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Asked: May 10, 20262026-05-10T23:12:48+00:00 2026-05-10T23:12:48+00:00

I am writing both sides of an ASP-webpage to ASP-webpage conversation in which the

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I am writing both sides of an ASP-webpage to ASP-webpage conversation in which the originating webpage pushes information to the receiving webpage which then processes it and sends back a response. The originating webpage must use the code below to start the converstation:

url = 'www.receivingwebsite.com\asp\receivingwebpage.asp information = 'UserName=Colt&PassWord=Taylor&Data=100' Set xmlhttp = server.Createobject('MSXML2.ServerXMLHTTP') xmlhttp.Open 'POST', url, false xmlhttp.setRequestHeader 'Content-Type', 'text/xml' xmlhttp.send information 

…and then somehow the ASP code in the receiving page has to be able to see the information that was sent. I have tried everything I can think of. The information is not in the request object’s querystring or form arrays (because the content-type is text/xml) and I’ve tried passing the entire request object to a domdocument via its load() and/or loadxml() methods.

No matter what I do, I can’t find the information but I know that it is being sent because when I change the content-type to application/x-www-form-urlencoded, I can see it in request.form array.

So where is my information when the content-type is text/xml?

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  1. 2026-05-10T23:12:49+00:00Added an answer on May 10, 2026 at 11:12 pm

    When you set the content-type to ‘text/xml’ you really need to send the information as an XML string, not a name-value list.

    url = 'www.receivingwebsite.com\asp\receivingwebpage.asp' information = '<Send><UserName>Colt</UserName><PassWord>Taylor</PassWord><Data>100</Data></Send>' Set xmlhttp = server.Createobject('MSXML2.ServerXMLHTTP') xmlhttp.Open 'POST', url, false xmlhttp.setRequestHeader 'Content-Type', 'text/xml'  xmlhttp.send information 

    Then, in your receiving ASP page, you would then capture the XML as follows:

    Dim xmlDoc Dim userName set xmlDoc=Server.CreateObject('Microsoft.XMLDOM') xmlDoc.async='false' xmlDoc.load(Request) userName = xmlDoc.documentElement.selectSingleNode('UserName').firstChild.nodeValue 
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