Question

我正在尝试在JScrollPane内实现对位置敏感的缩放。 JScrollPane包含一个具有自定义paint的组件，该组件将在其分配的任何空间内绘制 - 因此缩放就像使用MouseWheelListener一样简单，可根据需要调整内部组件的大小。 / p>

但我也希望放大（或退出）一个点，以便在生成的放大（或 - ）视图中将该点保持为中心（这就是我所说的'位置敏感'缩放），类似于缩放如何在谷歌地图中工作。我确信之前已经做了很多次 - 有没有人知道在Java Swing下做“权利”的方式？使用Graphic2D转换而不是使用JScrollPanes？

会更好吗？

示例代码如下：

package test;

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

public class FPanel extends javax.swing.JPanel {

private Dimension preferredSize = new Dimension(400, 400);    
private Rectangle2D[] rects = new Rectangle2D[50];

public static void main(String[] args) {        
    JFrame jf = new JFrame("test");
    jf.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    jf.setSize(400, 400);
    jf.add(new JScrollPane(new FPanel()));
    jf.setVisible(true);
}    

public FPanel() {
    // generate rectangles with pseudo-random coords
    for (int i=0; i<rects.length; i++) {
        rects[i] = new Rectangle2D.Double(
                Math.random()*.8, Math.random()*.8, 
                Math.random()*.2, Math.random()*.2);
    }
    // mouse listener to detect scrollwheel events
    addMouseWheelListener(new MouseWheelListener() {
        public void mouseWheelMoved(MouseWheelEvent e) {
            updatePreferredSize(e.getWheelRotation(), e.getPoint());
        }
    });
}

private void updatePreferredSize(int n, Point p) {
    double d = (double) n * 1.08;
    d = (n > 0) ? 1 / d : -d;
    int w = (int) (getWidth() * d);
    int h = (int) (getHeight() * d);
    preferredSize.setSize(w, h);
    getParent().doLayout();
    // Question: how do I keep 'p' centered in the resulting view?
}

public Dimension getPreferredSize() {
    return preferredSize;
}

private Rectangle2D r = new Rectangle2D.Float();
public void paint(Graphics g) {
    super.paint(g);
    g.setColor(Color.red);
    int w = getWidth();
    int h = getHeight();
    for (Rectangle2D rect : rects) {
        r.setRect(rect.getX() * w, rect.getY() * h, 
                rect.getWidth() * w, rect.getHeight() * h);
        ((Graphics2D)g).draw(r);
    }       
  }
}

Answer 1

经过测试，似乎有效......

private void updatePreferredSize(int n, Point p) {
    double d = (double) n * 1.08;
    d = (n > 0) ? 1 / d : -d;

    int w = (int) (getWidth() * d);
    int h = (int) (getHeight() * d);
    preferredSize.setSize(w, h);

    int offX = (int)(p.x * d) - p.x;
    int offY = (int)(p.y * d) - p.y;
    setLocation(getLocation().x-offX,getLocation().y-offY);

    getParent().doLayout();
}

<强>更新

以下是一个解释：点p是鼠标相对于FPanel的位置。由于您要缩放面板的大小，p的位置（相对于面板的大小）将按相同的因子缩放。通过从缩放位置减去当前位置，可以获得调整面板大小时“移位”的点数。然后，只需将滚动窗格中的面板位置向相反方向移动相同的量即可将p放回鼠标光标下。

Answer 2

这是@Kevin K解决方案的次要重构：

private void updatePreferredSize(int wheelRotation, Point stablePoint) {
    double scaleFactor = findScaleFactor(wheelRotation);
    scaleBy(scaleFactor);
    Point offset = findOffset(stablePoint, scaleFactor);
    offsetBy(offset);
    getParent().doLayout();
}

private double findScaleFactor(int wheelRotation) {
    double d = wheelRotation * 1.08;
    return (d > 0) ? 1 / d : -d;
}

private void scaleBy(double scaleFactor) {
    int w = (int) (getWidth() * scaleFactor);
    int h = (int) (getHeight() * scaleFactor);
    preferredSize.setSize(w, h);
}

private Point findOffset(Point stablePoint, double scaleFactor) {
    int x = (int) (stablePoint.x * scaleFactor) - stablePoint.x;
    int y = (int) (stablePoint.y * scaleFactor) - stablePoint.y;
    return new Point(x, y);
}

private void offsetBy(Point offset) {
    Point location = getLocation();
    setLocation(location.x - offset.x, location.y - offset.y);
}

Answer 3

您的MouseWheelListener还必须找到光标，将其移动到JScrollPane的中心，并调整要查看的内容的xmin / ymin和xmax / ymax。

Answer 4

我认为这样的smt应该有用......


private void updatePreferredSize(int n, Point p) {
    double d = (double) n * 1.08;
    d = (n > 0) ? 1 / d : -d;
    int w = (int) (getWidth() * d);
    int h = (int) (getHeight() * d);
    preferredSize.setSize(w, h);

    // Question: how do I keep 'p' centered in the resulting view?

    int parentWdt = this.getParent( ).getWidth( ) ;
    int parentHgt = this.getParent( ).getHeight( ) ;

    int newLeft = p.getLocation( ).x - ( p.x - ( parentWdt / 2 ) ) ;
    int newTop = p.getLocation( ).y - ( p.y - ( parentHgt / 2 ) ) ;
    this.setLocation( newLeft, newTop ) ;

    getParent().doLayout();
}

编辑：改变了一些事情。

如何在JScrollPane中实现位置敏感的缩放？

4 个答案: