以随机顺序显示CardLayout的卡片吗? [英] Display cards of CardLayout in random order?

问题描述

我想在CardLayout中随机显示卡或屏幕.我需要有关如何完成此操作的指导.我应该使用什么策略?

I want to have a random order for displaying the cards or screens in my CardLayout. I need guidance on how to accomplish this. What is strategy I should use?

我尝试使用下面的代码，但它的顺序固定.我希望能够选择我喜欢的任何顺序.

I tried using the code below, but it is in a fixed order. I want to be able to choose whichever order I like.

编辑！

对不起，按照随机顺序，我不是说改组.但是，很高兴知道.我希望程序的用户能够输入一些输入.根据输入的值，显示特定的屏幕/卡.

Sorry, by random order I did not mean shuffling. But, it is good to know. I want the user of the program to be able to enter some input. Depending on the value of the input, a particular screen/card is displayed.

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

public class CardLayoutExample extends JFrame {

private int currentCard = 1;
private JPanel cardPanel;
private CardLayout cl;

public CardLayoutExample() {

    setTitle("Card Layout Example");
    setSize(300, 150);
    cardPanel = new JPanel();

    cl = new CardLayout();
    cardPanel.setLayout(cl);
    JPanel p1 = new JPanel();
    JPanel p2 = new JPanel();
    JPanel p3 = new JPanel();
    JPanel p4 = new JPanel();
    JLabel lab1 = new JLabel("Card1");
    JLabel lab2 = new JLabel("Card2");
    JLabel lab3 = new JLabel("Card3");
    JLabel lab4 = new JLabel("Card4");
    p1.add(lab1);
    p2.add(lab2);
    p3.add(lab3);
    p4.add(lab4);

    cardPanel.add(p1, "1");
    cardPanel.add(p2, "2");
    cardPanel.add(p3, "3");
    cardPanel.add(p4, "4");
    JPanel buttonPanel = new JPanel();
    JButton b1 = new JButton("Previous");
    JButton b2 = new JButton("Next");
    buttonPanel.add(b1);
    buttonPanel.add(b2);
    b1.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {
            if (currentCard > 1) {
                currentCard -= 1;
                cl.show(cardPanel, "" + (currentCard));
            }
        }
    });

    b2.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {
            if (currentCard < 4) {
                currentCard += 1;
                cl.show(cardPanel, "" + (currentCard));
            }
        }
    });
    getContentPane().add(cardPanel, BorderLayout.NORTH);
    getContentPane().add(buttonPanel, BorderLayout.SOUTH);
}

public static void main(String[] args) {
    CardLayoutExample cl = new CardLayoutExample();
    cl.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    cl.setVisible(true);
}
}

推荐答案

这是直接跳转到卡片的简单方法.

Here is a simple way to jump directly to a card.

final JButton jumpTo = new JButton("Jump To");
buttonPanel.add(jumpTo);
jumpTo.addActionListener( new ActionListener(){
    @Override
    public void actionPerformed(ActionEvent ae) {
        String[] names = {"1","2","3","4"};
        String s = (String)JOptionPane.showInputDialog(
            jumpTo,
            "Jump to card",
            "Navigate",
            JOptionPane.QUESTION_MESSAGE,
            null,
            names,
            names[0]);
        if (s!=null) {
            cl.show(cardPanel, s);
        }
    }
} );

显然，这将需要对其余代码进行一些更改.这是SSCCE.

Obviously this will require some changes to the rest of the code. Here is an SSCCE.

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

public class CardLayoutExample extends JFrame {

private int currentCard = 1;
private JPanel cardPanel;
private CardLayout cl;

public CardLayoutExample() {

    setTitle("Card Layout Example");
    setSize(300, 150);
    cardPanel = new JPanel();

    cl = new CardLayout();
    cardPanel.setLayout(cl);
    JPanel p1 = new JPanel();
    JPanel p2 = new JPanel();
    JPanel p3 = new JPanel();
    JPanel p4 = new JPanel();
    JLabel lab1 = new JLabel("Card1");
    JLabel lab2 = new JLabel("Card2");
    JLabel lab3 = new JLabel("Card3");
    JLabel lab4 = new JLabel("Card4");
    p1.add(lab1);
    p2.add(lab2);
    p3.add(lab3);
    p4.add(lab4);

    cardPanel.add(p1, "1");
    cardPanel.add(p2, "2");
    cardPanel.add(p3, "3");
    cardPanel.add(p4, "4");
    JPanel buttonPanel = new JPanel();
    JButton b1 = new JButton("Previous");
    JButton b2 = new JButton("Next");
    buttonPanel.add(b1);
    buttonPanel.add(b2);
    b1.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {
            if (currentCard > 1) {
                currentCard -= 1;
                cl.show(cardPanel, "" + (currentCard));
            }
        }
    });

    b2.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {
            if (currentCard < 4) {
                currentCard += 1;
                cl.show(cardPanel, "" + (currentCard));
            }
        }
    });

    final JButton jumpTo = new JButton("Jump To");
    buttonPanel.add(jumpTo);
    jumpTo.addActionListener( new ActionListener(){
        @Override
        public void actionPerformed(ActionEvent ae) {
            String[] names = {"1","2","3","4"};
            String s = (String)JOptionPane.showInputDialog(
                jumpTo,
                "Jump to card",
                "Navigate",
                JOptionPane.QUESTION_MESSAGE,
                null,
                names,
                names[0]);
            if (s!=null) {
                cl.show(cardPanel, s);
            }
        }
    } );

    getContentPane().add(cardPanel, BorderLayout.NORTH);
    getContentPane().add(buttonPanel, BorderLayout.SOUTH);
}

public static void main(String[] args) {
    CardLayoutExample cl = new CardLayoutExample();
    cl.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    cl.setVisible(true);
}
}

顺便说一句-我的评论您提示用户输入卡号的代码部分在哪里?" 实际上是尝试&进行沟通.为尽快获得更好的帮助，请发布 SSCCE .

BTW - my comment "Where is the part of the code where you prompt the user for a card number?" was actually a very subtle way to try & communicate.. For better help sooner, post an SSCCE.

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