Re: [xml] best way copy of xmlXPathObjectPtr->node
- From: Mark_Vakoc peoplesoft com
- To: xml gnome org
- Subject: Re: [xml] best way copy of xmlXPathObjectPtr->node
- Date: Tue, 20 Jan 2004 09:56:56 -0700
Something like:
xmlNodeSetPtr doSomething(xmlDocPtr doc, const xmlChar* xpath) {
xmlNodeSetPtr ret = NULL;
xmlXPathObjectPtr obj;
......
obj = xmlXPathEval(xpath,ctxt);
if (obj != NULL && obj->type == XPATH_NODESET) {
ret = obj->nodesetval;
obj->nodesetval = NULL;
}
if (obj != NULL)
xmlXPathFreeObject(obj);
......
return ret;
}
oliverst online d
e To: xml gnome org
Sent by: cc:
xml-admin gnome o Subject: [xml] best way copy of xmlXPathObjectPtr->node
rg
01/20/2004 07:40
AM
Hey!
I wanted to do a function you give a xmlDocPtr and an XPath expression,
that returns a xmlNodeSetPtr, so you don't have to create any
xmlXPathContextPtr. The problem is, that nodesetval is being free'd,
when I call xmlXPathFreeObject() and I haven't found a function in
libxml2 to copy a xmlNodeSetPtr. Is there a better way to get a copy
beside loop through all the nodes and put them in a new xmlNodeSetPtr
one-by-one?
TIA
Oliver
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