从其他网站的呼叫接收XML来ServerXMLHTTP的帖子在传统的ASP [英] receiving xml from another website's call to ServerXMLHTTP post in classic asp

问题描述

我写一个ASP的网页到ASP-网页对话中，始发网页推信息到接收的网页，然后进行处理，并发送回一个响应的两侧。始发网页必须使用下面的code键启动converstation：

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

...然后以某种方式在接收页在ASP code的，以便能够看到，被发送的信息。我已经试过所有我能想到的。这些信息是不是在请求对象的查询字符串或形式阵列（因为内容类型为文本/ XML ）和我试过路过整个请求对象到 DOM文档通过其负荷（）和/或的loadXML（）方法。

...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.

不管我做什么，我无法找到的资料，但我知道它正在发送，因为当我改变内容类型为应用程序/ x-WWW的形式urlen codeD ，我可以看到它在的Request.Form 阵列。

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.

那么，是当内容类型为文本/ XML ？

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

推荐答案

当您设置的内容类型为text / xml的你真的需要发送的信息作为XML字符串，而不是一个名称值列表。

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

然后，在你收到的ASP页面，你会然后捕获XML如下：

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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