Grouping when using 2 different XML files as sources? - xslt

Question

what i want to do is a typical grouping that can usualy be done using xsl:key, but it become more complicated as data to groups are in 2 differents files. How process ?
Here is an example of what i want to do, can i request your help ? must be xslt-1.0 compliant.
bookreference.xml :
<t>
<book isbn="1">
<category>SF</category>
</book>
<book isbn="2">
<category>SF</category>
</book>
<book isbn="3">
<category>SF</category>
</book>
<book isbn="4">
<category>Comedy</category>
</book>
<book isbn="5">
<category>Comedy</category>
</book>
</t>
mylibrary.xml :
<t>
<book isbn="1">
<price>10</price>
</book>
<book isbn="2">
<price>10</price>
</book>
<book isbn="3">
<price>20</price>
</book>
<book isbn="4">
<price>5</price>
</book>
</t>
output wanted:
SF : 3 book(s) - Total : 40$
Comedy : 2 book(s) - Total : 5$

Answer 1

As already suggested in my comment, you can first merge two documents into a result tree fragment, then use exsl:node-set to get a node-set on which you can then apply Muenchian grouping:
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:exsl="http://exslt.org/common"
exclude-result-prefixes="exsl"
version="1.0">
<xsl:param name="price-url" select="'test2011113002.xml'"/>
<xsl:variable name="doc2" select="document($price-url)"/>
<xsl:output method="text"/>
<xsl:variable name="rtf">
<xsl:apply-templates select="//book" mode="merge"/>
</xsl:variable>
<xsl:template match="book" mode="merge">
<xsl:copy>
<xsl:variable name="isbn" select="#isbn"/>
<xsl:copy-of select="#* | node()"/>
<xsl:for-each select="$doc2">
<xsl:copy-of select="key('k1', $isbn)/price"/>
</xsl:for-each>
</xsl:copy>
</xsl:template>
<xsl:key name="k1" match="book" use="#isbn"/>
<xsl:key name="k2" match="book" use="category"/>
<xsl:template match="/">
<xsl:apply-templates select="exsl:node-set($rtf)/book[generate-id() = generate-id(key('k2', category)[1])]"/>
</xsl:template>
<xsl:template match="book">
<xsl:variable name="current-group" select="key('k2', category)"/>
<xsl:value-of select="concat($current-group/category, ': ', count($current-group), ' - Total : ', sum($current-group/price), '
')"/>
</xsl:template>
</xsl:stylesheet>

Answer 2

Good question, +1.
There is no need for complicated (more than necessary) grouping or for any extension functions:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="kBookByCat" match="book"
use="category"/>
<xsl:key name="kPriceByIsbn" match="price"
use="../#isbn"/>
<xsl:variable name="vMyLib" select=
"document('file:///c:/temp/delete/mylibrary.xml')"/>
<xsl:template match=
"book[generate-id()
=
generate-id(key('kBookByCat', category)[1])
]
">
<xsl:variable name="vBooksinCat" select=
"key('kBookByCat', category)"/>
<xsl:value-of select="category"/> : <xsl:text/>
<xsl:value-of select="count($vBooksinCat)"/>
<xsl:text> book(s) - Total : $</xsl:text>
<xsl:for-each select="$vMyLib">
<xsl:value-of select="sum(key('kPriceByIsbn', $vBooksinCat/#isbn))"/>
<xsl:text>
</xsl:text>
</xsl:for-each>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>
When this transformation is applied on the first of the two provided data fragments (corrected to well-formed XML document):
<t>
<book isbn="1">
<category>SF</category>
</book>
<book isbn="2">
<category>SF</category>
</book>
<book isbn="3">
<category>SF</category>
</book>
<book isbn="4">
<category>Comedy</category>
</book>
<book isbn="5">
<category>Comedy</category>
</book>
</t>
and having the second data fragment (corrected to well-formed XML document) saved as C:\Temp\Delete\mylibrary.xml,
The wanted, correct result is produced:
SF : 3 book(s) - Total : $40
Comedy : 2 book(s) - Total : $5