來自for循環的打印元素

Question

我有一個很奇怪的問題，我有一個方法noOfPlayers，詢問遊戲中的玩家數量，一旦我有遊戲中的玩家數量，我要求他們的每個名字反過來。一旦我得到了玩家的名字，就會創建一個框架，要求他們指定他們想要選擇的計數器。當他們點擊紫色寶石（出於測試目的）時，它應該在控制檯中打印出玩家的名字，但for循環似乎不起作用。任何想法如何讓環路正常工作？

public class setupPlayers extends JFrame implements ActionListener { 
    int intOfPlayers, purpleClick = 0, orangeClick = 0, iceClick = 0, greenClick = 0; 
    ArrayList<Player> arrayOfPlayers = new ArrayList<Player>(); 
    JButton purpleGemBTN, greenGemBTN, iceCubeBTN, orangeGemBTN; 
    JFrame organisationPanel; 
    JPanel titleChoiceCounter, counterSelector; 
    ImageIcon finalCounter; 
    private static Dialog d; 

    public setupPlayers() {} 

    public void noOfPlayers() { 
     try { 
      String inputValue = JOptionPane.showInputDialog("Please input the number of players"); 
      intOfPlayers = Integer.parseInt(inputValue); 
      if (intOfPlayers > 4) { 
       JOptionPane.showMessageDialog(null, "Only 1-4 can play!", "Error!", JOptionPane.ERROR_MESSAGE); 
       noOfPlayers(); 
       intOfPlayers = 0; 
      } 

      for (int z = 0; z < intOfPlayers; z++) { 
       String playerName = JOptionPane.showInputDialog("Player " + (z + 1) + " please input your name"); 
       chooseCounter(); 
       arrayOfPlayers.add(new Player(playerName, (z + 1), null, 0)); 
      } 
     } catch (NumberFormatException e) { 
      JOptionPane.showMessageDialog(null, "You did not enter the number of players, please enter the number of players", "Error!", JOptionPane.ERROR_MESSAGE); 
      noOfPlayers(); 
     } 
    } 

    public void chooseCounter() { 
     Frame window = new Frame(); 

     ImageIcon purpleGemImg = new ImageIcon("C:\\Users\\Anonymous\\Documents\\pink.png"); 
     ImageIcon greenGemImg = new ImageIcon("C:\\Users\\Anonymous\\Documents\\yellow.png"); 
     ImageIcon orangeGemImg = new ImageIcon("C:\\Users\\Anonymous\\Documents\\brown.png"); 
     ImageIcon iceCubeImg = new ImageIcon("C:\\Users\\Anonymous\\Documents\\white.png"); 

     d = new Dialog(window, "Please select your counter", true); 
     d.setLayout(new GridLayout(2, 2)); 
     d.setLocation(400, 300); 
     d.setSize(500, 500); 

     purpleGemBTN = new JButton("purple", purpleGemImg); 
     greenGemBTN = new JButton(greenGemImg); 
     orangeGemBTN = new JButton(orangeGemImg); 
     iceCubeBTN = new JButton(iceCubeImg); 

     purpleGemBTN.addActionListener(this); 
     greenGemBTN.addActionListener(this); 
     iceCubeBTN.addActionListener(this); 
     orangeGemBTN.addActionListener(this); 

     d.add(purpleGemBTN); 
     d.add(greenGemBTN); 
     d.add(orangeGemBTN); 
     d.add(iceCubeBTN); 

     d.setVisible(true); 
    } 

    public static void main(String[] args) { 
     setupPlayers spObj = new setupPlayers(); 
    } 

    public void actionPerformed(ActionEvent e) { 
     JButton pressed = new JButton(); 
     pressed = (JButton) e.getSource(); 
     if (pressed.getText().equals("purple")) { 
      for (int z = 0; z < arrayOfPlayers.size() - 1; z = z) { 
       String currentPlayer = arrayOfPlayers.get(z).playerNme; 
       System.out.println(currentPlayer); 
      } 
      d.setVisible(false); 
     } 
    } 
} 

Answer 1

所以，從以前的回答您的意見，我猜你的問題更關係到你是如何獲得遞歸的名字。

這是一對快速的方法，似乎可以實現你想要的而無需遞歸。

private ArrayList<Player> arrayOfPlayers = new ArrayList<>(); 
private int intOfPlayers; 

public void noOfPlayers() { 

    while (true) { 
     String inputValue = JOptionPane.showInputDialog("Please input the number of players"); 

     if (inputValue != null) { // Text was entered, cancel not clicked 
      try { 
       intOfPlayers = Integer.parseInt(inputValue); 
       if (intOfPlayers > 4 || intOfPlayers < 1) { 
        JOptionPane.showMessageDialog(null, "Only 1-4 can play!", "Error!", JOptionPane.ERROR_MESSAGE); 
       } else { 
        break; // stop asking for numbers 
       } 
      } catch (NumberFormatException e) { 
       JOptionPane.showMessageDialog(null, "Please enter a number!", "Error!", JOptionPane.ERROR_MESSAGE); 
       e.printStackTrace(); // never ignore errors, even if obvious 
      } 
     } else { 
      System.out.println("Quitting from number players input"); 
      System.exit(0); // Canceled the dialog, so quit the program 
     } 

    } 

    for (int z = 0; z < intOfPlayers; z++) { 
     String playerName = JOptionPane.showInputDialog("Player " + (z + 1) + " please input your name"); 
     if (playerName != null) { 
      // chooseCounter(playerName); 
      arrayOfPlayers.add(new Player(playerName, (z + 1), null, 0)); 
     } else { 
      System.out.println("Quitting from player " + (z + 1) + " name input"); 
      System.exit(0); // Canceled the dialog, so quit the program 
     } 
    } 

    printPlayerNames(); 
} 

private void printPlayerNames() { 
    for (int z = 0; z < arrayOfPlayers.size(); z++) { 
     String currentPlayer = arrayOfPlayers.get(z).playerNme; 
     System.out.println(currentPlayer); 
    } 
} 

Answer 2

嘗試寫這樣的循環：

for (int z=0; z<arrayOfPlayers.size()-1;z=z){... 

像這樣：

for (int z=0; z<arrayOfPlayers.size();z++){... 

增加z++和arrayOfPlayers.size()-1刪除-1因爲0指數z開始：

Answer 3

你不通過for循環迭代z++不z=z：

for (int z=0; z<arrayOfPlayers.size()-1;z++){ 
    String currentPlayer = arrayOfPlayers.get(z).playerNme; 
    System.out.println(currentPlayer); 

Answer 4

或對每個循環

for(Player p : arrayOfPlayers) 
    { 
     String name = p.playerNme; 
     System.out.println(name); 
    } 

Answer 5

更好，因爲你有for (int z=0; z<arrayOfPlayers.size()-1;z=z)，你會無限循環，因爲z將永遠等於0。

嘗試將z=z更改爲z++，以使迭代結束一次z = arrayOfPlayers.size()-1。

你的代碼將是：

for (int z=0; z<arrayOfPlayers.size();z++){ 
    String currentPlayer = arrayOfPlayers.get(z).playerNme; 
    System.out.println(currentPlayer); 
}