Question

我目前正试图通过使用GUI制作彩票计划。我无法弄清楚为什么方法getTicketNumbers（）没有返回任何东西。它只是打印[]。

同样如此：

String output = "Name: " + name + "\nNumbers: " + ticketNumbers + "\n\n";

tickNumbers输出[]，但名称成功打印出名称。

我在构造函数中的Ticket类中添加了一个System.out.print，以确认正在成功传递ArrayList，它是：

public Ticket(ArrayList<Integer> ticketNumbers, String name) {
    this.ticketNumbers = ticketNumbers;
    this.name = name;
    System.out.print("Are the numbers being passed:" + ticketNumbers + "\n");
}

方法toString成功打印名称但是再次打印名为ticketNumbers的ArrayList。

private void enterLottoButtonActionPerformed(java.awt.event.ActionEvent evt) {                                                 
    if (nameInput.getText().equalsIgnoreCase("")) {
        JOptionPane.showMessageDialog(null, "Please Enter Your Name", "Error", JOptionPane.ERROR_MESSAGE);
    } else if (ticketNumbers.size() < 4) {
        JOptionPane.showMessageDialog(null, "Please Enter Four Numbers", "Error", JOptionPane.ERROR_MESSAGE);
    } else {
        ticketList.add(new Ticket(ticketNumbers, nameInput.getText()));

        JOptionPane.showMessageDialog(null, "You Successfully Entered! \n\nName: " + nameInput.getText() + "\nNumbers: " + ticketNumbers.toString());

        ticketNumbers.clear();
        numbersTextField.setText("");
        nameInput.setText("");     
        numberOfPeopleLabel.setText("           People Entered: " + ticketList.size());
    }

门票类：

import java.util.ArrayList;
import java.util.Collections;
import javax.swing.JOptionPane;

public class Ticket {
    private String name;
    private ArrayList<Integer> ticketNumbers = new ArrayList<>();

public Ticket(ArrayList<Integer> ticketNumbers, String name) {
    this.ticketNumbers = ticketNumbers;
    this.name = name;
    System.out.print("Are the numbers being passed:" + ticketNumbers + "\n");
}

public String getName() {
    return name;
}

public ArrayList<Integer> getTicketNumbers() {
    return ticketNumbers;
}

public ArrayList<Integer> getSortedTicketNumbers() {
    Collections.sort(ticketNumbers);
    return ticketNumbers;
}

@Override
public String toString() {
    String output = "Name: " + name + "\nNumbers: " + ticketNumbers + "\n\n";

    return output;
}
}

如果需要，请完整编写代码： http://pastebin.com/7i8VWQLk和http://pastebin.com/iRd49Nc7

Answer 1

看起来您正在复制对列表的引用，因此当您在一个地方清除它（或修改它）时，它会随处更改/清除。

尝试制作防御性副本。像：

ArrayList<Integer> tickets = new ArrayList<>(ticketNumbers)

Answer 2

ticketNumbers.clear()

此行导致问题。正如您所说，在构造Ticket对象期间，数据已在此行中正确传递：ticketList.add(new Ticket(ticketNumbers, nameInput.getText()));。但你事后清理了它们。您需要克隆ticketList。无论最佳实践如何，您都可以在Ticket课程内完成。

private List<Integer> clonedTicketNumbersList = new ArrayList<Integer>();

public Ticket(ArrayList<Integer> ticketNumbers, String name) {
    this.ticketNumbers = ticketNumbers;
    this.name = name;
    for(Integer ticketNum : ticketNumbers) {
      clonedTicketNumbersList.add(ticketNum );
    }
    System.out.print("Are the numbers being passed:" + ticketNumbers + "\n");
}

@Override
public String toString() {
    String output = "Name: " + name + "\nNumbers: " + clonedTicketNumbersList + "\n\n";

    return output;
}

JAVA - 发布打印ArrayList

2 个答案: