在经典asp中从另一个网站对ServerXMLHTTP post的调用接收xml [英] receiving xml from another website's call to ServerXMLHTTP post in classic asp

问题描述

我正在将 ASP 网页的两面都写到 ASP 网页对话中，其中原始网页将信息推送到接收网页，然后接收网页对其进行处理并发回响应.原始网页必须使用以下代码开始转换:

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.comasp
eceivingwebpage.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

...然后接收页面中的 ASP 代码必须能够看到发送的信息.我已经尝试了我能想到的一切.信息不在请求对象的 querystring 或表单数组中(因为 content-type 是 text/xml)，我尝试将整个请求对象传递给 domdocument 通过其 load() 和/或 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.

无论我做什么，我都找不到信息，但我知道它正在发送，因为当我将内容类型更改为 application/x-www-form-urlencoded 时，我可以在 request.form 数组中看到它.

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.

那么当内容类型为 text/xml 时，我的信息在哪里?

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

推荐答案

当您将内容类型设置为text/xml"时，您确实需要将信息作为 XML 字符串发送，而不是名称-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.comasp
eceivingwebpage.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，在您的接收 ASP 页面中，您将捕获 XML，如下所示:

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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