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Manish Dhelia

Subject: how to remove the XML elements dynamically
Author: Manish Dhelia
Date: 28 Mar 2007 03:40 AM

Following is my XML data:

<?xml version="1.0" encoding="UTF-8" ?>
<?xml-stylesheet type="text/xsl" href="movies.xsl"?>
<ul>
<li type="square">Screen</li> <br/>
<li>
<ul>
<li>sub bullet</li>
</ul>
</li>
</ul>

Problem:How can I remove the second "li" tag dynamically using XSLT in the above XML data keeping the other tags under it intact.

and following is my XSLT file saved as movies.xsl that I'm working on:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="1.0" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">

<xsl:variable name="var" select="//movies/movie/name"/>

<xsl:for-each select="//ul">
<xsl:apply-templates select="li" />
</xsl:for-each>

</xsl:template>
<xsl:template match="strong|p|u|em|ul|li|b|br|h1|h2|h3|h4|h5|h6|hr|i|ol|q|s|sub|sup|small|table|tr|td|thead|tfoot|th|tbody|big|tt|span|blockquote">
<xsl:element name="{name()}">
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

Please reply with the solution.

(Deleted User)

Subject: how to remove the XML elements dynamically
Author: (Deleted User)
Date: 30 Mar 2007 08:02 AM

Hi Manish,
if your reference to the "second 'li'" node has to be interpreted as "the 'li' node that is nested inside another 'li' node" you can try:
- adding a mode to the copier template
- removing 'li' from the list of matches for that template
- propagating the mode to the apply-templates
- creating a new template processing just the 'li' node with a mode=1 that will not create the element, but just invoke apply-templates
- creating a new template processing just the 'li' node with a mode=0 that will create the element, and then invoke apply-templates with mode=1

Hope this helps,
Alberto

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