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RE: namespaces in xslt

2005-04-01 05:16:16


Hi,
     By "detailed navigational expressions" if you mean using indices like
[1], [2] etc in my xpath, I was only using it bcos I could be sure that
<root> will have a child named <tr> and <tr>, a child named <slideshow>
(just now realized).  However, I will try to follow your advise in future.

Cheers,
Omprakash.V











                                                                                
                                   
                    "Michael Kay"                                               
                                   
                    <mike(_at_)saxonic        To:     
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                    a.com>               cc:     (bcc: omprakash.v/Polaris)     
                                   
                                         Subject:     RE: [xsl] namespaces in 
xslt                                 
                    04/01/2005                                                  
                                   
                    05:35 PM                                                    
                                   
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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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