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Antw: Merging two xml files

Subject: Antw: Merging two xml files
From: agnes.kielen@xxxxxxx
Date: Tue, 13 Jan 2004 10:15:18 +0100
xsl two xml files
Hi Arul,

<?xml version="1.0"?>
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">	
  <xsl:output method="xml" indent="yes"/>	
  <!-- load the merge file -->	
  <xsl:variable name="emps" select="document('employeeAddresses.xml')"/>
  <xsl:template match="/">		
     <employees>			
	<xsl:for-each select="employees/employee">			<!-- cache the employee's ID -->				<xsl:variable name="id" select="@id"/>				<!-- copy the pincode of the second XML -->			<employee pincode="{$emps//employee[@id = $id]/@pincode}">
        <!-- copy the child nodes -->
        <xsl:copy-of select="*"/>	
	<!-- copy the address -->
        <xsl:copy-of select="$emps//employee[@id = $id]/address"/>
    </employee>				
   </xsl:for-each>			
</employees>	
</xsl:template>
</xsl:stylesheet> 

I changed the 
<xsl:variable name="id"><xsl:value-of select="@id"/></xsl:variable>
in <xsl:variable name="id" select="@id"/>
This is shorter to write, easier to read and you don't create a RTF.
Furhtermore I changed child::* in *.

I select the correct nodes from employeeAddress.xml with the help of the following pattern: $emps//employee[@id = $id].
In english this means something like select those employee from the second file from which the id-attribute is equal to the current id in the main xml.


Cheers,
Agnes







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


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