Re: [xsl] Hello and quick question

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

What's worse, the expression:

/ROOT/Client/Contacts/Contact[isReferrer='0'][1]/ContactFirstName

will give results that are counterintuitive, since the [1] predicate 
operates not on the expression as a whole, but only on that particular step 
of that expression,

"Contact[isReferrer='0'][1]"

which (as all XSLT users certainly know!) is short for

child::Contact[isReferrer='0'][position()=1]

which means you'll get, not the first such Contact node in the document, 
but every Contact node (that meets the isReferrer criterion) *that is the 
first such child of its parent*.

To get around this, you can say

(/ROOT/Client/Contacts/Contact[isReferrer='0'])[1]/ContactFirstName

(Grouping the entire expression indicates the predicate [1] applies to the 
expression as a whole.)

But I actually rather like Corey's solution

string(/ROOT/Client/Contacts/Contact[isReferrer='0']/ContactFirstName)

...although were I to use it in a stylesheet, I'd certainly explain it in a 
comment -- as something non-obvious that might be expected to break if, 
say, the rules of how node-sets are converted into strings should change....

Cheers,
Wendell

At 11:36 PM 1/6/2003, Joerg's wrote:
> Maybe you can better use some possible optimizations of the proceesor by 
> adding [1] into the expression:
>
> /ROOT/Client/Contacts/Contact[isReferrer='0'][1]/ContactFirstName
>
> Furthermore you sometimes need a node set (even it's a single node) and 
> not a string, e.g. <xsl:apply-templates 
> select="/ROOT/Client/Contacts/Contact[isReferrer='0'][1]/ContactFirstName"/>, 
> that won't work with string().
>
> Simon has a little error in his expression. He selects the first Contact 
> element, independent of [isReferrer='0'] or not. If this first Contact 
> element does not have a child 'isReferrer' with text value '0', his node 
> set would be empty.


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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