显示CardLayout卡以随机顺序?(Display cards of CardLayout in

2019-08-17 08:56发布

我想为我的CardLayout显示卡或屏幕随机顺序。 我需要如何做到这一点的指导。 什么是战略,我应该使用?

我试着用下面的代码,但它是在一个固定的顺序。 我希望能够选择我喜欢哪个顺序。

编辑!

对不起,按随机顺序我不是故意洗牌。 但是,它是很好的了解。 我希望该程序的用户可以输入一些。 根据输入的值,则显示一个特定的屏幕/卡。

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);
}
}

Answer 1:

这里是直接跳转到一个卡的简单方法。

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,这将需要一些改变代码的其余部分。 这里是一个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“哪里是你提示用户输入卡号代码的一部分吗?” 实际上,试图与沟通。 对于更好地帮助越早,后期的一个非常微妙的方式SSCCE 。



Answer 2:

把CartLayouts在一个列表,随机播放列表,添加到列表的顺序包含布局。



文章来源: Display cards of CardLayout in random order?