RE: [xsl] Sorting a NodeSet Contained Within a Variable

[Bryan Tulloch <b dot tulloch at solicitec dot com>]

Here is the code that builds my variable:

<xsl:variable name="casenumber.tf">
<xsl:for-each
select="//User[not(./User_Name=preceding::User_Name)]/User_Name">
<xsl:variable name="this" select="." />
<xsl:value-of
select="count(//User[./User_Name=$this][not(./Case_Code=preceding::Case
_Code)]/Case_Code)" />
</xsl:for-each>
</xsl:variable>


<xsl:variable name="casenumber" select="msxml:node-set($casenumber.tf)"  />

When I prompt for the value of the variable, I get the output I expect - a
series of figures. But when I try to use the variable as a sort parameter,
nothing happens - or, rather, I simply find my initial template being
invoked - it's as if I was pressing the Refresh button.

HELP!!

Bryan

-----Original Message-----
From: Oliver Becker [mailto:obecker@informatik.hu-berlin.de]
Sent: Wednesday, July 11, 2001 6:24 PM
To: xsl-list@lists.mulberrytech.com
Subject: RE: [xsl] Sorting a NodeSet Contained Within a Variable


Hi Bryan,

> Thanks for your guidance, but this isn't working 100% for me.
> I can sort the node-set if it is comprised purely of nodes from my xml
> source tree, but not if it consists entirely of the results of a series of
> calculations (using count()) done on nodes from the source tree. In other
> words, I want to put those results into a nodeset and sort them.

Hmm, then you should probably show us how your variable is built.
In (pure) XSLT 1.0 you can't create a node-set as a result of a
transformation.
Fortunately many vendors invented a node-set() extension function.

So I assume the problem is that your variable actually doesn't contain
a node-set, but rather a value of a different type.

Cheers,
Oliver


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