Question

实际上我有c＃中的代码用一个POST方法做一个httpwebrequest，发送一个XML并接收一个XML答案并保存它，这里是代码：

public static XmlDocument PostXMLTransaction(string v_strURL)
{
//Declare XMLResponse document
XmlDocument XMLResponse = null;

//Declare an HTTP-specific implementation of the WebRequest class.
HttpWebRequest objHttpWebRequest;

//Declare an HTTP-specific implementation of the WebResponse class
HttpWebResponse objHttpWebResponse = null;

//Declare a generic view of a sequence of bytes
Stream objRequestStream = null;
Stream objResponseStream = null;

//Declare XMLReader
XmlTextReader objXMLReader;

//Creates an HttpWebRequest for the specified URL.
objHttpWebRequest = (HttpWebRequest)WebRequest.Create(v_strURL);

try
{
//---------- Start HttpRequest 

//Set HttpWebRequest properties
byte[] bytes;
bytes = System.Text.Encoding.UTF8.GetBytes("C:\\Documents\ \Aplicacion1\\Aplicacion1\\absisSIHttpProfileTO.xm l");
objHttpWebRequest.Method = "POST";
objHttpWebRequest.ContentLength = bytes.Length;
objHttpWebRequest.ContentType = "text/xml; encoding='utf-8'";

//Get Stream object 
objRequestStream = objHttpWebRequest.GetRequestStream();

//Writes a sequence of bytes to the current stream 
objRequestStream.Write(bytes, 0, bytes.Length);

//Close stream
objRequestStream.Close();

//---------- End HttpRequest

//Sends the HttpWebRequest, and waits for a response.
objHttpWebResponse = (HttpWebResponse)objHttpWebRequest.GetResponse();

//---------- Start HttpResponse
if (objHttpWebResponse.StatusCode == HttpStatusCode.OK)
{
//Get response stream 
objResponseStream = objHttpWebResponse.GetResponseStream();

//Load response stream into XMLReader
objXMLReader = new XmlTextReader(objResponseStream);

//Declare XMLDocument
XmlDocument xmldoc = new XmlDocument();
xmldoc.Load(objXMLReader);
xmldoc.Save(@"C:\Documents\Aplicacion1\Aplicacion1\myxml.xml");

//Set XMLResponse object returned from XMLReader
XMLResponse = xmldoc;

//Close XMLReader
objXMLReader.Close();
}

//Close HttpWebResponse
objHttpWebResponse.Close();
}
catch (WebException we)
{
//TODO: Add custom exception handling
throw new Exception(we.Message);
}
catch (Exception ex)
{
throw new Exception(ex.Message);
}
finally
{
//Close connections
objRequestStream.Close();
objResponseStream.Close();
objHttpWebResponse.Close();

//Release objects
objXMLReader = null;
objRequestStream = null;
objResponseStream = null;
objHttpWebResponse = null;
objHttpWebRequest = null;
}

//Return
return XMLResponse;
}

现在我的问题是如何在一个httpwebrequest中发送多个XML作为参数？

像： public static XmlDocument PostXMLTransaction（xmldocument xml1，xmldocument xml2）

Answer 1

嗯，这真的依赖于你的协议。另一方的服务器需要知道如何通过1个连接处理多个消息。如果服务器是一个http网络服务器而不是你的幸运，因为该协议已经定义了重用连接的机制。（虽然客户端和服务器都必须同意使用它。）

为此，您可以使用HttpWebRequest的KeepAlive属性，请参阅：http://msdn.microsoft.com/en-us/library/system.net.httpwebrequest.keepalive.aspx

如何将多个xml发布到WebService

1 个答案: