Your root element has only one child, it is a <tr> element. So root/*[2]
selects nothing. The <slideshow> is a child of the <tr>.
You seem to be trying to write detailed navigational expressions to reach
specific nodes in your source document. This isn't very robust, you will
have to make lots of changes when the source changes. It's much better to
write template rules to process each kind of element, and then recurse down
the tree using xsl:apply-templates.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: omprakash(_dot_)v(_at_)polaris(_dot_)co(_dot_)in
[mailto:omprakash(_dot_)v(_at_)polaris(_dot_)co(_dot_)in]
Sent: 01 April 2005 12:54
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] namespaces in xslt
Hi,
Iam having diffculty following what's happening:
I have xml like this
<table>
<th>
<root>
<tr>
<slideshow xmlns="urn:hl7-org/v3" title="Sample Slide Show"
date="Date of
publication" author="Yours Truly">
<td><a href="file:///N65538"/>
</td>
</slideshow>
</tr>
</root>
</th>
</table>
And my xsl:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl
="http://www.w3.org/1999/XSL/Transform"; xmlns="urn:hl7-org/v3" >
<xsl:output method="xml" indent="yes" />
<xsl:template match="/">
<xsl:for-each select="//root/*[1]">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates select="@*|node()">
<xsl:with-param name="rootnode" select="$rootnode"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
When I select //own:root/*[1] in the <xsl:for-each>, I get
the following
<?xml version="1.0" encoding="UTF-8"?>
<tr xmlns="http://XYZ"/>
but when I make it //root/*[2], I get nothing. But if you see, the
slideshow is really at position 2 so I need to use 2.
Cheers,
Omprakash.V
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