Re: [xsl] XPathAPI: How to associate namespaces

Subject: Re: [xsl] XPathAPI: How to associate namespaces
From: me@xxxxxxxxxxxx
Date: Tue, 06 May 2003 09:17:12 -0700 (PDT)
> Dear list,
> I found the recent posting attached, but the thread
> ended there. This is
> about Xalan-J 2.x, so I am not too sure if it belongs
> here, but I'll try.
> Q: I have a document with the root element being in a
> default namespace:
> <?xml version="1.0" encoding="UTF-8"?>
> <Report xmlns="";;>
>     <Type>memo</Type>
>     <Type>rep</Type>
>     <Link name="xyz">123</Link>
> </Report>
> I would like to extract just the Link from that
> document trying to use 
> XPathAPI.selectNodeList(myDoc,"/report/link")

I don't know anything about XPathAPI, but by just
looking at the two items you posted you are using the
wrong case. Meaning "/report/link" is not the same as
"/Report/Link" - just as Torsten is not the same as
torsten (perhaps you have been spoiled :)

> will return an empty result set.
> Using
> XPathAPI.selectNodeList(myDoc,"/abc:report/abc:link")
> will result in a transformer exception telling me that
> the namespace abc
> needs to be assiciated somewhere.
> Where can I do that? Inside the XPath expression? I
> read the spec; it look
> to me as if XPointer has a syntax for that while XPath
> hasn't. But I could
> not
> find any method to call in order to do this either.

google says:


public static NodeList selectNodeList(
 Node contextNode, java.lang.String str,               
                      Node namespaceNode) throws

Use an XPath string to select a nodelist. XPath
namespace prefixes are resolved from the namespaceNode.

contextNode - The node to start searching from.str - A
valid XPath string.namespaceNode - The node from which
prefixes in the XPath will be resolved to
A NodeIterator, should never be null.Throws:
TransformerException -  

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