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Phil ParkerSubject: Replacing carriage returns and line breaks in XML with elements
Author: Phil Parker
Date: 24 Mar 2005 05:56 PM
From an authoring system, users are creating line breaks and carriage returns
within content that is then enclosed within in a single element upon export to
and XML file.

What is the best approach using XSLT, going from XML to XML, once the
line breaks and carriage returns are in the XML file to convert those
to elements.

For example:
<root>
<para>Here is beginning content
line break
Here is ending content
</para>
</root>

change to

<root>
<para>Here is beginning content</para>
<para>Here is ending content</para>
</root>

Phil
Posttop
(Deleted User) Subject: Replacing carriage returns and line breaks in XML with elements
Author: (Deleted User)
Date: 24 Mar 2005 07:22 PM
Try the the following piece of xslt. Please note it only does the <para> transform. I assume you know how to do other element.

<?xml version='1.0' ?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="para">
<xsl:call-template name="split">
<xsl:with-param name="str" select="text()"/>
</xsl:call-template>
</xsl:template>

<xsl:template name="split">
<xsl:param name="str"/>
<xsl:if test="string-length($str) != 0">
<para><xsl:value-of select="substring-before($str, '&#xA;')"/>
</para>
<xsl:call-template name="split">
<xsl:with-param name="str" select="substring-after($str, '&#xA;')"/>
</xsl:call-template>
</xsl:if>
</xsl:template>

</xsl:stylesheet>

song

   
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