Thread: xsl question
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Old September 14th, 2005, 01:22 PM
bluisana bluisana is offline
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I still am not able to get exactly what I want.

I did change my xml format to the suggestion from above and I tried many variations on the xsl that both of you guys posted. The xsl that you guys posted only returned the top link after the LINKS element with the matching name attribute. I want to display all of the LINK elements.

It seemed logical to me that this would do exactly what I want.

 <xsl:template match="/">
    <html>
    <body>
    <xsl:apply-templates select="HEADER/LINKS[@name = $LINK]" />
    <xsl:for-each select="LINK">
    <a>
      <xsl:attribute name="HREF"><xsl:value-of select="TITLE"/>test
      </xsl:attribute>
      </a>
      </xsl:for-each>
    </body>
     </html>
  </xsl:template>

The xsl above sends out all of the TITLES and all of the URLS and it doesn't format them into html href links.

Waht is going on here?

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