key() is defined to return its results in document order, with duplicates
eliminated. If you want a different order, you have to call it multiple
times, once for each key value, as you suggest.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Elliot Schlegelmilch [mailto:elliot(_at_)bozemanpass(_dot_)com]
Sent: 12 November 2005 21:10
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] key() returning nodes in undesired order.
I'm attempting to simplify and reduce the size of my xml. Instead of
repeating items many many times, I'm providing a reference.
Being a good
beginner xsl programmer, I made a key for each 'b' so i can refer to
them simply by key('all-b', /set/x/a/reference) and have a very handy
node set of b's.
<xsl:key name="all-b" match="/set/y/b" use="reference"/>
This part all works fine. However, the problem is when the references
refer to them in an order which differs than how they are
contained in
y. So "key('all-b', /set/x/a/reference)" returns the B nodes
I'm after,
but not in the order which they are in A. Is there such a way to get
them in the order which I desire?
Unfortunately, having the xml in the 'correct order' before I process
isn't an option, and neither is adding sort criteria for each
B so I can
sort the node set before I use it.
The only way I can think to do it now is individually, like:
<xsl:for-each select="/set/x/a/reference">
<xsl:variable name="x" select="key('all-b', .)"/>
...
</xsl:for-each>
<set>
<x>
<a>
<reference>reference2</reference>
<reference>reference1</reference>
</a>
</x>
<y>
<b>
<reference>reference1</reference>
<c>other data</c>
</b>
<b>
<reference>reference2</reference>
<c>other data</c>
</b>
</y>
</set>
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