I am newbie to XSLT.

I am having one XML document which root element have five attributes.

for example
<root xmlns="org:v3" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="" id="root" value="1">.
I have to convert this to another XML document . I m using XSLT for this and I am using Xpath in the "xsl:apply-templates select=" statement.

My .xsl statement is as follows.

<xsl:apply-templates select="root/document/firstnode"/>

The above .xsl statement is not working. If i remove the attributes from the root element, the above statement is working fine.

I don't know why its not working. I have tried lot of possibilities with select="root[@*]/document/firstnode" ,
select="root[0]/document/firstnode". But couldn't make it out.

Please help me.

One more doubt:

I am using stylesheet reference in the XML file like <?xml-stylesheet href="stylesheet.xsl" type="text/xsl"?>.
This statement is missing in the output XML file. What i have to do for this?

Thanks and Regards,
N.Raja

Suggest you read up on namespaces:

http://www.dpawson.co.uk/xsl/sect2/N5536.html

especially numbers 13 and 23.



--

Joe (Microsoft MVP - XML)

I got the answer for my first question

the answer is

root[1]/document/firstnode.

I m not clear of the second one. Please help me. My i/p XML file is having the stylesheet reference and my output file is missing that reference.

<?xml-stylesheet href="stylesheet.xsl" type="text/xsl"?>.
it should come to the next line to <?xml version="1.0" encoding="utf-8"?>

You need to understand that XPath expressions work on a representation of the XML document called the XPath Data Model. In this model:

* namespaces are not attributes (so your root element has 3 attributes, not 5)

* the <?xml-stylesheet?> ends up as a processing-instruction node. If you want to copy processing instructions, you need to write a template rule for them:

<xsl:template match="processing-instruction()">
  <xsl:copy/>
</xsl:template>

* the XML declaration is not represented by a node in the data model at all. The serializer looks after the job of creating a correct XML declaration in the output document.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference