我有一个带有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();
}
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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.
您可以将对象序列化为字符串,将字符串加载到a中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();
}
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