How do you link two elements together when they have the same name in xslt? - xslt

Question

I have these trees, one with this structure /cars/car and the second /maker/cars/car. The first one has a reference to the id of the second list of cars.
<xsl:template match="t:cars/t:car">
<tr>
<td>
<xsl:if test="position()=1">
<b><xsl:value-of select="../#name"/><xsl:text> </xsl:text></b>
</xsl:if>
</td>
</tr>
I have this, it was filled in with a for loop I learn after a bit that I could't do it.
This is what it was before:
<xsl:template match="t:cars/t:car">
<tr>
<td>
<xsl:if test="position()=1">
<b><xsl:value-of select="../#name"/><xsl:text> </xsl:text></b>
</xsl:if>
<xsl:for-each select="/t:root/t:maker/t:car">
<xsl:if test="t:root/t:maker/#id = #ref">
<xsl:value-of select="#title"/>
</xsl:if>
</xsl:for-each>
</td>
</tr>
sample:
auto>
<maker type="toyota">
<car name="prius" id="1"/>
</maker>
<cars name="My Collection">
<car ref="1" />
</cars>

Answer 1

This simple transformation:
<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="kCarById" match="maker/car" use="#id"/>
<xsl:template match="/*">
<table>
<xsl:apply-templates/>
</table>
</xsl:template>
<xsl:template match="cars/car">
<tr>
<td>
<b>
<xsl:value-of select="key('kCarById', #ref)/#name"/>
</b>
</td>
</tr>
</xsl:template>
</xsl:stylesheet>
when applied on this XML document (the provided one, just extended a little):
<auto>
<maker type="toyota">
<car name="prius" id="1"/>
</maker>
<maker type="honda">
<car name="accord" id="2"/>
</maker>
<maker type="benz">
<car name="mercedes" id="3"/>
</maker>
<cars name="My Collection">
<car ref="2" />
<car ref="3" />
</cars>
</auto>
produces the wanted, correct result:
<table>
<tr>
<td>
<b>accord</b>
</td>
</tr>
<tr>
<td>
<b>mercedes</b>
</td>
</tr>
</table>
Explanation: Appropriate use of keys.