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

Subject: Re: [xsl] relative . not working in complex xpath
From: "mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Mon, 21 Jul 2014 11:00:52 -0000
"." 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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