Nima ,
Remember also to free the returned string using xmlFree() as it is
allocated on heap via xmlNodeListGetString()
xmlChar* pszNodeContent = NULL;
pszNodeContent =
xmlNodeListGetString(m_ptrXMLDoc,ptrCurrXMLNode->children,1);
if (pszNodeContent)
{
printf("Value of <pic_file> = %s", (const
char*)pszNodeContent);
xmlFree(pszNodeContent);
pszNodeContent = NULL;
}
-- Lav
________________________________
From: Lav Mehrotra
Sent: Tuesday, June 10, 2008 11:56 AM
To: 'Nima Tiran'
Cc: [email protected]
Subject: RE: [xml] Simple parsing XML question...
Nima ,
Plz use the following API()
xmlNodeListGetString(xmlDocPtr doc , xmlNodePtr list , int inline )
Get the (xmlDocPtr doc) from xmlParseFile(const char* filename) API or
xmlParseMemory(const char* buffer , int size)
Get the Current (xmlNodePtr list) for ur node <pic_file> by traversing
the DOM Document Tree , until u encounter a node with xmlNodePtr->name
as same as pic_file ( Take the help of xmlStrcmp() API ) . Say this
node is xmlNodePtr pXmlNodePicFile
Once u have this pXmlNodePicFile u cannot blindly use
pXmlNodePicFile->content as the TEXT Content MyPicFileName.pdf is also a
child node of the pXmlNodePicFile
So pass the 1st child of pXmlNodePicFile to xmlNodeListGetString() API
like this
xmlNodeListGetString(doc, pXmlNodePicFile->children , 1)
Regds,
Lav
________________________________
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of
Nima Tiran
Sent: Tuesday, June 10, 2008 3:11 AM
To: [email protected]
Subject: [xml] Simple parsing XML question...
Hi all,
I'm newbie here and got a simple question:
I'm parsing an XML file and having a node like:
<pic_file>MyPicFileName.pdf</pic_file>
in XML file, just wondering how can I retrieve "MyPicFileName" from
this node?
Thnx
Nima
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