在post请求中发送文件+参数

我正在使用此代码将参数发送到网页并从中获取正确的响应。

System.Net.WebClient oWeb = new System.Net.WebClient(); oWeb.Proxy = System.Net.WebRequest.DefaultWebProxy; oWeb.Proxy.Credentials = System.Net.CredentialCache.DefaultCredentials; oWeb.Headers.Add("Content-Type", "application/x-www-form-urlencoded"); byte[] bytArguments = System.Text.Encoding.ASCII.GetBytes("value1=123&value2=xyz"); byte[] bytRetData = oWeb.UploadData("http://website.com/file.php", "POST", bytArguments); response = System.Text.Encoding.ASCII.GetString(bytRetData); 

但是现在我想发送一个类似（ .doc ）的文件到它+上面的参数（ value1 ， value2 ），但我不知道该怎么做。

Answer 1

使用WebClient.QueryString传递与请求关联的名称/值对。

 NameValueCollection parameters = new NameValueCollection(); parameters.Add("value1", "123"); parameters.Add("value2", "xyz"); oWeb.QueryString = parameters; var responseBytes = oWeb.UploadFile("http://website.com/file.php", "path to file"); string response = Encoding.ASCII.GetString(responseBytes); 

Answer 2

  public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc) { string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x"); byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n"); HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url); wr.ContentType = "multipart/form-data; boundary=" + boundary; wr.Method = "POST"; wr.KeepAlive = true; wr.Credentials = System.Net.CredentialCache.DefaultCredentials; Stream rs = wr.GetRequestStream(); string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}"; foreach (string key in nvc.Keys) { rs.Write(boundarybytes, 0, boundarybytes.Length); string formitem = string.Format(formdataTemplate, key, nvc[key]); byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem); rs.Write(formitembytes, 0, formitembytes.Length); } rs.Write(boundarybytes, 0, boundarybytes.Length); string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n"; string header = string.Format(headerTemplate, paramName, file, contentType); byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header); rs.Write(headerbytes, 0, headerbytes.Length); FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.Read); byte[] buffer = new byte[4096]; int bytesRead = 0; while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0) { rs.Write(buffer, 0, bytesRead); } fileStream.Close(); byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n"); rs.Write(trailer, 0, trailer.Length); rs.Close(); WebResponse wresp = null; try { wresp = wr.GetResponse(); Stream stream2 = wresp.GetResponseStream(); StreamReader reader2 = new StreamReader(stream2); result = reader2.ReadToEnd(); } catch (Exception ex) { System.Windows.MessageBox.Show("Error occurred while converting file", "Error!"); if (wresp != null) { wresp.Close(); wresp = null; } } finally { wr = null; } } 

从SO复制但不记得它的链接。 这就是它的使用方式

  NameValueCollection nvc = new NameValueCollection(); nvc.Add("parm1", "value1"); nvc.Add("parm2", "value2"); nvc.Add("parm3", "value3"); HttpUploadFile("http://www.example.com/upload.php",@filepath, "file", "text/html", nvc); 

这里@filepath是你要上传的文件的路径：c：\ file_to_upload.doc和file是php中使用的字段的名称$ _Files [‘file’]