Hi, Brian,

    Please try this.

    DOM_NodeList nodeList = doc.getElementsByTagName("HostName");   // Not
"Server"
    for( unsigned int i=0; i<nodeList.getLength(); i++)
    {
         DOM_Node child=nodeList.item(i);
         if (child.getNodeType() == DOM_Node::ELEMENT_NODE)
             cout<<"\nValues are" <<child.getFirstChild().getNodeValue
().transcode()<<endl;
                                                                   // get
the textNode of "HostName"
    }//for


Regards,

Peiyong Zhang
____________________________________________
XML Parsers Development
IBM Toronto Laboratory email: [EMAIL PROTECTED]
Phone: (416)448-4088; Fax: (416)448-4414; T/L: 778-4088



"Jones, Brian O." <[EMAIL PROTECTED]> on 03/27/2001 11:05:21
AM

Please respond to [EMAIL PROTECTED]

To:   "'[EMAIL PROTECTED]'" <[EMAIL PROTECTED]>
cc:
Subject:  RE: Parser problem


Peiyongz,

When I tried to get the child node of the values in the node list I
received
an error.

That should be the parent node, which is the nodelist?? (i.e.
parent.getFirstChild())

> -----Original Message-----
> From:         [EMAIL PROTECTED] [SMTP:[EMAIL PROTECTED]]
> Sent:         Tuesday, March 27, 2001 10:57 AM
> To:           [EMAIL PROTECTED]
> Subject:           Re: Parser problem
>
>
> Hi, Brian,
>
>      Is the ###doc### a typo?
>
> for(DOM_Node child=doc.getFirstChild(); child!=NULL;
> child=child.getNextSibling() )
> {
>      if (child.getNodeType()== DOM_Node::ELEMENT_NODE)
>      {
>           int chTypeNode = child.getNodeType();
>           DOMString chNodeName = child.getNodeName();
>          char* chNodeNameVal = chNodeName.transcode();
>
>           DOM_Node children = ###doc###.getFirstChild();
>                                  ???
>           DOM_Node children = child.getFirstChild();
>
>           DOMString chNdValue = children.getNodeValue();
>           char* chNdNameVal = chNdValue.transcode();
>          //I want to print out each value in the children nodes
>         cout << "\nValues are" << chNdNameVal << endl;
>      }
>      else
>      {
>           //do nothing
>      }
> }
>
> Regards,
>
> Peiyong Zhang
> ____________________________________________
> XML Parsers Development
> IBM Toronto Laboratory email: [EMAIL PROTECTED]
> Phone: (416)448-4088; Fax: (416)448-4414; T/L: 778-4088
>
>
>
> "Jones, Brian O." <[EMAIL PROTECTED]> on 03/27/2001 09:55:58
> AM
>
> Please respond to [EMAIL PROTECTED]
>
> To:   "'xercesDev'" <[EMAIL PROTECTED]>
> cc:
> Subject:  Parser problem
>
>
> Hello all,
>
> I am having no luck getting values from the children nodes.  Any
> suggestions
> on what to do?
>
>
>
> here is a code subset of what I did...
>           DOM_Document doc = parser->getDocument();
>           DOM_NodeList nodeList = doc.getElementsByTagName("Server");
>           for(DOM_Node child=doc.getFirstChild(); child!=NULL;
> child=child.getNextSibling() )
>           {
>              if (child.getNodeType()== DOM_Node::ELEMENT_NODE)
>              {
>
>                      int chTypeNode = child.getNodeType();
>                      DOMString chNodeName = child.getNodeName();
>                      char* chNodeNameVal = chNodeName.transcode();
>                   DOM_Node children = doc.getFirstChild();
>                      DOMString chNdValue = children.getNodeValue();
>                      char* chNdNameVal = chNdValue.transcode();
>                   //I want to print out each value in the children nodes
>                      cout << "\nValues are" << chNdNameVal << endl;
>              }
>              else
>              {
>                //do nothing
>              }
>           }
>
>
> My XML file looks something like this...
>
> <HWCONFIG>
>     <Server>
>         <HostName>
>             PEPSI
>         </HostName>
>         <Channel>
>             <Type>
>                 TNT
>             </Type>
>         </Channel>
>     </Server>
> </HWCONFIG>
> I am trying to get the values PEPSI and TNT out. And having no luck
>
> If you have any suggestions it would be greatly appreciated...
> Thanks in Advance.
>
>
> Brian Jones
> Software Engineer
> (Email) [EMAIL PROTECTED]
>
>
>
>
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>
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