Question

I have two source files which I need to combine on a row by row basis. I am happy reading the files into a variable and I am happy with the logic but the syntax has me stumped. For each row in file 1 I need to loop round each row in file 2 and output the two variables concatenated together:

File 1:

<rows>
    <row>1</row>
    <row>2</row>
    <row>3</row>
    <row>4</row>
</rows>

File 2:

<rows>
    <row>a</row>
    <row>b</row>
</rows>

Required output:

<rows>
    <row>1/a</row>
    <row>1/b</row>
    <row>2/a</row>
    <row>2/b</row>
    <row>3/a</row>
    <row>3/b</row>
    <row>4/a</row>
    <row>4/b</row>
<rows>

My (poor) attempt at getting the XSLT to work:

<rows>
    <xsl:apply-templates select="document('file1.xml')/rows/row" />
</rows>

<xsl:template match="row">
    <xsl:apply-templates select="document('file2.xml')/rows/row" />
</xsl:template>  

<xsl:template match="row">
    <row><xsl:value-of select="???" />/<xsl:value-of select="???" /></row>
</xsl:template>

(These files are simplified versions of what I actually have)

How do I make one template match one ‘row’ value and the other match another (both source files use the same structure). And how do I set those ‘???’ values?

Answer 1

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:variable name="vDoc2">
    <rows>
        <row>a</row>
        <row>b</row>
    </rows>
 </xsl:variable>

 <xsl:template match="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="/*">
  <rows>
    <xsl:apply-templates/>
  </rows>
 </xsl:template>

 <xsl:template match="row">
   <xsl:apply-templates select="$vDoc2/*/row" mode="doc2">
     <xsl:with-param name="pValue" select="."/>
   </xsl:apply-templates>
 </xsl:template>

 <xsl:template match="row" mode="doc2">
   <xsl:param name="pValue" />

   <row><xsl:sequence select="concat($pValue, '/', .)"/></row>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on the provided first XML document:

<rows>
    <row>1</row>
    <row>2</row>
    <row>3</row>
    <row>4</row>
</rows>

the wanted, correct result is produced:

<rows>
   <row>1/a</row>
   <row>1/b</row>
   <row>2/a</row>
   <row>2/b</row>
   <row>3/a</row>
   <row>3/b</row>
   <row>4/a</row>
   <row>4/b</row>
</rows>