嵌套的 if 语句在 Java 中无法正常工作。

huangapple
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英文:

Stacked if statements not working in java

问题

在以下的代码中，当我执行它时，只有第一个 if 块会起作用（当我按下其他选项的按钮时，没有弹出窗口）。我做错了什么？这是一个用于创建一个可以在二进制、十进制和十六进制之间转换数字的应用程序的程序。

  1. public class Menu implements ActionListener {
  2.     public static void main(String[] args) {
  3.         new Menu().createMenu();
  4.     }
  6.     JFrame frame = new JFrame();
  7.     JPanel panel = new JPanel();
  8.     JButton binToHex = new JButton("Bin to Hex");
  9.     JButton binToDec = new JButton("Bin to Dec");
  10.     JButton hexToBin = new JButton("Hex to Bin");
  11.     JButton hexToDec = new JButton("Hex to Dec");
  12.     JButton decToHex = new JButton("Dec to Hex");
  13.     JButton decToBin = new JButton("Dec to Bin");
  15.     private void createMenu() {
  16.         frame.setVisible(true);
  17.         frame.add(panel);
  18.         panel.add(binToHex);
  19.         panel.add(binToDec);
  20.         panel.add(hexToBin);
  21.         panel.add(hexToDec);
  22.         panel.add(decToHex);
  23.         panel.add(decToBin);
  24.         frame.pack();
  25.         binToHex.addActionListener(this);
  26.     }
  28.     public void actionPerformed(ActionEvent e) {
  29.         String input = new String();
  30.         if (e.getSource() == binToHex) {
  31.             input = JOptionPane.showInputDialog(null, "Enter a binary number");
  32.             Binary.convertToHexadecimal();
  33.         } 
  34.         
  35.         if (e.getSource() == binToDec) {
  36.             input = JOptionPane.showInputDialog(null, "Enter a binary number");
  37.             Binary.convertToDecimal();
  38.         } 
  39.         
  40.         if (e.getSource() == hexToBin) {
  41.             input = JOptionPane.showInputDialog(null, "Enter a hexadecimal number");
  42.             Hexadecimal.convertToBinary();
  43.         }
  44.         
  45.         if (e.getSource() == hexToDec) {
  46.             input = JOptionPane.showInputDialog(null, "Enter a hexadecimal number");
  47.             Hexadecimal.convertToDecimal();
  48.         }
  49.         
  50.         if (e.getSource() == decToHex) {
  51.             input = JOptionPane.showInputDialog(null, "Enter a decimal number");
  52.             Decimal.convertToHexadecimal(input);
  53.         }
  54.         
  55.         if (e.getSource() == decToBin) {
  56.             input = JOptionPane.showInputDialog(null, "Enter a decimal number");
  57.             Decimal.convertToBinary(input);
  58.         }
  59.         
  60.     }
  61. }

如果你遇到了只有第一个 if 块起作用的问题，这可能是因为其他选项的按钮没有绑定正确的动作监听器，或者在按钮被按下时没有执行相应的转换操作。请确保每个按钮都正确地绑定了 ActionListener，并在相应的 if 块中执行相应的转换操作。

英文:

In the following code, when I execute it, only the first if block will work (when I press the button for the other options, no pop-ups happen). What am I doing wrong? This is a program designed to make an app that can convert numbers between binary, decimal, and hexadecimal.

  1. public class Menu implements ActionListener {
  2. public static void main(String[] args) {
  3. new Menu().createMenu();
  4. }
  5. JFrame frame = new JFrame();
  6. JPanel panel = new JPanel();
  7. JButton binToHex = new JButton("Bin to Hex");
  8. JButton binToDec = new JButton("Bin to Dec");
  9. JButton hexToBin = new JButton("Hex to Bin");
  10. JButton hexToDec = new JButton("Hex to Dec");
  11. JButton decToHex = new JButton("Dec to Hex");
  12. JButton decToBin = new JButton("Dec to Bin");
  13. private void createMenu() {
  14. frame.setVisible(true);
  15. frame.add(panel);
  16. panel.add(binToHex);
  17. panel.add(binToDec);
  18. panel.add(hexToBin);
  19. panel.add(hexToDec);
  20. panel.add(decToHex);
  21. panel.add(decToBin);
  22. frame.pack();
  23. binToHex.addActionListener(this);
  24. }
  25. public void actionPerformed(ActionEvent e) {
  26. String input = new String();
  27. if (e.getSource() == binToHex) {
  28. input = JOptionPane.showInputDialog(null, "Enter a binary number");
  29. Binary.convertToHexadecimal();
  30. } 
  31. if (e.getSource() == binToDec) {
  32. input = JOptionPane.showInputDialog(null, "Enter a binary number");
  33. Binary.convertToDecimal();
  34. } 
  35. if (e.getSource() == hexToBin) {
  36. input = JOptionPane.showInputDialog(null, "Enter a hexadecimal number");
  37. Hexadecimal.convertToBinary();
  38. }
  39. if (e.getSource() == hexToDec) {
  40. input = JOptionPane.showInputDialog(null, "Enter a hexadecimal number");
  41. Hexadecimal.convertToDecimal();
  42. }
  43. if (e.getSource() == decToHex) {
  44. input = JOptionPane.showInputDialog(null, "Enter a decimal number");
  45. Decimal.convertToHexadecimal(input);
  46. }
  47. if (e.getSource() == decToBin) {
  48. input = JOptionPane.showInputDialog(null, "Enter a decimal number");
  49. Decimal.convertToBinary(input);
  50. }
  51. }

}

答案1

得分: 3

你忘记在所有的 JButton 实例上调用 addActionListener 方法：

  1. binToHex.addActionListener(this); // 你已经有这个
  2. binToDec.addActionListener(this);
  3. hexToBin.addActionListener(this);
  4. hexToDec.addActionListener(this);
  5. decToHex.addActionListener(this);
  6. decToBin.addActionListener(this);
英文:

You forgot to call addActionListener on all your JButton instances:

  1. binToHex.addActionListener(this); // you have this one
  2. binToDec.addActionListener(this);
  3. hexToBin.addActionListener(this);
  4. hexToDec.addActionListener(this);
  5. decToHex.addActionListener(this);
  6. decToBin.addActionListener(this);

huangapple
  • 本文由 发表于 2020年10月6日 06:45:53
  • 转载请务必保留本文链接：https://go.coder-hub.com/64217162.html
  • if-statement
  • java
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