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Editorial Team
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Asked: 2026-06-05T20:55:57+00:00

Here is my XML: <section> <para>part1 <breakline/> part2 <list> <para>list1 </para> <para>list2 </para> </list>

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Here is my XML:

<section>
  <para>part1 <breakline/> part2
    <list>
      <para>list1 </para>
      <para>list2 </para>
    </list> 
  </para>
<para>before_line_break <breakline/> after</para>
<para>para3</para>
</section>

Expected output:

part1 
part2
before_line_break 
after
para3

I want “part2” but not “list1” and “list2”

An equivalent HTML file would be:

<html>
  <body>
    part1 <br/>
    part2 <br/>
    before_line_break <br/>
    after <br/>
    para3 <br/>
  </body>
</html>

I have tried many things in my xsl file. Here is just an example (that doesn’t work of course):

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
  <body>
    <xsl:for-each select="section/para">
      <xsl:value-of select="."/>
      <br/>
    </xsl:for-each>
  </body>
</html>
</xsl:template>
</xsl:stylesheet>

How should I change my XSL file to get the expected output?
Thanks in advance for your help.

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1 Answer

  1. Editorial Team

    Take this just as starting point:

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

    <xsl:template match="/">
        <html>
            <body>
                <xsl:apply-templates select="section/para"/>
            </body>
        </html>
    </xsl:template>

    <xsl:template match="para">
        <xsl:apply-templates select="text()|breakline"/>
    </xsl:template>

    <xsl:template match="text()">
        <xsl:value-of select="."/>
        <!-- print br if text() is at end of para -->
        <xsl:if test="position()=last()">
            <br/>
        </xsl:if>
    </xsl:template>

    <!-- replace breakline with br -->
    <xsl:template match="breakline">
        <br/>
    </xsl:template>

    <!-- mute list -->
    <xsl:template match="list/para"/>

</xsl:stylesheet>
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