Question

你好，我有一个有粘滞便笺的程序。你点击它们来捡起它们然后再次点击将它们放在某个地方。我的问题是当两个或多个粘滞便笺在彼此之上时，两者都被拾起我只想要拿起最上面的一个。我怎么能解决这个问题，到目前为止是我的代码：

  public class PhoneMsg {

public int x, y, id, hour, minute;
public boolean drag = false;
public String name, lastname, msg, msg2, msg3;

public Rectangle rx = new Rectangle(x + 290, y, 20, 20);
public Rectangle rdrag = new Rectangle(x, y, 310, 20);

public boolean remove;

private Image img;

public PhoneMsg(int x, int y, String name, String lastname, int hour, int minute, int id) {
    this.x = x;
    this.y = y;
    this.name = name;
    this.lastname = lastname;
    this.hour = hour;
    this.minute = minute;
    this.id = id;

    rdrag = new Rectangle(x, y, 310, 20);
    rx = new Rectangle(x + 290, y, 20, 20);
    genMsg();
}

public void tick() {
    rx = new Rectangle(x + 290, y, 20, 20);
    rdrag = new Rectangle(x, y, 310, 20);

    if (rx.intersects(Comp.mx, Comp.my, 1, 1)) {
        if (Comp.ml) {
            remove = true;
            for (int i = 0; i < play.ph.pp.toArray().length; i++) {
                // play.ph.pp.get(i).canreadtxt = true;
            }

        }
    }

    // dragging
    if (drag) {
        x = Comp.mx - 140;
        y = Comp.my - 10;
    }

    if (msg == null) {
        genMsg();
    }
}

    public void render(Graphics g) {
    Graphics2D g2 = (Graphics2D) g;

    RenderingHints rh = new RenderingHints(RenderingHints.KEY_TEXT_ANTIALIASING, RenderingHints.VALUE_TEXT_ANTIALIAS_GASP);
    g2.setRenderingHints(rh);

    // display
    g.setColor(new Color(19, 165, 34));
    g.fillRect(x, y, 310, 150);
    g.setColor(new Color(27, 53, 16));
    g.drawRect(x, y, 310, 150);
    g.setColor(new Color(27, 53, 16));
    g.drawRect(x, y, 310, 20);

    // Exit part
    g.setColor(new Color(27, 53, 16));
    g.drawRect(rx.x, rx.y, rx.width, rx.height);

    g.setColor(Color.red);
    g.setFont(new Font("italic", Font.BOLD, 12));
    g.drawString("X", rx.x + 7, rx.y + 15);

    // name
    g.setColor(Color.black);
    g.setFont(new Font("italic", Font.BOLD, 12));
    g.drawString("" + name + " " + lastname + "'s Recent Messages", x + 5, y + 15);

    // details
    String msg11 = String.format("%02d:%02d", hour, minute);
    String msg21 = String.format("%02d:%02d", hour, minute);
    String msg31 = String.format("%02d:%02d", hour, minute);

    if(play.hud.wifi >= 1){
    g.setColor(Color.CYAN);
    g.setFont(new Font("italic", Font.BOLD, 12));
    g.drawString("  " + msg, x + 2, y + 38);

    g.setColor(Color.white);
    g.setFont(new Font("italic", Font.BOLD, 12));
    g.drawString("  " + msg2, x + 2, y + 58);

    g.setColor(Color.cyan);
    g.setFont(new Font("italic", Font.BOLD, 12));
    g.drawString("  " + msg3, x + 2, y + 78);
    }else if(play.hud.wifi <= 0){
        g.setColor(Color.red);
        g.setFont(new Font("italic", Font.PLAIN, 18));
        g.drawString("Lost Connection", x +90, y +85);

        g.setColor(Color.red);
        g.setFont(new Font("italic", Font.PLAIN, 18));
        g.drawString("_________________", x +70, y +88);

    }

}

} 这是鼠标监听器中允许你拿起便签的部分：

     // dragging msg's
        for (int i1 = 0; i1 < play.ph.pm.toArray().length; i1++) {
            if (play.ph.pm.get(i1).drag == false && play.holding == false) {
                if (play.ph.pm.get(i1).rdrag.contains(Comp.mx, Comp.my)) {
                    play.ph.pm.get(i1).drag = true;
                    play.holding = true;

                }
            } else {
                play.ph.pm.get(i1).drag = false;
                play.holding = false;

            }
        }

Answer 1

如果您退出for循环，您可以获取代码在重叠时获得的第一个便笺。

if (play.ph.pm.get(i1).rdrag.contains(Comp.mx, Comp.my)) {
    play.ph.pm.get(i1).drag = true;
    play.holding = true;
    break; // Found sticky note, exits loop
}

如果这与你的else语句冲突，只需创建一个标志。

 // dragging msg's
 boolean gotCard = false;
 for (int i1 = 0; i1 < play.ph.pm.toArray().length; i1++) {
     if (play.ph.pm.get(i1).drag == false && play.holding == false && gotNote == false) { // Added gotNote check
         if (play.ph.pm.get(i1).rdrag.contains(Comp.mx, Comp.my)) {
             play.ph.pm.get(i1).drag = true;
             play.holding = true;
             gotNote = true; // Set flag to true
         }
     } else {
         play.ph.pm.get(i1).drag = false;
         play.holding = false;
     }
 }

但是你应该有一个z-index实现，以便能够分辨出哪个便签在顶部，然后按照z-index对你的数组进行排序。

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