Question

We need to parse an XML file with XSLT into a CVS file. The code below works but only if the fields in the XML are always constant.

The fields in the our XML file will always vary. How can I change my XSLT file so that it dynamically includes in the CSV file all the fields from the XML file?

XML:

<?xml version='1.0' encoding='iso-8859-1'?> <data>     <row>         <customerID>06104539-463E-4B1A-231-34342343434</customerID>         <contactID>23434-99F2-4325-B228-6F343483469389FB</contactID>         <firstName>Jim</firstName>         <lastName>Smith</lastName>     </row>     <row>         <customerID>223434-463E-4B1A-231-A1E7EA248796</customerID>         <contactID>6675767-99F2-4325-B234328-6F83469389FB</contactID>         <specialID>112332</specialID>         <firstName>John</firstName>         <middleName>S.</middleName>         <lastName>Jones</lastName>     </row> </data> 

XSLT:

<?xml version='1.0' encoding='ISO-8859-1'?>  <xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>  <xsl:template match='/'>     customerID,contactID<br/>     <xsl:for-each select='data/row'>         <xsl:value-of select='customerID'/>,<xsl:value-of select='contactID'/><br/>     </xsl:for-each> </xsl:template>  </xsl:stylesheet> 

Here is the ASP.NET file that parses the files above:

<%@ Page Language='c#' %> <%@ import Namespace='System.Xml' %> <%@ import Namespace='System.Xml.Xsl' %> <%@ import Namespace='System.Xml.XPath' %> <%@ import Namespace='System.IO' %> <%@ import Namespace='System.Text' %> <script runat='server'>      public void Page_Load(Object sender, EventArgs E) {          string xmlPath = Server.MapPath('test2.xml');         string xslPath = Server.MapPath('test2.xsl');          StreamReader reader = null;;         XmlTextReader xmlReader = null;          FileStream fs = new FileStream(xmlPath, FileMode.Open, FileAccess.Read);         reader = new StreamReader(fs,Encoding.UTF7);         xmlReader = new XmlTextReader(reader);         XPathDocument doc = new XPathDocument(xmlReader);          XslTransform xslDoc = new XslTransform();         xslDoc.Load(xslPath);          xslDoc.Transform(doc,null, Response.Output);          reader.Close();         xmlReader.Close();      }  </script> 

Answer 1

I remembered doing something similar a ways back, and looking through my archive work, lo-and-behold:

<?xml version='1.0' encoding='ISO-8859-1'?>  <xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>      <xsl:key name='muench' match='/data/row/*' use='local-name()'/>      <xsl:template match='/'>         <xsl:for-each select='/data/row/*[generate-id() = generate-id(key('muench',local-name())[1])]'>             <xsl:if test='not(position()=1)'>,</xsl:if><xsl:value-of select='local-name()'/>         </xsl:for-each>         <br/>         <xsl:for-each select='/data/row'>             <xsl:variable name='current' select='.'/>             <xsl:for-each select='/data/row/*[generate-id() = generate-id(key('muench',local-name())[1])]'>                 <xsl:variable name='field' select='local-name()'/>                 <xsl:if test='not(position()=1)'>,</xsl:if><xsl:value-of select='$current/*[local-name()=$field]'/>                          </xsl:for-each>             <br/>         </xsl:for-each>     </xsl:template>  </xsl:stylesheet> 

All hail the ugly but functional.

I’m seriously not proud of this, but I’ve updated for your data and tested it and it works.