At 2009-10-16 13:04 -0700, M C wrote:
How do I transform this input document:
<div>
<p>string 1<lb/>string 2</p>
</div>
to this output document?
<div>
<p>string 1</p>
<p>string 2</p>
</div>
This is a very limited specification, so I tried to be as general as
possible in the example below. You use only text, but what if you
have other formatting? I've tried to accommodate that.
And I don't see the issue as an empty element issue, but as a content
grouping issue.
The principle I've used below is to create groups starting with each
line-break element, and put out each group using the parent element.
I hope this helps.
. . . . . . . . . . Ken
p.s. registration is still open for public XSLT/XQuery hands-on class
November 2-6, 2009
T:\ftemp>type morgan.xml
<div>
<p>string 1<lb/>string 2</p>
</div>
T:\ftemp>xslt2 morgan.xml morgan.xsl
<?xml version="1.0" encoding="UTF-8"?><div>
<p>string 1</p><p>string 2</p>
</div>
T:\ftemp>type morgan.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
<xsl:template match="*[lb]">
<!--remember where we are-->
<xsl:variable name="this" select="."/>
<!--determine all of the groups of nodes-->
<xsl:for-each-group select="node()" group-starting-with="lb">
<!--return to the node being copied-->
<xsl:for-each select="$this">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<!--but only copy that content not including the break-->
<xsl:apply-templates select="current-group()[not(self::lb)]"/>
</xsl:copy>
</xsl:for-each>
</xsl:for-each-group>
</xsl:template>
<xsl:template match="@*|node()"><!--identity for all other nodes-->
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
T:\ftemp>
--
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