Hi all,

I need to perform a transformation from one XML format to other XML format. Before that, I decided to implement the concept on a dummy XML. So I wrote the following XML:

------- XML ------------

<?xml version="1.0" ?>
<?xml-stylesheet type="text/xsl" href="books.xsl" ?>
<books>
<book>
<title>Beg ASP3.0</title>
<ISBN>1-861003-38-2</ISBN>
<authors>
<author_name>Sanket</author_name>
<author_name>Keyur</author_name>
<author_name>Deepak</author_name>
<author_name>Jeevan</author_name>
<author_name>Utanka</author_name>
</authors>
</book>
</books>

------------- XSL for the above XML ------------------

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" omit-xml-declaration="yes"/>
<xsl:template match="/">
<books>
    <book>
        <BOOK_TITLE><xsl:value-of select="//books/book/title"/></BOOK_TITLE>
        <BOOK_ISBN><xsl:value-of select="//books/book/ISBN"/></BOOK_ISBN>
    </book>
    <authors>
    </authors>
</books>
</xsl:template>
</xsl:stylesheet>
------------------------------

But, when I view it in I.E 6.0, it shows plane html with only Book's name and ISBN. When I view the source, the source is no different from the original XML. But, when I export it to an excel sheet, there I can view the modified XML.

 Can anybody please tell me why is this happening?

It's just the way IE does things. It treats the transform result as html, so ignores the tags and just displays the text. Also, View Source always shows the original xml, not the transformation result.

You need something a bit more useful to develop and test your stylesheets. If you're looking for a good freebie, try Cooktop, http://xmlcooktop.com/

Phil, I want to display db fields in columns and I am using this code:
dim evalRS
Set evalRS=server.createobject("ADODB.recordset")
sqltext = "SELECT ClassNo, EvaluationNo FROM manager_class_query"
evalRS.Open sqltext, "dsn=ClassEval"
dim RecordArray
dim n
Const cField = 0
If Not evalRS.EOF Then
RecordArray = evalRS.GetRows()
Response.Write("<table>")
For n = 0 to 25
Response.Write("<tr>")
Response.Write("<td>")
Response.Write(RecordArray(cField, n))
Response.Write("</td>")
Response.Write("<td>")
Response.Write(RecordArray(cField, n+26))
Response.Write("</td>")
Response.Write("</tr>")
Next
Response.Write("</table>")

END IF

This results in 2 columns of the ClassNo field. I need to display another field EvaluationNo.

Could you help me with this? Thank you.

I don't see what this has to do with XSLT, but don't you just need: