or make new window if user pushes ok then  
       lbInsturctions.setText("Enter color number " + num + ":");  
       main.add(lbInsturctions); 
       main.add(txtInput);

       txtInput.addActionListener(new colorTester());       
    }

    public class colorTester implements ActionListener
    {

      public void actionPerformed (ActionEvent e)
      {  


         //Need to delete or make new window if user pushes ok then  
         lbInsturctions.setText("Enter color number " + num + ":"); 

         //grabs the users input to see if it is corect
         String guess= "";
         guess = txtInput.getText();

         System.out.println(guess);

         //Checks to see if the users input is the same as the initalizaton
         if (color[num+1].equalsIgnoreCase(guess) || num > 6)
         {
            System.out.println("You got it!");  
            ++num;

            lbInsturctions.setText("Enter color number " + num + ":");
            txtInput.setText("");
         }

         //if the User input is wrong
         else
         {
            System.out.println("It's a good thing your not graded!");
            txtInput.setVisible(false);
            lbInsturctions.setText("It's a good thing this is not graded!");
          }

          if (num == 5)
          {
            lbInsturctions.setText("You memory is perfect good job!");
            txtInput.setVisible(false);
          }

        }




   }


}//end of program

Answer 1

这与刷新缓冲区无关。

您在此处获得用户输入：guess = txtInput.getText();，它位于 intializeGame 方法中。这意味着您在创建时从 txtInput JTextField 获取文本，在用户有机会在字段中输入任何内容之前。我认为您习惯于编写线性控制台程序，您可以立即获得用户的输入，但这不是事件驱动的 GUI 的工作方式。相反，您必须获取用户的输入并对其做出反应on event，这里可能是某个按钮的 ActionListener。也许您的代码需要一个“提交”JButton 或类似的东西，并在其 ActionListener 中，从 JTextField 中提取输入并对其做出响应。这样做，您的代码就有更好的工作机会。

其他问题：

• 您似乎从未将 txtInput JTextField 添加到 GUI 中。
• homeScreen JLabel 也是如此

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编辑问题底部发布的新代码也有同样的问题。