XSLT variable in predicate not working - variables

Question

my xml looks something like this:
<units>
<unit>
<unitnum></unitnum>
<Year></Year>
<Qty></Qty>
</unit>
</units>
I create a key to capture all the possible years. Like this:
<xsl:key name="yr" match="//Year/text()" use="." />
Then I for-each loop through the keys to get unique years. My ultimate goal is to sum the quantites like this:
<xsl:for-each select="//Year/text()[generate-id()=generate-id(key('yr',.)[1])]">
<li>
<xsl:variable name="varjobyear" select="."/>
<xsl:value-of select="$varjobyear"/> - sum -
<xsl:value-of select="sum(//unit[Year=$varjobyear]/Qty)"/>
</li>
</xsl:for-each>
For some reason my output looks like this:
2010 - sum - 0
2011 - sum - 0
2012 - sum - 0
But what I want is for those 0's to be the actual sum of the quantities for each year.
My best guess is that there is something wrong with the predicate [Year=$varjobyear], but I can't figure out what it is. I have also tried [Year=string($varjobyear)] and also [Year=number($varjobyear)].
What am I doing wrong? Thanks!

Answer 1

You want something like this:
<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="kYearByVal" match="Year" use="."/>
<xsl:template match="Year[generate-id()=generate-id(key('kYearByVal',.)[1])]">
<li>
<xsl:value-of select="concat(., ' - sum - ', sum(key('kYearByVal',.)/../Qty))"/>
</li>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>
When this transformation is applied on the following XML document:
<units>
<unit>
<unitnum>12</unitnum>
<Year>1922</Year>
<Qty>3</Qty>
</unit>
<unit>
<unitnum>13</unitnum>
<Year>1922</Year>
<Qty>8</Qty>
</unit>
<unit>
<unitnum>14</unitnum>
<Year>1999</Year>
<Qty>5</Qty>
</unit>
<unit>
<unitnum>15</unitnum>
<Year>1999</Year>
<Qty>7</Qty>
</unit>
<unit>
<unitnum>16</unitnum>
<Year>2003</Year>
<Qty>3</Qty>
</unit>
</units>
the wanted, correct result is produced:
<li>1922 - sum - 11</li>
<li>1999 - sum - 12</li>
<li>2003 - sum - 3</li>