RE: constructing the Node Sets

hi -

I have tried this out in my xsl, but it doesn't seem to be working.

Here is a sample A.xml and B.xsl.  (count of a: 4, count of b: 3, count of
c: 0) is the output of the html.

I expected that the intersection of a and b will produce a count of 2, but
it is coming out as 0.  (Because the obid="A" and obid="B" are common to
both a and b).

Can you pl. tell us where i am doing things wrong?

thanks
- siva jasthi



A.xml
-----

<?xml version="1.0" encoding="ISO-8859-1"?>

<?xml-stylesheet type="text/xsl" href="B.xsl"?>

<Root>
  <Class name="Asm" obid="A"/>
  <Class name="Asm" obid="B"/>
  <Class name="Asm" obid="C"/>
  <Class name="Asm" obid="D"/>
  <Class name="Cmp" obid="A"/>
  <Class name="Cmp" obid="B"/>
  <Class name="Cmp" obid="X"/>
</Root>


B.xsl
-----
<?xml version="1.0" encoding="ISO-8859-1"?>

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

<xsl:template match="/">

  <xsl:variable name="a" select="/Root/Class[@name='Asm']/@obid"/>
  <xsl:variable name="b" select="/Root/Class[@name='Cmp']/@obid"/>

  <xsl:variable name="c" select="$a[ count( . | $b ) = count( $b ) ]"/>

count of a: <xsl:value-of select="count($a)" />
count of b: <xsl:value-of select="count($b)" />
count of c: <xsl:value-of select="count($c)" />


</xsl:template>

</xsl:stylesheet>





-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Roger Glover
Sent: Tuesday, January 28, 2003 12:39 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  constructing the Node Sets


Idea copied from Michael Kay's "XSLT Programmer's Reference", 2nd ed., Wrox
Press, 2001, p. 425 ...

<xsl:variable name="C" select="$A[ count( . | $B ) = count( $B ) ]"/>

The select expression selects the node subset C of node set A, composed of
any member node of A which, when added to node set B, produces a node set
the same size as B (in other words, the node was already in node set B).

The usage of A and B in the select expression can be reversed with no side
effects except possibly:
    o   a change in the list ordering of nodes in node set C
    o   a change in the speed of execution of the expression

-- Roger Glover

Siva Jasthi wrote
> I have constructed two variables (A and B) each of which contains a Node
> Set (through select="XPath Expression").
>
> A = [Node1, Node2, Node3, Node4, Node5]
> B = [Node3, Node4, Node5, Node6, Node7, Node8]
>
> I now would like to construct another variable C with the Nodes that exist
> both in A and B.  (So as to produce a third table with the Nodes that are
> common to both A and B).
>
> C= [Node3, Node4, Node5]
>
> Is there any way to construct C from A and B?.



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