Subject: RE: Question: Should namespace::* return anything? From: Khun Yee Fung <kyeefung@xxxxxxxxxx> Date: Fri, 17 Dec 1999 16:25:24 -0500 |
I modified the code for the tool to give this behaviour. Now my problem is this: I ran the expression: //namespace::*. What should the result be? >From the spec: Elements never share namespace nodes: if one element node is not the same node as another element node, then none of the namespace nodes of the one element node will be the same node as the namespace nodes of another element node. For instance, if the XML document is as follow: <?xml version='1.0?> <top> <second/> </top> and the expression is "//namespace::*". Are there two nodes (albeit looking identical) in the final nodeset or there is one node in the final nodeset, or there is no node in the final nodeset? Of course, the final option is, the expression is not valid. Thanks. Khun Yee -----Original Message----- From: Ray Waldin [mailto:rwaldin@xxxxxxxxxxx] Sent: Friday, December 17, 1999 2:54 PM To: xsl-list@xxxxxxxxxxxxxxxx Subject: Re: Question: Should namespace::* return anything? Khun Yee Fung wrote: > How about namespace::*? Is it the same as namespace::node() or the result is > always an empty nodeset? My understanding is that namespace::* returns an empty node-set for all context nodes except when the context node is an element. For element nodes it should always return a node-set with at least one namespace node, the node representing the "built-in" namespace (prefix="xml", uri="http://www.w3.org/XML/1998/namespace"), plus a node for each namespace that's been declared in the element itself or any of it's ancestors. -Ray XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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