selecting nodes - syntax?

Hello everyone.

I have problem with syntax when I want to select nodes.
Here is example XML:

<main>
	<list>
		<a code="1" priority="10" .../>
		<a code="2" priority="7" .../>
		<a code="3" priority="8" .../>
		<a code="4" priority="3" .../>
		<a code="5" priority="5" .../>
	</list>
	<elem>
		<b>
			...
			<a code="1"/>
			<a code="3"/>
			...
		</b>
		<b>
			...
			<a code="4"/>
			<a code="5"/>
			<a code="3"/>
			...
			<!--  hundreds of elements <b> -->
		</b>
		...
	</elem>
</main>

Every element <a> have lots of attributes (XSL-FO properties). There is cca.
20 different elements <a>.
Element <b> could have several elements <a>. Because of hundreds of elements
<b> I don't want that there is copy of <a> in every <b>. I would like to
have only attribute @code in element <a> that is same as @code in <a> with
all properties.
I want to select, for every <b> element, element <a> with lowest priority.

I tried with something like this:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

	<xsl:template match="/">
		<xsl:apply-templates select="main/elem/b"/>
	</xsl:template>

	<xsl:template match="b">
		<xsl:variable name="selected">
			<xsl:for-each select="/main/a[@code = ./a/@code]"> <!-- I guess that
problem is that node . isn't <b> -->
				<xsl:sort data-type="number" select="@priority" order="ascending"/>
				<!-- debug --> debug[<xsl:value-of select="."></xsl:value-of>]
				<xsl:if test="position()=1">
					<xsl:value-of select="./@code"/>
				</xsl:if>
			</xsl:for-each>
		</xsl:variable>
		selected[<xsl:value-of select="$selected"/>]
	</xsl:template>
</xsl:stylesheet>


I don't know syntax to select element <a> with all properties.

What is correct syntax?

Any help is appreciated.
Thanks in advance,
Denis

P.S.
Sorry for my lousy english :-)



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