Question

我有一个带2个参数的函数：1 = XML文件，2 = XSLT文件，然后执行转换并返回生成的HTML。

这是功能：

/// <summary>
/// Will apply an XSLT style to any XML file and return the rendered HTML.
/// </summary>
/// <param name="xmlFileName">
/// The file name of the XML document.
/// </param>
/// <param name="xslFileName">
/// The file name of the XSL document.
/// </param>
/// <returns>
/// The rendered HTML.
/// </returns>
public string TransformXml(string xmlFileName, string xslFileName)
{
    var xtr = new XmlTextReader(xmlFileName)
                  {
                      WhitespaceHandling = WhitespaceHandling.None
                  };
    var xd = new XmlDocument();
    xd.Load(xtr);

    var xslt = new System.Xml.Xsl.XslCompiledTransform();
    xslt.Load(xslFileName);
    var stm = new MemoryStream();
    xslt.Transform(xd, null, stm);
    stm.Position = 1;
    var sr = new StreamReader(stm);
    xtr.Close();
    return sr.ReadToEnd();
}

我想更改函数不接受XML的文件，而只是一个对象。该对象与xslt完全兼容，如果它被序列化为文件。但我不想首先将它序列化为文件。

所以回顾一下：保持xslt来自文件，但xml输入应该是我传递的对象，并且希望在没有任何文件系统交互的情况下生成xml。

Answer 1

您可以将对象序列化为字符串，将字符串加载到XmlDocument，然后执行转换：

public string TransformXml(object data, string xslFileName)
{

    XmlSerializer xs = new XmlSerializer(data.GetType());
    string xmlString;
    using (StringWriter swr = new StringWriter())
    {
        xs.Serialize(swr, data);
        xmlString = swr.ToString();
    }

    var xd = new XmlDocument();
    xd.LoadXml(xmlString);

    var xslt = new System.Xml.Xsl.XslCompiledTransform();
    xslt.Load(xslFileName);
    var stm = new MemoryStream();
    xslt.Transform(xd, null, stm);
    stm.Position = 0;
    var sr = new StreamReader(stm);
    return sr.ReadToEnd();
}

Answer 2

这是一个将对象转换为XDocument的函数（如果你还没有使用XDocument，可以为XmlDocument更改它）。当然，如果对象不可序列化，这将抛出异常。

public static XDocument ConvertToXml<T>(this T o)
{
    StringBuilder builder = new StringBuilder();
    StringWriter writer = new StringWriter(builder);
    XmlSerializer serializer = new XmlSerializer(typeof(T));
    serializer.Serialize(writer,o);
    StringReader reader = new StringReader(builder.ToString());
    return XDocument.Load(reader);
}

，这是XmlDocument的一个

public static XmlDocument ConvertToXml<T>(this T o)
{
    StringBuilder builder = new StringBuilder();
    StringWriter writer = new StringWriter(builder);
    XmlSerializer serializer = new XmlSerializer(typeof(T));
    serializer.Serialize(writer,o);
    StringReader reader = new StringReader(builder.ToString());
    XmlDocument doc = new XmlDocument();
    doc.Load(reader);
    return doc;
}

Answer 3

未经过测试，但您可以使用XPathDocument获取Stream，并且由于XPathDocument实现了IXPathNavigable，因此可以将其用于转换：

 public string TransformXml(Stream xmlFile, string xslFileName) 
 { 
      var doc = new XPathDocument(xmlFile);
      var xslt = new System.Xml.Xsl.XslCompiledTransform(); 
      xslt.Load(xslFileName); 
      var stm = new MemoryStream(); 
      xslt.Transform(doc, null, stm); 
      stm.Position = 1; 
      var sr = new StreamReader(stm); 
      return sr.ReadToEnd(); 
  }

Answer 4

查看this article，其中描述了创建一个XPathNavigator，它可以导航对象图的属性，这是一个非常强大的XPath和XSLT组合。

C＃如何在内存对象中执行实时xslt转换？

4 个答案: