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!