RE: [xsl] constructing the Node Sets

["Roger Glover" <glover_roger@xxxxxxxxx>]

David Carlisle wrote:

> > I expected that the intersection of a and b will produce a count of 2,
>
> ah. In that case you asked the wrong question.
>
> Your node sets are disjoint, no node has both  @name='Asm' and @name='Cmp'
> so the intersection of the sets $a and $b is empty.
>
> You don't want the intesection of the sets, you want all elements of $a
> that have a string value equal to some member of $b, that's easier than
> the question you asked,
>
> $a[.=$b]

Err...  I think that you mean:
    $a[@obid = $b/@obid]

Right?  The standard for "equality" is that the "obid" attributes have the
same value.  Your expression will result in "$a" all over again, since all
the elements in "$a" and "$b" have empty string values.


> except that if you have the possibility of duplicate obid values you may
> need to remove duplicates as well.
>
> Note that
> $a[ count( . | $b ) = count( $b ) ]
> is the same as
> $b[ count( . | $a ) = count( $a ) ]
> it really is the nodes in both $a and $b,

Well, except possibly for order, as I noted earlier.


> but
>
> $a[.=$b]

    $a[@obid = $b/@obid]


> is not the same
> as
>
>
> $b[.=$a]

    $b[@obid = $a/@obid]


> one is the set of nodes in $a the other is the set in $b.

Right, in the first expression all the "Class" elements will have a "name"
attribute with the value "Asm".  In the second expression all the "Class"
elements will have a "name" attribute with the value "Cmp".


-- Roger Glover
   glover_roger@xxxxxxxxx



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