Java 从for循环打印元素
我遇到了一个非常奇怪的问题,我有一个方法,询问游戏中的玩家数量,一旦我有了游戏中的玩家数量,我会依次询问他们的名字。一旦给了我玩家的名字,就会创建一个框架,要求他们指定他们想要选择的计数器。当他们点击紫色宝石(用于测试目的)时,它应该在控制台中打印出玩家的名字,但是for循环似乎不起作用。知道我该如何让循环正常工作吗Java 从for循环打印元素,java,swing,loops,for-loop,Java,Swing,Loops,For Loop,我遇到了一个非常奇怪的问题,我有一个方法,询问游戏中的玩家数量,一旦我有了游戏中的玩家数量,我会依次询问他们的名字。一旦给了我玩家的名字,就会创建一个框架,要求他们指定他们想要选择的计数器。当他们点击紫色宝石(用于测试目的)时,它应该在控制台中打印出玩家的名字,但是for循环似乎不起作用。知道我该如何让循环正常工作吗 public class setupPlayers extends JFrame implements ActionListener { int intOfPlayers,
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);
}
}
}
公共类setupPlayers扩展JFrame实现ActionListener{
int intOfPlayers,purpleClick=0,orangeClick=0,iceClick=0,greenClick=0;
ArrayList ArrayOffLayers=新的ArrayList();
JButton purpleGemBTN、greenGemBTN、iceCubeBTN、orangeGemBTN;
JFrame组织小组;
JPanel titleChoiceCounter,副选举人;
ImageIcon finalCounter;
专用静态对话框d;
公共设置播放器(){}
公共空间层(){
试一试{
String inputValue=JOptionPane.showInputDialog(“请输入玩家数量”);
intOfPlayers=Integer.parseInt(inputValue);
如果(输入FPLayers>4){
JOptionPane.showMessageDialog(null,“只有1-4个可以播放!”、“错误!”、JOptionPane.Error\u消息);
noOfPlayers();
intOfPlayers=0;
}
对于(int z=0;z
尝试为循环编写以下内容:
for (int z=0; z<arrayOfPlayers.size()-1;z=z){...
for(intz=0;z尝试编写以下for循环:
for (int z=0; z<arrayOfPlayers.size()-1;z=z){...
for(int z=0;z您没有迭代for循环z++
notz=z
:
for (int z=0; z<arrayOfPlayers.size()-1;z++){
String currentPlayer = arrayOfPlayers.get(z).playerNme;
System.out.println(currentPlayer);
for(int z=0;z您没有迭代for循环z++
notz=z
:
for (int z=0; z<arrayOfPlayers.size()-1;z++){
String currentPlayer = arrayOfPlayers.get(z).playerNme;
System.out.println(currentPlayer);
for(int z=0;z或更好)用于每个循环
for(Player p : arrayOfPlayers)
{
String name = p.playerNme;
System.out.println(name);
}
或者更好地使用for-each循环
for(Player p : arrayOfPlayers)
{
String name = p.playerNme;
System.out.println(name);
}
正如你的for(intz=0;z和for(intz=0;z一样,从你对前面答案的评论来看,我猜你的问题更多地与你如何递归地获得名字有关
下面是一对快速的方法,它们似乎可以在没有递归的情况下实现您想要的
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);
}
}
private ArrayList arrayOfPlayers=new ArrayList();
私有层;
公共空间层(){
while(true){
String inputValue=JOptionPane.showInputDialog(“请输入玩家数量”);
如果输入了(inputValue!=null){//Text,则不单击取消
试一试{
intOfPlayers=Integer.parseInt(inputValue);
如果(intOfPlayers>4 | | intOfPlayers<1){
JOptionPane.showMessageDialog(null,“只有1-4个可以播放!”、“错误!”、JOptionPane.Error\u消息);
}否则{
break;//不要再问数字了
}
}捕获(数字格式){
showMessageDialog(null,“请输入一个数字!”,“Error!”,JOptionPane.Error\u消息);
e、 printStackTrace();//永远不要忽略错误,即使错误很明显
}
}否则{
System.out.println(“退出数字播放器输入”);
System.exit(0);//取消了对话框,因此退出程序
}
}
对于(int z=0;z