RE: recursive counter not incrementing

> David
> Ultimately i am trying to check if the items in $s1 can also 
> be found in $s2.  
> For example - 
> $s1 would consist of 2 items, $s1/item/Milk and $s1/item/Sugar
> $s2 would consist of 10 different items, Milk and Sugar are 
> among them.

In predicate logic that's

 for each ITEM1 in $s1
   there exists ITEM2 in $s2
     such that ITEM1 = ITEM2

or in XPath 2:

 every $I1 in $s1 satisfies 
    some $I2 in $s2 satisfies
       $I1 eq $I2

In XPath 1 the equivalent of "some" can be achieved with

$s2[predicate]

and the equivalent of "every" can be achieved with

not($s1[not(predicate)])

but without range variables, you can't combine the two conditions into a
single expression.

However, there's another tool in the kitbag: the "existential equals". The
"=" operator, and its friends, compare two sets, rather than just
singletons. A=B is true if some pair of items from A and B are equal

So you can write

not($s1[not(item = $s2/item)])

which I think tests the condition you are looking for.

Michael Kay
http://www.saxonica.com/

  
> I've accomplished this using the for-each loop you suggested, however
> i need a way to signal that ALL items have been found after the loop,
> and not only some of them.
> 
> I'm thinking something like:
> 
> <xsl:for-each select="$s1/item[.=$s2/item]">
>    <!--now, try to place the items that it found matches for, into a
> nodeset and compare it to the original $s1 itemset.  If the results
> from the for-each loop and the $s1 itemset contain the same items,
> output COMPLETE MATCH-->
>        
> However, i'm not sure how to accomplish getting every matched item it
> finds into a nodeset of some sort that i can use for a comparison
> later on.  Any help would be appreciated...I'm a beginner and
> frustration is setting in, but so far you guys have helped me a good
> deal. Thanks.
> 
> Chris
> 
> On Wed, 8 Dec 2004 22:41:37 GMT, David Carlisle 
> <davidc@xxxxxxxxx> wrote:
> > 
> > It's not clear why you are using recursion rather than just using
> > postion() or count to get this number, but perhaps that's 
> just because
> > you have cut the example down, I didn't follow all the logic of your
> > stylesheet but can comment on some lines:
> > 
> >                <xsl:with-param name="s1"
> >                select="$s1/item[following-sibling]"/>
> > 
> > That selects all items that have a child called 
> following-sibling so it
> > will be empty, I don't know which nodes you mean to select here but
> > perhaps
> >   select="following-sibling::*"
> > 
> >      <xsl:for-each select="$s1/item">
> >         <xsl:variable name="sub1" select="."/>
> > 
> >         <xsl:for-each select="$s2/item">
> >           <xsl:if test="$sub1 = . ">
> > 
> > 
> > This is rather strange, it loops through all pairs of items in s1
> > and s1 but only does anything if the items are equal
> > so you could just select those pairs with a single xpath:
> >   <xs:for-each select="$s2/item[.=$s1/item]">
> > 
> > which is more compact to write and at least gives the 
> system a chance of
> > being a bit more efficient.
> > 
> > It seems that you want to use an integer counter as your 
> main control
> > and repeatedly pass the whole of yor set $s1 and keep 
> accessing $s1[$i]
> > this isn't the usual XSL way, normally you would just 
> simply pass the
> > current node as a parameter, and then move control to teh following
> > sibling, without ever needing your integer position variable.
> > 
> > David
> > 
> > 
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