Re: [libxml++] default namespace

[Darko M. <da...@uv...>]

pa...@at... wrote:
 > The reading I've done tells me that with XML1.1 there is no way to 
use >default namespaces in XPath searches.  Ok, so now I'd like to know 
the >work-around.  One thing I found in another post was this:

That is true.

 > If I understand the above code correctly, it is assigning a prefix to 
 >the default namespace, then using that prefix in XPath searches.  Is 
 >that correct?  I AM PERFECTLY HAPPY resolving my problem this way (if 
 >it will work).  But I want to find out if this can be done through the 
 >libxml++ wrapper - and if this over-all approach is Kosher.

Yes it is the only way. I included changes to my local version of 
libxml++ in node.cc. In node class I added namespaceprefix and 
namespaceref items. Then in node::find method I added this before the 
search is to be executed:

if (!nsprefix_.empty() && !nsref_.empty()) {
  xmlXPathRegisterNs( ctxt,
  reinterpret_cast<const xmlChar*>(nsprefix_.c_str()),
  reinterpret_cast<const xmlChar*>(nsref_.c_str()));
   };
}

Allso it was needed to add special new method for setting these two 
values in node class that look like this:

void Node::set_namespace_with_ref(const std::string& ns_prefix, const 
std::string& ns_ref) {
   xmlNs* ns = xmlSearchNs( cobj()->doc, cobj(), 
(xmlChar*)ns_prefix.c_str() );
   if(!ns) {
       nsprefix_ = ns_prefix;
       nsref_ = ns_ref;
   } else {
       nsprefix_.clear();
       nsref_.clear();
   }
}

With this everything works just fine.

Darko