我必须使用pdfbox绘制一个饼图。
让数据为：
Subject Mark in Percentage Mark in Degrees Cumulative Degrees设半径和中心为100像素和（250,400）。
让我们将初始线平行于x轴。 绘制初始行语句将是： contentStream.drawLine（250,400,350,400）;
我坚持： a）在圆上找到距离初始线一定程度的点的x，y坐标，以绘制半径 b）使用贝塞尔曲线在两点之间绘制弧。
任何有关解决问题的帮助都将受到高度赞赏！
根据角度在圆上找到x，y坐标是学校数学，即sin（）和cos（），棘手的部分是用Bézier曲线绘制弧。
这是一些绘制您要求的饼图的代码。请注意，createSmallArc()只能在最大90°的角度下工作。如果你想要更多，你要么必须通过绘制几个弧来修改代码，直到你回到（0,0），或者只是绘制几个切片。
（createSmallArc()是由Hans Muller，许可证：Creative Commons Attribution 3.0。所做的更改：在java中实现原始AS代码。算法由Aleksas Riškus提供）
public class PieChart
{
public static void main(String[] args) throws IOException
{
PDDocument doc = new PDDocument();
PDPage page = new PDPage();
doc.addPage(page);
PDPageContentStream cs = new PDPageContentStream(doc, page);
cs.transform(Matrix.getTranslateInstance(250, 400));
cs.setNonStrokingColor(Color.yellow);
drawSlice(cs, 100, 0, 80);
cs.fill();
cs.setNonStrokingColor(Color.red);
drawSlice(cs, 100, 80, 150);
cs.fill();
cs.setNonStrokingColor(Color.green);
drawSlice(cs, 100, 150, 215);
cs.fill();
cs.setNonStrokingColor(Color.blue);
drawSlice(cs, 100, 215, 305);
cs.fill();
cs.setNonStrokingColor(Color.ORANGE);
drawSlice(cs, 100, 305, 360);
cs.fill();
cs.close();
doc.save("piechart.pdf");
doc.close();
}
private static void drawSlice(PDPageContentStream cs, float rad, float startDeg, float endDeg) throws IOException
{
cs.moveTo(0, 0);
List<Float> smallArc = createSmallArc(rad, Math.toRadians(startDeg), Math.toRadians(endDeg));
cs.lineTo(smallArc.get(0), smallArc.get(1));
cs.curveTo(smallArc.get(2), smallArc.get(3), smallArc.get(4), smallArc.get(5), smallArc.get(6), smallArc.get(7));
cs.closePath();
}
/**
* From https://hansmuller-flex.blogspot.com/2011/10/more-about-approximating-circular-arcs.html
*
* Cubic bezier approximation of a circular arc centered at the origin,
* from (radians) a1 to a2, where a2-a1 < pi/2. The arc's radius is r.
*
* Returns a list with 4 points, where x1,y1 and x4,y4 are the arc's end points
* and x2,y2 and x3,y3 are the cubic bezier's control points.
*
* This algorithm is based on the approach described in:
* Aleksas Riškus, "Approximation of a Cubic Bezier Curve by Circular Arcs and Vice Versa,"
* Information Technology and Control, 35(4), 2006 pp. 371-378.
*/
private static List<Float> createSmallArc(double r, double a1, double a2)
{
// Compute all four points for an arc that subtends the same total angle
// but is centered on the X-axis
double a = (a2 - a1) / 2;
double x4 = r * Math.cos(a);
double y4 = r * Math.sin(a);
double x1 = x4;
double y1 = -y4;
double q1 = x1*x1 + y1*y1;
double q2 = q1 + x1*x4 + y1*y4;
double k2 = 4/3d * (Math.sqrt(2 * q1 * q2) - q2) / (x1 * y4 - y1 * x4);
double x2 = x1 - k2 * y1;
double y2 = y1 + k2 * x1;
double x3 = x2;
double y3 = -y2;
// Find the arc points' actual locations by computing x1,y1 and x4,y4
// and rotating the control points by a + a1
double ar = a + a1;
double cos_ar = Math.cos(ar);
double sin_ar = Math.sin(ar);
List<Float> list = new ArrayList<Float>();
list.add((float) (r * Math.cos(a1)));
list.add((float) (r * Math.sin(a1)));
list.add((float) (x2 * cos_ar - y2 * sin_ar));
list.add((float) (x2 * sin_ar + y2 * cos_ar));
list.add((float) (x3 * cos_ar - y3 * sin_ar));
list.add((float) (x3 * sin_ar + y3 * cos_ar));
list.add((float) (r * Math.cos(a2)));
list.add((float) (r * Math.sin(a2)));
return list;
}
}
如果您只想使用PDFBox将某些图表绘制成PDF并且不想自己完成所有数学运算，那么您可以使用我的PDFBox Graphics2D adapter。这允许您使用任何基于Graphics2D的java库来绘制图表（例如JFreeChart）。它将创建一个XForm，然后您可以在PDF中自由放置。
如果你真的想自己这样做，你仍然可以使用Java2D API来帮助形状。
Arc2D.Float arc = new Arc2D.Float(x,y,w,h,start,extend, Arch2D.OPEN);
AffineTransform tf = new AffineTransform();
// You may need to setup tf to correctly position the drawing.
float[] coords = new float[6];
PathIterator pi = arc.getPathIterator(tf);
while (!pi.isDone()) {
int segment = pi.currentSegment(coords);
switch (segment) {
case PathIterator.SEG_MOVETO:
if (isFinite(coords, 2))
contentStream.moveTo(coords[0], coords[1]);
break;
case PathIterator.SEG_LINETO:
if (isFinite(coords, 2))
contentStream.lineTo(coords[0], coords[1]);
break;
case PathIterator.SEG_QUADTO:
if (isFinite(coords, 4))
contentStream.curveTo1(coords[0], coords[1], coords[2], coors[3]);
break;
case PathIterator.SEG_CUBICTO:
if (isFinite(coords, 6))
contentStream.curveTo(coords[0], coords[1], coords[2], coords[3], coords[4], coords[5]);
break;
case PathIterator.SEG_CLOSE:
contentStream.closePath();
break;
}
pi.next();
}
contentStream.fill();