转换XAML的PathGeometry到WPF的PathGeometry [英] Convert XAML PathGeometry to WPF PathGeometry

问题描述

我想这是由线段的的PathGeometry。

所以，我用这首代码，但它的错误。

 的PathGeometry TEMP =（的PathGeometry）Geometry.Parse（
< PathGeometry.Figures>+ 
<的PathFigure StartPoint可以= \193.5， 283.5\isClosed返= \True\>中+ 
< PathFigure.Segments>+ 
<线段点= \418.5,283.5\/ >中+ 
<线段点= \418.5,508.5\/>中+ 
<线段点= \193.5,508.5\/> + 
<线段点= \193.5,283.5\/>中+ 
< /PathFigure.Segments>+ 
< /&的PathFigure GT; + 
< /PathGeometry.Figures>）; 
  

如果我用这个第二个代码，这不是错误，但它不包括线段的。结果将PolyLineSegment但我想线段。

 的PathGeometry TEMP =（的PathGeometry）Geometry.Parse（
M29 ，329L30,331L31,334L33,336L34,338L36,341L38,343L39,345L41,348L44,352L46,353L47,355L48,356L49,357L49,357L50,358L50,358L51,357L50,356L51,354L51,350L53,342L54,334L58,320L60,315L61 ，311L63,308L63,306L64,304L65,303L65,302L66,301L66,301L66,301L66,301L66,301L66,301L66,301）; 
  

我如何转换XAML的PathGeometry到WPF的PathGeometry？

感谢

解决方案

您解析的XAML是不正确的代码，你需要使用XAML读取器和投结果到所需的类型。例如：

  System.Windows.Shapes.Path的新路径=（System.Windows.Shapes.Path）System.Windows.Markup。 XamlReader.Parse（<路径的xmlns =HTTP：//schemas.microsoft.com/winfx/2006/xaml/presentation'宽= '20'高度= '80'拉伸=补补='＃FF000000'数据='M 20,25.2941L 20,29.4118L 15.9091,29.4118L 15.9091,40L 12.2727,40L 12.2727,29.4118L 2.54313e-006,29.4118L 2.54313e-006,25.6985L 13.4872,7.62939e-006L 15.9091,7.62939e -006L 15.9091,25.2941L 20,25.2941 ZM 12.2727,25.2941L 12.2727,5.28493L 2.09517,25.2941L 12.2727,25.2941 ZM 20,65.2941L 20,69.4118L 15.9091,69.4118L 15.9091,80L 12.2727,80L 12.2727,69.4118L -5.08626电子006,69.4118L -5.08626e-006,65.6985L 13.4872,40L 15.9091,40L 15.9091,65.2941L 20,65.2941 ZM 12.2727,65.2941L 12.2727,45.2849L 2.09517,65.2941L 12.2727,65.2941 Z'的Horizo​​ntalAlignment ='左' VerticalAlignment ='顶'保证金='140,60,0,0/>中）; 
 LayoutRoot.Children.Add（的新路径）; 
  

如果您正在使用的代码背后，是有要分析一个XAML代码片段的原因是什么？您可以按如下编程创建路径：

 路径path =新路径（）; 
的PathGeometry几何=新的PathGeometry（）; 
的PathFigure图=新的PathFigure（）; 
 figure.StartPoint =新的点（10,10）; 
 figure.Segments.Add（新线段（）
 {
点=新的点（20，20）
}）; 
 
 //例如添加更多的细分这里
 
 geometry.Figures.Add（图） 
 path.Data =几何; 
  

一个路径由一几何形状，它是由数字，这是由段的<！ / p>

如果你想使用简化路径数据在后面的代码，你可以使用通用值转换器：

< A HREF =http://www.scottlogic.co.uk/blog/colin/2010/07/a-universal-value-converter-for-wpf/> http://www.scottlogic.co.uk/博客/科林/ 2010/07 / A-普世价值的转换器换WPF /

I want the PathGeometry that consist of LineSegment.

So, I use this first code but it's error.

PathGeometry temp = (PathGeometry)Geometry.Parse(
            "<PathGeometry.Figures>" +
                "<PathFigure StartPoint=\"193.5,283.5\" IsClosed=\"True\">" +
                    "<PathFigure.Segments>" +
                        "<LineSegment Point=\"418.5,283.5\" />" +
                        "<LineSegment Point=\"418.5,508.5\" />" +
                        "<LineSegment Point=\"193.5,508.5\" />" +
                        "<LineSegment Point=\"193.5,283.5\" />" +
                    "</PathFigure.Segments>" +
                "</PathFigure>" +
            "</PathGeometry.Figures>");

If I use this second code, it's not error but it doesn't consist of LineSegment. The result will be PolyLineSegment but I want LineSegment.

PathGeometry temp = (PathGeometry)Geometry.Parse(
                "M29,329L30,331L31,334L33,336L34,338L36,341L38,343L39,345L41,348L44,352L46,353L47,355L48,356L49,357L49,357L50,358L50,358L51,357L50,356L51,354L51,350L53,342L54,334L58,320L60,315L61,311L63,308L63,306L64,304L65,303L65,302L66,301L66,301L66,301L66,301L66,301L66,301L66,301");

How do I convert XAML PathGeometry to WPF PathGeometry?

Thanks

解决方案

Your code for parsing the XAML is incorrect, you need to use a XAML reader and cast the result to the required type. e.g.:

System.Windows.Shapes.Path newPath = (System.Windows.Shapes.Path)System.Windows.Markup.XamlReader.Parse("<Path xmlns='http://schemas.microsoft.com/winfx/2006/xaml/presentation'  Width='20' Height='80' Stretch='Fill' Fill='#FF000000' Data='M 20,25.2941L 20,29.4118L 15.9091,29.4118L 15.9091,40L 12.2727,40L 12.2727,29.4118L 2.54313e-006,29.4118L 2.54313e-006,25.6985L 13.4872,7.62939e-006L 15.9091,7.62939e-006L 15.9091,25.2941L 20,25.2941 Z M 12.2727,25.2941L 12.2727,5.28493L 2.09517,25.2941L 12.2727,25.2941 Z M 20,65.2941L 20,69.4118L 15.9091,69.4118L 15.9091,80L 12.2727,80L 12.2727,69.4118L -5.08626e-006,69.4118L -5.08626e-006,65.6985L 13.4872,40L 15.9091,40L 15.9091,65.2941L 20,65.2941 Z M 12.2727,65.2941L 12.2727,45.2849L 2.09517,65.2941L 12.2727,65.2941 Z ' HorizontalAlignment='Left' VerticalAlignment='Top' Margin='140,60,0,0'/>");
LayoutRoot.Children.Add(newPath);

If you are using code-behind, is there any reason you want to parse a XAML snippet? You can programmatically create a path as follows:

Path path = new Path();
PathGeometry geometry = new PathGeometry();
PathFigure figure = new PathFigure();
figure.StartPoint = new Point(10,10); 
figure.Segments.Add(new LineSegment()
{
  Point = new Point (20, 20)
});

// e.g. add more segments here

geometry.Figures.Add(figure);
path.Data = geometry;

A path is composed of a geometry, which is composed of figures, which are composed of segments!

If you want to use the simplified path data in code behind you could use a universal value converter:

http://www.scottlogic.co.uk/blog/colin/2010/07/a-universal-value-converter-for-wpf/

这篇关于转换XAML的PathGeometry到WPF的PathGeometry的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！