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RE: the nearest ancestor with the attribute
- To: <xsl-list at lists dot mulberrytech dot com>, "'Wendell Piez'" <wapiez at mulberrytech dot com>
- Subject: RE: [xsl] the nearest ancestor with the attribute
- From: "Michael Kay" <mhkay at iclway dot co dot uk>
- Date: Wed, 17 Jan 2001 09:19:36 -0000
- Reply-To: xsl-list at lists dot mulberrytech dot com
> You're absolutely right that the ancestor axis is a reverse axis. If
> you were using:
>
> ancestor::*[1]/@attr
>
> you'd be fine (aside from the fact that you'd be stuffed if the node's
> parent didn't have a 'attr' attribute). However, the last axis used in
> the XPath is the 'attribute' axis, which takes you back to document
> order again. To get the last one, use:
>
> ancestor::*/@attr[last()]
>
> At least I *think* that's right ;)
No, I don't think it is. This expression will return a set consisting of the
last @attr attribute of each ancestor element. An element can only have one
@attr attribute so saying you want its last one is pointless.
An axis returns a list of nodes, which is always in either forwards or
reverse document order. This affects the meaning of position() and last()
when applied to a predicate used in a Step based on that axis. The final
result of a path expression, however, is always a set of nodes, with no
intrinsic order. Many operations on a node-set access the nodes in document
order. This includes a filter expression, so:
(ancestor::*/@attr)[last()] will select the last in document order (i.e. the
innermost available attribute).
Mike Kay
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