【发布时间】:2014-11-12 23:42:50
【问题描述】:
我正在尝试绘制菱形(用于合成 UML 关系)。现在,我正在制作这样的三角形:
但我想做这样的事情:
我用这段代码做三角形:
private void drawArrowHead(Graphics2D g2, Point tip, Point tail,
Color color) {
g2.setPaint(color);
double dy = tip.y - tail.y;
double dx = tip.x - tail.x;
double theta = Math.atan2(dy, dx);
double x, y, rho = theta + phi;
Point p1 = new Point();
Point p2 = new Point();
p1.setLocation(tip.x - barb * Math.cos(rho), tip.y - barb * Math.sin(rho));
rho = theta - phi;
p2.setLocation(tip.x - barb * Math.cos(rho), tip.y - barb * Math.sin(rho));
int[] xPoints = new int[5];
int[] yPoints = new int[5];
xPoints[0] = tip.x;
xPoints[1] = p1.x;
xPoints[2] = p2.x;
yPoints[0] = tip.y;
yPoints[1] = p1.y;
yPoints[2] = p2.y;
g2.setPaint(Color.BLACK);
Shape shape = new Polygon(xPoints, yPoints, 3);
g2.fill(shape);
//tip.x - barb * Math.cos(rho);
//y = tip.y - barb * Math.sin(rho);
}
有人知道如何用它制作钻石吗?谢谢:)
import java.awt.BasicStroke;
import java.awt.Color;
import java.awt.Cursor;
import java.awt.EventQueue;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Panel;
import java.awt.Point;
import java.awt.Polygon;
import java.awt.RenderingHints;
import java.awt.Shape;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.MouseAdapter;
import java.awt.event.MouseEvent;
import java.awt.geom.Line2D;
import java.awt.geom.Point2D;
import java.awt.geom.Rectangle2D;
import java.util.ArrayList;
import java.util.List;
import javax.swing.JFrame;
import javax.swing.JMenuBar;
import javax.swing.JMenuItem;
import javax.swing.JPanel;
import javax.swing.UIManager;
import javax.swing.UnsupportedLookAndFeelException;
public class DrawArrows {
public static void main(String[] args) {
new DrawArrows();
}
public DrawArrows() {
EventQueue.invokeLater(new Runnable() {
@Override
public void run() {
try {
UIManager.setLookAndFeel(UIManager
.getSystemLookAndFeelClassName());
} catch (ClassNotFoundException | InstantiationException
| IllegalAccessException
| UnsupportedLookAndFeelException ex) {
ex.printStackTrace();
}
JFrame frame = new UMLWindow();
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setBounds(30, 30, 1000, 700);
frame.getContentPane().setBackground(Color.white);
frame.setVisible(true);
frame.setLocationRelativeTo(null);
}
});
}
public static class UMLWindow extends JFrame {
Shapes shapeList = new Shapes();
Panel panel;
private static final long serialVersionUID = 1L;
public UMLWindow() {
addMenus();
panel = new Panel();
}
public void addMenus() {
getContentPane().add(shapeList);
setSize(300, 200);
setLocationRelativeTo(null);
setDefaultCloseOperation(EXIT_ON_CLOSE);
JMenuItem lineMenuItem = new JMenuItem("New Line");
lineMenuItem.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent event) {
System.out.println("adding line");
shapeList.addLine();
}
});
JMenuBar menubar = new JMenuBar();
menubar.add(lineMenuItem);
setJMenuBar(menubar);
}
}
public static class Shapes extends JPanel {
private static final long serialVersionUID = 1L;
private List<Line2D.Double> lines = new ArrayList<Line2D.Double>();
private Boolean drawing = false;
private Point lineStartingPoint = new Point();
private Point lineEndingPoint = new Point();
private Line2D.Double linePath;
double phi = Math.toRadians(40);
int barb = 20;
public Shapes() {
MyMouseAdapter myMouseAdapter = new MyMouseAdapter();
addMouseListener(myMouseAdapter);
addMouseMotionListener(myMouseAdapter);
this.setOpaque(true);
this.setBackground(Color.WHITE); // set canvas color
}
public void addLine() {
drawing = true;
repaint();
}
@Override
protected void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2 = (Graphics2D) g;
g2.setStroke(new BasicStroke(2));
if (drawing) {
g2.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
g2.setStroke(new BasicStroke(2));
g2.drawLine(lineStartingPoint.x, lineStartingPoint.y,
lineEndingPoint.x, lineEndingPoint.y);
drawArrowHead(g2, lineEndingPoint, lineStartingPoint,
Color.BLACK);
}
for (Line2D line : lines) {
g2.setColor(Color.BLACK);
Point sw = new Point((int) line.getX1(), (int) line.getY1());
Point ne = new Point((int) line.getX2(), (int) line.getY2());
g2.draw(line);
drawArrowHead(g2, ne, sw, Color.BLACK);
}
}
public Rectangle2D drawRect(int x, int y) {
return new Rectangle2D.Double(x - 4, y - 4, 8, 8);
}
private void drawArrowHead(Graphics2D g2, Point tip, Point tail,
Color color) {
g2.setPaint(color);
double dy = tip.y - tail.y;
double dx = tip.x - tail.x;
double theta = Math.atan2(dy, dx);
double x, y, rho = theta + phi;
Point p1 = new Point();
Point p2 = new Point();
p1.setLocation(tip.x - barb * Math.cos(rho), tip.y - barb * Math.sin(rho));
rho = theta - phi;
p2.setLocation(tip.x - barb * Math.cos(rho), tip.y - barb * Math.sin(rho));
int[] xPoints = new int[5];
int[] yPoints = new int[5];
xPoints[0] = tip.x;
xPoints[1] = p1.x;
xPoints[2] = p2.x;
yPoints[0] = tip.y;
yPoints[1] = p1.y;
yPoints[2] = p2.y;
g2.setPaint(Color.BLACK);
Shape shape = new Polygon(xPoints, yPoints, 3);
g2.fill(shape);
//tip.x - barb * Math.cos(rho);
//y = tip.y - barb * Math.sin(rho);
}
class MyMouseAdapter extends MouseAdapter {
int currentIndex;
Point2D.Double startPoint = new Point2D.Double();
Point2D.Double endPoint = new Point2D.Double();
Boolean resizing = false;
@Override
public void mousePressed(MouseEvent e) {
if (drawing) {
lineStartingPoint = e.getPoint();
lineEndingPoint = lineStartingPoint;
}
}
@Override
public void mouseDragged(MouseEvent e) {
if (drawing) {
lineEndingPoint = e.getPoint();
repaint();
System.out.println(lines.size());
}
}
@Override
public void mouseReleased(MouseEvent e) {
if (drawing) {
drawLine(e);
}
drawing = false;
}
@Override
public void mouseMoved(MouseEvent e) {
}
public void drawLine(MouseEvent e) {
drawing = false;
lineEndingPoint = e.getPoint();
linePath = new Line2D.Double(lineStartingPoint.getX(),
lineStartingPoint.getY(), lineEndingPoint.getX(),
lineEndingPoint.getY());
lines.add(linePath);
repaint();
}
}
}
}
【问题讨论】:
-
老实说,创建一个自定义形状,然后根据需要对其进行平移/变换并绘制它,这比尝试使用两个断开连接的点数组更简单...
-
好吧,我更关心如何计算第 4 点的位置以使其成为钻石。但我会接受你的建议:)