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Not sure if this will help you, but this VBScript code works fine:
dim oDoc : set oDoc = server.createobject("MSXML.DomDocument")
oDoc.LoadXML "<applet><transport name='jrmp'/><transport
name='codebase'/></applet>"
Dim nsr: set nsr = oDoc.DocumentElement
Dim nodeList : set nodeList=nsr.selectNodes("/applet/transport")
Dim node: set node= nodeList.item(0)
Dim trans_val1 : trans_val1 = node.getAttribute("name")
Dim node1 : set node1=nodeList.item(1)
Dim trans_val2 : trans_val2 = node1.getAttribute("name")
response.write "<br>Name of item(1)=" & trans_val1
response.write "<br>Name of item(2)=" & trans_val2
So, the xpath part of the code works as I would expect. Maybe namespaces or
some other difference would explain your problem.
-----Original Message-----
From: Nishi Prafull [mailto:nishiprafull@g...]
Sent: 24 January 2005 23:06
To: John Rivett-Carnac
Cc: xml-dev@l...
Subject: Re: Java NPE at node.getAttribute( )
Hi John:
I tried your suggestion:
NodeList nodeList=xmlDocument.selectNodes("/applet/transport", nsr);
XMLElement node=(XMLElement)nodeList.item(0);
String trans_val1 = node.getAttribute("name");
But I still get a NPE at node.getAttribute("name").
Can you suggest anything?
Thanks.
On Mon, 24 Jan 2005 22:00:06 -0000, John Rivett-Carnac
<jbrc@b...> wrote:
>
> I think you should get the "transport" nodes in a node collection,
> then get the name attribute of each transport node:
>
> NodeList nodeList=xmlDocument.selectNodes("/applet/transport", nsr);
> XMLElement node=(XMLElement)nodeList.item(0);
> String trans_val1 = node.getAttribute("name");
>
>
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