xml to json using xslt

问题:

am trying to convert the XML to Json Conversion by applying the xslt transformation using following code in C#. Am getting following error can any one suggest me on below? Code:

 Li = p.GetRegisterEntry();
        var std = Li.Where(s => s.Id == id).FirstOrDefault();
        string xml = std.contentxml.Value;
        doc.LoadXml(xml);            
        string XSLT = std.TemplateXSLT.Value;
        Xslt.LoadXml(XSLT);            
        XslCompiledTransform xslt = new XslCompiledTransform();
        xslt.Load(Xslt.CreateNavigator());           
        // Transform our Xml-ified JSON
        var outputDocument = new XmlDocument();
        var stream = new MemoryStream();
        xslt.Transform(doc, null, stream);
        var sr = new StreamReader(stream);
        var myStr = sr.ReadToEnd();
        stream.Position = 0;
        outputDocument.Load(stream);
        // Convert back to JSON
        string jsonText = JsonConvert.SerializeXmlNode(outputDocument);

am getting the following error at outputDocument.Load(stream); An exception of type 'System.Xml.XmlException' occurred in System.Xml.dll but was not handled in user code Additional information: There are multiple root elements. Line 2, position 2.

can any one please help me to crack this?

here are my XML and XSTL Files XML:

   <NewDataSet>
      <RegisterEntry type="CM2" desc="Request to Change Name, Address and /or Singapore Address for Service for Agent, Applicant/Proprietor and/or other interested Parties">
        <EventDate>23/09/2051</EventDate>
        <DataItems>
          <LodgementDate>26/08/2022</LodgementDate>
          <DecisionDate>21/12/2031</DecisionDate>
          <Particulars>
            <Particular>
              <TransactionType>Test Data</TransactionType>
              <Details>Test Data</Details>
            </Particular>
          </Particulars>
        </DataItems>
      </RegisterEntry>
    </NewDataSet>

XSLT :

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="html" />
  <xsl:template match="RegisterEntry">
    <p>      Test Data Parties<xsl:value-of select="DataItems/TransactionType" /></p>
    <div id="content">
      <table>
        <tr>
          <th valign="top" align="left">            Lodgement Date          </th>
          <td>
            <xsl:value-of select="DataItems/LodgementDate" />
          </td>
        </tr>
        <tr>
          <th valign="top" align="left">            Decision Date          </th>
          <td>
            <xsl:value-of select="DataItems/DecisionDate" />
          </td>
        </tr>
        <xsl:for-each select="DataItems/Particulars/Particular">
          <tr>
            <th valign="top" align="left" style="width:50%">    Test Data <xsl:value-of select="TransactionType" /></th>
            <td>
              <xsl:value-of select="Details" />
            </td>
          </tr>
        </xsl:for-each>
      </table>
    </div>
  </xsl:template>
</xsl:stylesheet>

回答1:

As Tim has pointed out, the problem is that your XSLT does not create a well-formed XML document you could load with XmlDocument but only a fragment. So if you need an XmlNode with the result you can use a different approach with e.g.

XmlDocument resultDoc = new XmlDocument();
XmlDocumentFragment resultFrag = resultDoc.CreateDocumentFragment();

using (XmlWriter xw = resultFrag.CreateNavigator().AppendChild())
{
  xslt.Transform(doc, null, xw);
  xw.Close();
}

string jsonText = JsonConvert.SerializeXmlNode(resultFrag);

Whether your libray JsonConvert is able to handle a document fragment I don't know however.