嗨，我有一个带有以下代码的 web 服务 IIS：
// Implements multipart/form-data POST in C# http://www.ietf.org/rfc/rfc2388.txt
// http://www.briangrinstead.com/blog/multipart-form-post-in-c
public static class FormUpload
{
private static readonly Encoding encoding = Encoding.UTF8;
public static HttpWebResponse MultipartFormDataPost(string postUrl, string userAgent, Dictionary<string, object> postParameters)
{
string formDataBoundary = String.Format("----------{0:N}", Guid.NewGuid());
string contentType = "multipart/form-data; boundary=" + formDataBoundary;
byte[] formData = GetMultipartFormData(postParameters, formDataBoundary);
return PostForm(postUrl, userAgent, contentType, formData);
}
private static HttpWebResponse PostForm(string postUrl, string userAgent, string contentType, byte[] formData)
{
HttpWebRequest request = WebRequest.Create(postUrl) as HttpWebRequest;
if (request == null)
{
throw new NullReferenceException("request is not a http request");
}
// Set up the request properties.
request.Method = "POST";
request.ContentType = contentType;
request.UserAgent = userAgent;
request.CookieContainer = new CookieContainer();
request.ContentLength = formData.Length;
// You could add authentication here as well if needed:
// request.PreAuthenticate = true;
// request.AuthenticationLevel = System.Net.Security.AuthenticationLevel.MutualAuthRequested;
// request.Headers.Add("Authorization", "Basic " + Convert.ToBase64String(System.Text.Encoding.Default.GetBytes("username" + ":" + "password")));
// Send the form data to the request.
using (Stream requestStream = request.GetRequestStream())
{
requestStream.Write(formData, 0, formData.Length);
requestStream.Close();
}
return request.GetResponse() as HttpWebResponse;
}
private static byte[] GetMultipartFormData(Dictionary<string, object> postParameters, string boundary)
{
Stream formDataStream = new System.IO.MemoryStream();
bool needsCLRF = false;
foreach (var param in postParameters)
{
// Thanks to feedback from commenters, add a CRLF to allow multiple parameters to be added.
// Skip it on the first parameter, add it to subsequent parameters.
if (needsCLRF)
formDataStream.Write(encoding.GetBytes("\r\n"), 0, encoding.GetByteCount("\r\n"));
needsCLRF = true;
if (param.Value is FileParameter)
{
FileParameter fileToUpload = (FileParameter)param.Value;
// Add just the first part of this param, since we will write the file data directly to the Stream
string header = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"; filename=\"{2}\";\r\nContent-Type: {3}\r\n\r\n",
boundary,
param.Key,
fileToUpload.FileName ?? param.Key,
fileToUpload.ContentType ?? "application/octet-stream");
formDataStream.Write(encoding.GetBytes(header), 0, encoding.GetByteCount(header));
// Write the file data directly to the Stream, rather than serializing it to a string.
formDataStream.Write(fileToUpload.File, 0, fileToUpload.File.Length);
}
else
{
string postData = string.Format("--{0}\r\nContent-Disposition: form-data; name=\"{1}\"\r\n\r\n{2}",
boundary,
param.Key,
param.Value);
formDataStream.Write(encoding.GetBytes(postData), 0, encoding.GetByteCount(postData));
}
}
// Add the end of the request. Start with a newline
string footer = "\r\n--" + boundary + "--\r\n";
formDataStream.Write(encoding.GetBytes(footer), 0, encoding.GetByteCount(footer));
// Dump the Stream into a byte[]
formDataStream.Position = 0;
byte[] formData = new byte[formDataStream.Length];
formDataStream.Read(formData, 0, formData.Length);
formDataStream.Close();
return formData;
}
public class FileParameter
{
public byte[] File { get; set; }
public string FileName { get; set; }
public string ContentType { get; set; }
public FileParameter(byte[] file) : this(file, null) { }
public FileParameter(byte[] file, string filename) : this(file, filename, null) { }
public FileParameter(byte[] file, string filename, string contenttype)
{
File = file;
FileName = filename;
ContentType = contenttype;
}
}
}
从这里获取：来自 C# 客户端的多部分表单
但是现在，在 mi android 应用程序中，我有这个：
HttpParams httpParameters = new BasicHttpParams();
int timeoutConnection = 300000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
int timeoutSocket = 300000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
HttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpPost postRequest = new HttpPost("http://172.21.1.87:9999/Service1.svc");
ByteArrayBody bab = new ByteArrayBody(ficheroAEnviar, "prueba.jpg");
MultipartEntity reqEntity = new MultipartEntity();
postRequest.addHeader("Content-Type", " multipart/form-data");
reqEntity.addPart("Dictionary", new FileBody(new File(fileUri.toString(), "application/zip")));
reqEntity.addPart("boundary", new StringBody("envio"));
postRequest.setEntity(reqEntity);
HttpResponse response = httpClient.execute(postRequest);
BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
String sResponse;
StringBuilder s = new StringBuilder();
postRequest.getAllHeaders();
while ((sResponse = reader.readLine()) != null) {
s = s.append(sResponse);
}
System.out.println("Response: " + s);
}
catch (Exception e) {
// handle exception here
Log.e(e.getClass().getName(), e.getMessage());
}
}
但它返回错误请求 400 错误。我可以使用 c# 代码通过 android 上传多方文件还是需要对某些代码进行任何更改？
有人有一个例子 en c# 和 android 来做这个吗？用于上传大约 10 或 15 MB 的视频。
谢谢