Almost, your xml is not xml (this might be the bulletin board though, and you use post if you are sending data on the send data:
Code:
Dim HttpReq As New MSXML2.XMLHTTP30
HttpReq.open "POST", "http://localhost/books.xml", False
xmltext ="<root>XML Here</root>"
Dim xmldom As New MSXML2.DOMDocument
xmldom.loadXML xmltext
HttpReq.send(xmldom)
MsgBox HttpReq.responseXML
Your url won't be "http://localhost/books.xml" though, this will just return the contents of the xml file and will ignore your sent data. If you want to modify the response according to the data sent you need to use some sort of server side processing, ASP for example.
Code:
Set oDom = CreateObject("msxml2.domdocument.4.0")
oDom.async = False
oDom.load Request
'Examine oDom and then send back xml:
Response.contentType = "text/xml"
'...
--
Joe (
Microsoft MVP - XML)