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Re: How would we get xml tags in xsl if we have same element name

From: Martin Honnen <mahotrash@-----.-->
To: NULL
Date: 7/12/2007 6:23:00 PM

kamkaro wrote:

>  output:
>  Title: Redundant Group
> 		 Identifier = Q9D6L8
> 		 db = dbj
> Title: Database Cross-references
> 		Identifier = Z98T23
> 		db = GENE

Here is an XSLT 1.0 stylesheet that transforms each field element into a 
HTML table:

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

   <xsl:output method="html" indent="yes"/>

   <xsl:template match="/">
     <html lang="en">
       <head>
         <title>Example</title>
       </head>
       <body>
         <xsl:apply-templates/>
       </body>
     </html>
   </xsl:template>

   <xsl:template match="field">
     <table>
       <tbody>
         <tr>
           <th>Title:</th>
           <td><xsl:value-of select="title"/></td>
         </tr>
         <tr>
           <th>Identifier = </th>
           <td><xsl:value-of 
select="content/table/row/cell/idebtifier"/></td>
         </tr>
         <tr>
           <th>db = </th>
           <td><xsl:value-of select="content/table/row/cell/db"/></td>
         </tr>
       </tbody>
     </table>
   </xsl:template>

</xsl:stylesheet>

So what you are looking for is just a template matching field elements 
and then in that template you can construct whatever output you need for 
those elements and their descendant elements.



-- 

	Martin Honnen --- MVP XML
	http://JavaScript.FAQTs.com/


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