如何使用 Java Swing 编写肮脏的丰富渐变绘制边框

Question

修订：问题是绘制一个四边边框，其中每边都以纯色基础颜色开始，并在边框的范围内向内淡化为白色。挑战在于使边界的交叉点看起来无缝。为此，可以绘制边界，然后使用三角形“混合”角。如果绘制 broder 的矩形没有重叠，则每个角可以使用两个三角形，或者如果两个平行的边界边延伸边界的整个长度（即矩形重叠），每个角一个三角形就足够了（如下所示）。

    private static final int GRADIENT_LENGTH = 29;
    private static final int BAR_LENGTH = 25;
    public static void paintGradientBorder(Graphics g, Color borderColor) {

    Graphics2D g2 = (Graphics2D) g.create();

    GradientPaint gradientColorWest = new GradientPaint(0, 0, borderColor,
            GRADIENT_LENGTH, 0, Color.WHITE);
    GradientPaint gradientColorEast = new GradientPaint(WINDOW_WIDTH - GRADIENT_LENGTH,
            0, Color.WHITE, WINDOW_WIDTH, 0, borderColor);
    GradientPaint gradientColorNorth= new GradientPaint(0, 0, borderColor, 0,
            GRADIENT_LENGTH, Color.WHITE);
    GradientPaint gradientColorSouth = new GradientPaint(0, WINDOW_HEIGHT - GRADIENT_LENGTH,
            Color.WHITE,0, WINDOW_HEIGHT, borderColor);

    //south bar
    g2.setPaint(gradientColorSouth);
    g2.fillRect(0, WINDOW_HEIGHT - BAR_LENGTH, WINDOW_WIDTH, BAR_LENGTH);
    //north bar
    g2.setPaint(gradientColorNorth);
    g2.fillRect(0, 0, WINDOW_WIDTH, BAR_LENGTH);
    //west bar
    g2.setPaint(gradientColorWest);
    g2.fillRect(0, BAR_LENGTH, BAR_LENGTH, WINDOW_HEIGHT - BAR_LENGTH * 2);
    //east bar
    g2.setPaint(gradientColorEast);
    g2.fillRect(WINDOW_WIDTH - BAR_LENGTH, BAR_LENGTH, WINDOW_WIDTH, WINDOW_HEIGHT - BAR_LENGTH * 2);

    //NORTH WEST CORNER
    //left triangle
    Polygon p = new Polygon();        
    p.addPoint(0, 0);
    p.addPoint(BAR_LENGTH, BAR_LENGTH);
    p.addPoint(0, BAR_LENGTH);
    g2.setPaint(gradientColorWest);
    g2.fillPolygon(p);        
    //NORTH EAST CORNER
    //right triangle
    p.reset();
    p.addPoint(WINDOW_WIDTH, 0);
    p.addPoint(WINDOW_WIDTH - BAR_LENGTH, BAR_LENGTH);
    p.addPoint(WINDOW_WIDTH, BAR_LENGTH);
    g2.setPaint(gradientColorEast);
    g2.fillPolygon(p);
    //SOUTH WEST CORNER
    //left triangle
    p.reset();
    p.addPoint(0, WINDOW_HEIGHT);
    p.addPoint(0,WINDOW_HEIGHT - BAR_LENGTH);
    p.addPoint(BAR_LENGTH, WINDOW_HEIGHT - BAR_LENGTH);
    g2.setPaint(gradientColorWest);
    g2.fillPolygon(p);
    //SOUTH EAST CORNER
    //right triangle
    p.reset();
    p.addPoint(WINDOW_WIDTH, WINDOW_HEIGHT);
    p.addPoint(WINDOW_WIDTH, WINDOW_HEIGHT - BAR_LENGTH);
    p.addPoint(WINDOW_WIDTH - BAR_LENGTH, WINDOW_HEIGHT - BAR_LENGTH);
    g2.setPaint(gradientColorEast);
    g2.fillPolygon(p);

    g2.dispose();
}

Answer 1

如果您不与矩形相交而是使用多边形 (GeneralPath) 会怎样？

GeneralPath topBox = new GeneralPath();
topBox.moveTo(0, 0);
// upper right
topBox.lineTo(width, 0);
// lower right; move diagonally down and to the left as in a picture frame
topBox.lineTo(width - (insetX / 2), 0 + (insetY / 2));
// lower left
topBox.lineTo((insetX / 2), 0 + (insetY / 2));
topBox.closePath();
g2.fill(topBox);

这样矩形就不会重叠，而是在不同的部分之间有很好的清晰边缘。

Answer 2

我不会创建和绘制 4 个矩形，而是创建一个 Shape，通过使用 Area 从外部矩形减去内部矩形来代表您的边框区域：

Area area = new Area(new Rectangle2D.Double(...));
Area inner = new Area(new Rectangle2D.Double(...));
area.subtract(inner);

g2.setPaint(new GradientPaint(...));
g2.fill(area);