2012-09-10 86 views

回答

1

指功能PostTrivialData这里是代码的多数据发布到URI:

public string UploadChatFile(Stream fileStream, string uri, string postData, string fileName) 
     { 
      string boundary = "----------------------------" + 
      DateTime.Now.Ticks.ToString("x"); 

      HttpWebRequest httpWebRequest2 = (HttpWebRequest)WebRequest.Create(uri); 
      httpWebRequest2.ContentType = "multipart/form-data; boundary=" + 
      boundary; 
      httpWebRequest2.Method = "POST"; 
      httpWebRequest2.KeepAlive = true; 


      Stream memStream = new System.IO.MemoryStream(); 
      byte[] boundarybytes = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "\r\n"); 

      memStream.Write(boundarybytes, 0, boundarybytes.Length); 

      string headerTemplate = string.Format("Content-Disposition: form-data; name=\"postdata\"\r\n{0}\r\n\r\n", postData); 

      byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(headerTemplate); 

      memStream.Write(headerbytes, 0, headerbytes.Length); 

      memStream.Write(boundarybytes, 0, boundarybytes.Length); 

      headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n"; 

      //string header = string.Format(headerTemplate, "file" + i, files[i]); 
      string header = string.Format(headerTemplate, "uplTheFile", fileName); 

      headerbytes = System.Text.Encoding.UTF8.GetBytes(header); 

      memStream.Write(headerbytes, 0, headerbytes.Length); 


      //FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read); 
      byte[] buffer = new byte[1024]; 

      int bytesRead = 0; 

      while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0) 
      { 
       memStream.Write(buffer, 0, bytesRead); 

      } 
      memStream.Write(boundarybytes, 0, boundarybytes.Length); 
      fileStream.Close(); 

      httpWebRequest2.ContentLength = memStream.Length; 
      Stream requestStream = httpWebRequest2.GetRequestStream(); 
      memStream.Position = 0; 
      byte[] tempBuffer = new byte[memStream.Length]; 
      memStream.Read(tempBuffer, 0, tempBuffer.Length); 
      memStream.Close(); 
      requestStream.Write(tempBuffer, 0, tempBuffer.Length); 
      requestStream.Close(); 

      WebResponse webResponse2 = httpWebRequest2.GetResponse(); 

      Stream stream2 = webResponse2.GetResponseStream(); 
      StreamReader reader2 = new StreamReader(stream2); 

      string responseString = reader2.ReadToEnd(); 


      return responseString; 
     } 

这将张贴图片,你也可以包含你的数据POSTDATA参数。您需要仔细处理这个PostData服务器,因为它与多部分内容数据结合在一起。

+0

我怎样才能发送其他表单域假设我有窗体与FName,LName,电话和照片的用户我需要发布所有这些4字段3表单字段和1图像。上面的代码确实提供了表单域。 –

+0

您可以将表单数据包含在headerTemplate中。你需要处理这个数据服务器端。另一种方法是包含URL作为查询字符串。 – Mohan

+0

感谢Mohan,上面的代码适用于独立文件上传。在我的情况下,我需要上传文件以及一些表单提交。如果你看看这个http://stackoverflow.com/questions/10974468/why-do-i-receive-this-error-the-remote-server-returned-an-error-417-expectat杰克逊拉克回答帖子为我工作,但我需要添加一个作为文件归档可以请你帮我更新该代码。 –

相关问题