RE: [xsl] Extracting link information from xml document and displaying it in browser as hyperlink

[bmcleere@xxxxxxxxxx]

Jarno,

- thank you very much. The following line of code worked:

<a href="{.}">
<xsl:value-of select="."/>
</a>
Much appreciated.

Brenda
xsl-list@xxxxxxxxxxxxxxxxxxxxxx wrote:

<  
<  Hi,
<  
<  >  <location> http://someserver.ie</location>. 
<  >  
<  >  I want to retrieve the URL information and display it as a 
<  >  hyperlink in the browser.
<  >  
<  >  Currently, I can view the URL information but it is not a hyperlink. 
<  >  
<  >  I thought of using the following:-
<  >  <xsl:template match="technical/location">
<  >  <a href="http://{@link}";><xsl:value-of select="."/></a>
<  >  </xsl:template>
<  >  
<  >  I will be searching different documents so hardwiring in the 
<  >  URL where it says "@link" would not provide a solution for 
<  >  me. I think that if I could declare a variable in xsl to 
<  >  store the retrieved data and then get at the contents of this 
<  >  variable within the <a href...> tag - this would provide a solution.
<  
<    <a href="{.}">
<      <xsl:value-of select="."/>
<    </a>
<  
<  if the source is as you showed above. If the location of the URL varies, try something like
<  
<    <a>
<      <xsl:attribute name="href">
<        <xsl:choose>
<          <xsl:when test="...">
<            ...
<          </xsl:when>
<          ...
<        </xsl:choose>
<      </xsl:attribute>
<      <xsl:value-of select="."/>
<    </a>
<  
<  Cheers,
<  
<  Jarno - Front Line Assembly: Virus (Counterized Mix)
<  
<   XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
<  
<  



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list