Question

有人知道Office Open XML x中y和wrapPolygon的含义和单位是什么？

“指定x轴上的坐标。此坐标的原点应由父XML元素指定。...此属性的可能值由ST_Coordinate简单类型定义（第5.1.12.16节）。“

听起来它应该描述从图片左上角的x = 0和y = 0开始的多边形。单位为EMU。

但这不可能是真的，因为如果在代码Change image layout or wrap in DOCX with Apache POI中使用，那么，如果我做了类似的事情：

...
  +"<wp:wrapTight wrapText=\"bothSides\">"
  +"<wp:wrapPolygon edited=\"0\">"
  +"<wp:start x=\"0\" y=\"0\"/>"
  +"<wp:lineTo x=\"0\" y=\""+height+"\"/>"
  +"<wp:lineTo x=\""+width+"\" y=\""+height+"\"/>"
  +"<wp:lineTo x=\""+width+"\" y=\"0\"/>"
  +"<wp:lineTo x=\"0\" y=\"0\"/>"
  +"</wp:wrapPolygon>"
  +"</wp:wrapTight>"
...

然后得到的包裹点远远超出图片。

而是描述方形多边形21600 x 21600

...
  +"<wp:wrapPolygon edited=\"0\">"
  +"<wp:start x=\"0\" y=\"0\"/>"
  +"<wp:lineTo x=\"0\" y=\"21600\"/>"
  +"<wp:lineTo x=\"21600\" y=\"21600\"/>"
  +"<wp:lineTo x=\"21600\" y=\"0\"/>"
  +"<wp:lineTo x=\"0\" y=\"0\"/>"
  +"</wp:wrapPolygon>"
...

导致包裹点位于图片的全宽x高度。

这与图片尺寸本身无关。它可以是所有可能尺寸的正方形或矩形尺寸的图片。

因此，“此坐标的原点应由父XML元素指定。”并且多边形从图片左上角的x = 0和y = 0开始，单位宽度和高度不能是EMU。如果图片是一个矩形，方形多边形21600 x 21600会导致矩形包裹点，即使多边形本身的含义也不清楚。

这是在某处记录的吗？

Answer 1

好吧，似乎没有人回答。所以我至少会提供一个例子说明为什么这个规则：“方形多边形21600 x 21600导致包装点的全宽x高度与图片大小无关。”可能有用。

如果需要设置更复杂的包裹点，例如椭圆，则仅需要将此椭圆描绘为方形21600 x 21600，与图片大小无关。这比根据真实图片大小计算包裹点要简单得多。因此，我们可以强制Word在图片周围紧紧包裹文字。

示例代码：

import java.io.FileOutputStream;
import java.io.FileInputStream;
import java.io.InputStream;

import org.apache.poi.xwpf.usermodel.*;

import org.apache.poi.util.Units;

import org.openxmlformats.schemas.wordprocessingml.x2006.main.CTDrawing;
import org.openxmlformats.schemas.drawingml.x2006.main.CTGraphicalObject;
import org.openxmlformats.schemas.drawingml.x2006.wordprocessingDrawing.CTAnchor;

public class WordPicturesWrapTight {

 private static CTAnchor getAnchorWithGraphic(CTGraphicalObject graphicalobject, 
                                              String drawingDescr, int width, int height,
                                              int left, int top) throws Exception {

  String anchorXML = 
   "<wp:anchor xmlns:wp=\"http://schemas.openxmlformats.org/drawingml/2006/wordprocessingDrawing\" "
  +"simplePos=\"0\" relativeHeight=\"0\" behindDoc=\"1\" locked=\"0\" layoutInCell=\"1\" allowOverlap=\"1\">"
  +"<wp:simplePos x=\"0\" y=\"0\"/>"
  +"<wp:positionH relativeFrom=\"column\"><wp:posOffset>"+left+"</wp:posOffset></wp:positionH>"
  +"<wp:positionV relativeFrom=\"paragraph\"><wp:posOffset>"+top+"</wp:posOffset></wp:positionV>"
  +"<wp:extent cx=\""+width+"\" cy=\""+height+"\"/>"
  +"<wp:effectExtent l=\"0\" t=\"0\" r=\"0\" b=\"0\"/>"
  +"<wp:wrapTight wrapText=\"bothSides\">"
  +"<wp:wrapPolygon edited=\"1\">"; //Set edited 1, so Word will not destroying the wrap points.

  //A square polygon 21600 x 21600 leads to wrap points in fully width x height independent of picture size.
  //So if the need is setting more complex wrap points, a ellipse for example, only depicting this ellipse 
  //to a square 21600 x 21600 is needed independent of picture size.

  long[] x = new long[5];
  long[] y = new long[5];
  for (int i = 0; i < 5; i++) {
   x[i] = i * 2700L; //2700 = 21600/2/4
   y[i] = Math.round(Math.sqrt(116640000d - Math.pow(i * 2700d, 2d))); //116640000 = (21600/2)^2
  }

  anchorXML += "<wp:start x=\""+(x[0]+10800)+"\" y=\""+(10800-y[0])+"\"/>";

  for (int i = 1; i < 5; i++) {
   anchorXML += "<wp:lineTo x=\""+(x[i]+10800)+"\" y=\""+(10800-y[i])+"\"/>";
  }
  for (int i = 3; i > -1; i--) {
   anchorXML += "<wp:lineTo x=\""+(x[i]+10800)+"\" y=\""+(10800+y[i])+"\"/>";
  }
  for (int i = 1; i < 5; i++) {
   anchorXML += "<wp:lineTo x=\""+(10800-x[i])+"\" y=\""+(10800+y[i])+"\"/>";
  }
  for (int i = 3; i > -1; i--) {
   anchorXML += "<wp:lineTo x=\""+(10800-x[i])+"\" y=\""+(10800-y[i])+"\"/>";
  }

  anchorXML += "</wp:wrapPolygon>"
  +"</wp:wrapTight>"
  +"<wp:docPr id=\"1\" name=\"Drawing 0\" descr=\""+drawingDescr+"\"/><wp:cNvGraphicFramePr/>"
  +"</wp:anchor>";

  CTDrawing drawing = CTDrawing.Factory.parse(anchorXML);
  CTAnchor anchor = drawing.getAnchorArray(0);
  anchor.setGraphic(graphicalobject);
  return anchor;  
 }

 public static void main(String[] args) throws Exception {

  XWPFDocument document = new XWPFDocument();

  XWPFParagraph paragraph = document.createParagraph();
  XWPFRun run = paragraph.createRun();

  InputStream in = new FileInputStream("ellipticSample.png");
  run.addPicture(in, Document.PICTURE_TYPE_PNG, "ellipticSample.png", Units.toEMU(100), Units.toEMU(60));
  in.close();  
  CTDrawing drawing = run.getCTR().getDrawingArray(0);
  CTGraphicalObject graphicalobject = drawing.getInlineArray(0).getGraphic();
  CTAnchor anchor = getAnchorWithGraphic(graphicalobject, "ellipticSample.png", 
                                         Units.toEMU(100), Units.toEMU(60), 
                                         Units.toEMU(100), Units.toEMU(16));
  drawing.setAnchorArray(new CTAnchor[]{anchor});
  drawing.removeInline(0);

  run = paragraph.createRun();
  run.setText("The picture is anchored wrap tight. This means text will wrap this according to a polygon described by wrap points. Seems a square polygon 21600 x 21600 leads to wrap points in fully width x height independent of picture size. So if the need is setting more complex wrap points, a ellipse for example, only depicting this ellipse to a square 21600 x 21600 is needed independent of the picture size. This is much more simple if more complex wrap points shall be set. But it's a shame, that this feature not seems documented. So can we rely on it or not?");

  document.write(new FileOutputStream("WordPicturesWrapTight.docx"));
  document.close();
 }
}

结果：

Office Open XML`wrapPolygon`中`x`和`y`的含义和单位

1 个答案: