Re: [xsl] relative . not working in complex xpath

["Philipp Kursawe phil.kursawe@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Ah I see. Unfortunately MS still offers no XPath2 or XQuery in its msxml
system components in Win8.1. Seems to me they have abandoned the format
long ago.


On Mon, Jul 21, 2014 at 1:01 PM, mike@xxxxxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> "." within a predicate refers to the node that the predicate is being
> applied to, so the context item for ./xs:extension is the same as the
> context node for @name.
>
> If you're using XSLT you can often get arround this using the current()
> function. In pure XPath 1.0, join queries like this are difficult and not
> always possible. Switch to XPath 2.0 or XQuery if possible.
>
>
>
> Michael Kay
>
> Saxonica
>
>
>
> > I have selected an element from an XSD file and want to find its base
> > class
> > to accumulate all xs:attributes.
> > So I first select the node in question with
> > //xs:complexType[@name='SomeType']
> >
> > Then I want to go on from there and collect all xs:attribute
> > node.selectNodesNS(".//xs:attribute |
> > //xs:complexType[@name=./xs:extension/@base]/@name]//xs:attribute", xsns)
> >
> > Unfortunately that only gives me the attributes of the "node" I perform
> > the
> > select from.
> > I get the correct results if I explicitly use the node in my query,
> > instead
> > of the relative expression of "." in "@name=./xs:extension/@base" like
> > this:
> >
> > node.selectNodesNS(".//xs:attribute | //xs:complexType[@name =
> >
> //xs:element[@type=//xs:complexType[@name='SomeType']//xs:extension/@base]/@name]//xs:attribute",
> > xsns)
> >
> > What is going on here?
> > Using XPath 1.0 with MSXML.
> >
> > Thanks!
> >
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