Re: [xsl] Display link text as a hyperlink

Subject: Re: [xsl] Display link text as a hyperlink
From: "Andrew Welch" <andrew.j.welch@xxxxxxxxx>
Date: Wed, 6 Aug 2008 16:04:28 +0100
2008/8/6 Brent Solly <ultra@xxxxxxxxx>:
> I have an xml file that contains this character sequence: 'www.gamefaqs.com' .
>
> Primary Problem:
> After I convert the 'www' text to hyperlink format, the xsl displays it as plain text, but I would like to display it has a hyperlink.
>
> Secondary:
> I am aware that the url may also contain subfolders like: www.gamefaqs.com/console/n64   OR different a suffix like www.gamefaqs.ca., but right now I'll focus on .com, but feel free to make suggestions :) .
>
>
> <?xml version="1.0" encoding="UTF-8"?>
> <rss>
>   <channel>
>      <item>
>         <description>More information on this game can be found at (www.gamefaqs.com)</description>
>      </item>
>   </channel>
> </rss>
>
> <?xml version="1.0" encoding="UTF-8"?>
>
> <xsl:stylesheet
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> version="2.0">
> <xsl:strip-space elements="*" />
> <xsl:output method="html" encoding="UTF-8"/>
>
>
> <xsl:template name="globalReplace">
>  <xsl:param name="outputString"/>
>  <xsl:param name="target"/>
>  <xsl:param name="replacement"/>
>    <xsl:choose>
>   <!--Begin Test -->
>   <xsl:when test="contains($outputString,$target)">
>
>      <xsl:value-of select=
>        "concat(substring-before($outputString,$target),
>               $replacement)"/>
>      <xsl:call-template name="globalReplace">
>        <xsl:with-param name="outputString"
>             select="substring-after($outputString,$target)"/>
>        <xsl:with-param name="target" select="$target"/>
>        <xsl:with-param name="replacement"
>             select="$replacement"/>
>      </xsl:call-template>
>    </xsl:when>
>   <!--End Test-->
>
>    <xsl:otherwise>
>      <xsl:value-of select="$outputString"/>
>    </xsl:otherwise>
>
>  </xsl:choose>
> </xsl:template>
> <xsl:template match="rss">
>  <xsl:call-template name="globalReplace">
>  <xsl:with-param name="outputString" select="."/>
>  <xsl:with-param name="target" select="'www'"/>
>  <xsl:with-param name="replacement" select="concat('&lt;a href=&#34;http://www',substring-before(substring-after(.,'www'),substring-after(.,'com')),'&#34;&gt;',substring-before(substring-after(.,'www.'),'.com'),'&lt;/a&gt;')"/>
>
>  </xsl:call-template>
> </xsl:template>
>
>
> </xsl:stylesheet>


whoooaaaaah there.... you've set off down the wrong road (like so many
others) and it will only end in tears and resentment.  It's completely
the wrong apporach.

To do a search and replace on a string use either replace() or
xsl:analyze-string:

<xsl:analyze-string select="." regex="$the-regex">
  <xsl:matching-substring>
    <a href="http://{.}";>http://<xsl:value-of select="."/></a>
  </xsl:matching-substring>
  <xsl:non-matching-substring>
    <xsl:value-of select="."/>
  </
</


-- 
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/

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