XSLT for links

bmains · February 8th, 2005, 03:44 PM

Hello,

What I want do is have a series of paragraphs like this:

<Content>
  <Paragraph>This is my content<Paragraph>
  <Paragraph>This is my content <Link href="http://p2p.wrox.com">Forums</Link><Paragraph>
</Content>

I want to render the paragraphs, and the links within the paragraphs. This is what I came up with XSLT so far:

<xsl:template match="Content">
    <xsl:for-each select="Paragraph">
        <xsl:if test="position() > 1">
            <br/><br/>
        </xsl:if>

        <xsl:value-of select="."/>
    </xsl:for-each>
</xsl:template>

But I can't figure out how to render the links. Any ideas?

Brian

joefawcett · February 9th, 2005, 06:25 AM

The following could be slimmed down a bit but is easy to alter if your content becomes more complicated. I changed from a 'for-each' to an apply-templates' but that's really a question of style really. The default template for text nodes which just copies them to the output is used for the main content, anything that needs special treatment has its own template.

Code:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/">
    <html>
      <head><title>Paragraphs</title></head>
      <body>
        <xsl:apply-templates select="Content"/>    
      </body>
    </html>    
  </xsl:template>

  <xsl:template match="Content">    
    <xsl:apply-templates select="Paragraph"/>  
  </xsl:template>

  <xsl:template match="Paragraph">
    <xsl:if test="position() != 1">
      <br/><br/>
    </xsl:if>
    <xsl:apply-templates/>
  </xsl:template>

  <xsl:template match="Link">
    <a href="@href"><xsl:value-of select="text()"/></a>
  </xsl:template>
</xsl:stylesheet>

--

Joe (Microsoft MVP - XML)

bmains · February 9th, 2005, 02:07 PM

Thanks, I'll have to try that. So the <xsl:apply-templates/> in the Paragraph template knows to check for whatever template for the element it has come across, such as the paragraph content or any inner elements?

Brian

joefawcett · February 10th, 2005, 07:32 AM

It processes all children of the current node.

--

Joe (Microsoft MVP - XML)

bmains · February 10th, 2005, 08:27 AM

<xsl:template match="/"> (at the top)

This is the only problem that I have. It won't recognize the root element with this, instead, I have to have a separate element for the root (named article). How can I do that...

Brian

joefawcett · February 10th, 2005, 08:38 AM

Change it to:

Code:

<xsl:template match="/article">

--

Joe (Microsoft MVP - XML)