Solving Problems by Querying XML - 9.5 Joins
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Problem
You want to relate elements in a document to other elements in the same or different document.
Solution
A join is the process of considering all pairs of element as being related (i.e., a Cartesian product) and keeping only those pairs that meet the join relationship (usually equality).
To demonstrate, I have adapted the supplier parts database found in Date’s An Introduction to Database Systems (Addison Wesley, 1986) to XML:
<database>
<suppliers>
<supplier id="S1" name="Smith" status="20" city="London"/>
<supplier id="S2" name="Jones" status="10" city="Paris"/>
<supplier id="S3" name="Blake" status="30" city="Paris"/>
<supplier id="S4" name="Clark" status="20" city="London"/>
<supplier id="S5" name="Adams" status="30" city="Athens"/>
</suppliers>
<parts>
<part id="P1" name="Nut" color="Red" weight="12" city="London"/>
<part id="P2" name="Bult" color="Green" weight="17" city="Paris"/>
<part id="P3" name="Screw" color="Blue" weight="17" city="Rome"/>
<part id="P4" name="Screw" color="Red" weight="14" city="London"/>
<part id="P5" name="Cam" color="Blue" weight="12" city="Paris"/>
<part id="P6" name="Cog" color="Red" weight="19" city="London"/>
</parts>
<inventory>
<invrec sid="S1" pid="P1" qty="300"/>
<invrec sid="S1" pid="P2" qty="200"/>
<invrec sid="S1" pid="P3" qty="400"/>
<invrec sid="S1" pid="P4" qty="200"/>
<invrec sid="S1" pid="P5" qty="100"/>
<invrec sid="S1" pid="P6" qty="100"/>
<invrec sid="S2" pid="P1" qty="300"/>
<invrec sid="S2" pid="P2" qty="400"/>
<invrec sid="S3" pid="P2" qty="200"/>
<invrec sid="S4" pid="P2" qty="200"/>
<invrec sid="S4" pid="P4" qty="300"/>
<invrec sid="S4" pid="P5" qty="400"/>
</inventory>
</database>
The join to be performed will answer the question, “Which suppliers and parts are in the same city (co-located)?”
You can use two basic techniques to approach this problem in XSLT. The first uses nestedfor-eachloops:
<xsl:template match="/">
<result>
<xsl:for-each select="database/suppliers/*">
<xsl:variable name="supplier" select="."/>
<xsl:for-each select="/database/parts/*[@city=current()/@city]">
<colocated>
<xsl:copy-of select="$supplier"/>
<xsl:copy-of select="."/>
</colocated>
</xsl:for-each>
</xsl:for-each>
</result>
</xsl:template>
The second approach usesapply-templates:
<xsl:template match="/">
<result>
<xsl:apply-templates select="database/suppliers/supplier" />
</result>
</xsl:template>
<xsl:template match="supplier">
<xsl:apply-templates select="/database/parts/part[@city = current()/@city]">
<xsl:with-param name="supplier" select="." />
</xsl:apply-templates>
</xsl:template>
<xsl:template match="part">
<xsl:param name="supplier" select="/.." />
<colocated>
<xsl:copy-of select="$supplier" />
<xsl:copy-of select="." />
</colocated>
</xsl:template>
Next: Joins With Many Members >>
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 This article is excerpted from chapter nine of the XSLT Cookbook, Second Edition, written by Sal Mangano (O'Reilly; ISBN: 0596009747). Check it out today at your favorite bookstore. Buy this book now.
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