我必须将旧的Classic ASP站点的一些逻辑转换为asp.net项目。我无法理解一些负责发布数据的功能。
以下是Classic ASP中的功能:
<%Function PostHTTP(strURL, strBody, strErrTemplate)
ON ERROR RESUME NEXT
Dim objHTTP, strResult
Set objHTTP = Server.CreateObject("Msxml2.ServerXMLHTTP.3.0")
If Err.Number <> 0 Then
strResult = Replace(strErrTemplate, "%1", Err.Number)
strResult = Replace(strResult, "%2", Err.Description)
strResult = Replace(strResult, "%3", "Init::" & Err.Source)
Set objHTTP = Nothing
PostHTTP = strstrResult
Exit Function
End If
With objHTTP
.Open "POST", strURL, False
.setRequestHeader "Content-Type", "application/x-www-form-urlencoded"
.setTimeouts 30000, 30000, 60000, 240000
.send strBody
If Err.Number <> 0 Then
strResult = Replace(strErrTemplate, "%1", Err.Number)
strResult = Replace(strResult, "%2", Err.Description)
strResult = Replace(strResult, "%3", "Post::" & Err.Source)
Else
strResult = .responseText
End If
End With
' Response.Write "strResult: " & strResult
'Response.End
If Err.Number > 0 Then
strResult = Replace(strErrTemplate, "%1", Err.Number)
strResult = Replace(strResult, "%2", Err.Description)
strResult = Replace(strResult, "%3", Err.Source)
ElseIf Len(strResult) = 0 Then
strResult = Replace(strErrTemplate, "%1", 2000)
strResult = Replace(strResult, "%2", "No response received from remote server.")
strResult = Replace(strResult, "%3", "PostHTTP")
End If
PostHTTP = strResult
Set objHTTP = Nothing
End Function
这在asp.net中会是什么样子?
ps:我已经尝试了自己的发布功能,但显然错过了一些东西,因为我的工作不起作用。
答案 0 :(得分:1)
如果使用WebClient类,POST的主要任务将非常简单。例如,
// Form URL and POST-DATA
...
using (WebClient wc = new WebClient())
{
wc.Headers[HttpRequestHeader.ContentType] = "application/x-www-form-urlencoded";
string strResult = wc.UploadString(strURL, strBody);
}
...
要进行更精细的控制,可以使用WebRequest类。
编辑:这是WebRequest
的示例代码,因为您似乎需要指定WebClient
无法实现的超时值
var request = WebRequest.Create(strUrl);
request.Method = "POST";
request.ContentType = "application/x-www-form-urlencoded";
request.Timeout = 240000; // set timeout
using (var writer = new StreamWriter(request.GetRequestStream()))
{
// write to the body of the POST request
writer.Write(strBody);
}