如何使用 HTTP POST multipart/form-data 将文件上传到服务器？
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【中文标题】如何使用 HTTP POST multipart/form-data 将文件上传到服务器？【英文标题】：How to upload file to server with HTTP POST multipart/form-data? 【发布时间】：2013-11-26 02:01:12 【问题描述】：我正在开发 Windows Phone 8 应用程序。我想通过 php Web 服务使用 MIME 类型 multipart/form-data 和一个名为“userid=SOME_ID”的字符串数据的 HTTP POST 请求上传 SQLite 数据库。
我不想使用 HttpClient、RestSharp 或 MyToolkit 等第 3 方库。我尝试了下面的代码，但它没有上传文件，也没有给我任何错误。它在 android、PHP 等中运行良好，因此在 Web 服务中没有问题。下面是我给定的代码（用于 WP8）。有什么问题？
我已经用谷歌搜索过，但我没有具体了解 WP8
async void MainPage_Loaded(object sender, RoutedEventArgs e)
var file = await Windows.ApplicationModel.Package.Current.InstalledLocation.GetFileAsync(DBNAME);
//Below line gives me file with 0 bytes, why? Should I use
//IsolatedStorageFile instead of StorageFile
//var file = await ApplicationData.Current.LocalFolder.GetFileAsync(DBNAME);
byte[] fileBytes = null;
using (var stream = await file.OpenReadAsync())
fileBytes = new byte[stream.Size];
using (var reader = new DataReader(stream))
await reader.LoadAsync((uint)stream.Size);
reader.ReadBytes(fileBytes);
//var res = await HttpPost(Util.UPLOAD_BACKUP, fileBytes);
HttpPost(fileBytes);
private void HttpPost(byte[] file_bytes)
HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create("http://www.myserver.com/upload.php");
httpWebRequest.ContentType = "multipart/form-data";
httpWebRequest.Method = "POST";
var asyncResult = httpWebRequest.BeginGetRequestStream((ar) => GetRequestStreamCallback(ar, file_bytes); , httpWebRequest);
private void GetRequestStreamCallback(IAsyncResult asynchronousResult, byte[] postData)
//DON'T KNOW HOW TO PASS "userid=some_user_id"
HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;
Stream postStream = request.EndGetRequestStream(asynchronousResult);
postStream.Write(postData, 0, postData.Length);
postStream.Close();
var asyncResult = request.BeginGetResponse(new AsyncCallback(GetResponseCallback), request);
private void GetResponseCallback(IAsyncResult asynchronousResult)
HttpWebRequest request = (HttpWebRequest)asynchronousResult.AsyncState;
HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asynchronousResult);
Stream streamResponse = response.GetResponseStream();
StreamReader streamRead = new StreamReader(streamResponse);
string responseString = streamRead.ReadToEnd();
streamResponse.Close();
streamRead.Close();
response.Close();
我也尝试在 Windows 8 中解决我的问题，但它也无法正常工作。
public async Task Upload(byte[] fileBytes)
using (var client = new HttpClient())
using (var content = new MultipartFormDataContent("Upload----" + DateTime.Now.ToString(System.Globalization.CultureInfo.InvariantCulture)))
content.Add(new StreamContent(new MemoryStream(fileBytes)));
//Not sure below line is true or not
content.Add(new StringContent("userid=farhanW8"));
using (var message = await client.PostAsync("http://www.myserver.com/upload.php", content))
var input = await message.Content.ReadAsStringAsync();
【问题讨论】：
【参考方案1】：基于@Wolf5 的回答这对我有用
var client = new WebClient();
client.Encoding = Encoding.UTF8;
var boundary = $"--------------------------DateTime.Now.Ticks:x";
client.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
client.Headers.Add("Cookie", cookie);
var start = $"--boundary\r\nContent-Disposition: form-data; name=\"file\"; filename=\"Path.GetFileName(fileName)\"\r\nContent-Type: image/jpeg\r\n\r\n";
var end = $"\r\n--boundary--\r\n";
var lst = new List<byte>();
lst.AddRange(client.Encoding.GetBytes(start));
lst.AddRange(File.ReadAllBytes(fileName));
lst.AddRange(client.Encoding.GetBytes(end));
var resp = client.UploadData($"ApiUrl/api/upload/image", "POST", lst.ToArray());
【讨论】：
【参考方案2】：使用 MultipartFormDataContent 的基本实现：-
HttpClient httpClient = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();
form.Add(new StringContent(username), "username");
form.Add(new StringContent(useremail), "email");
form.Add(new StringContent(password), "password");
form.Add(new ByteArrayContent(file_bytes, 0, file_bytes.Length), "profile_pic", "hello1.jpg");
HttpResponseMessage response = await httpClient.PostAsync("PostUrl", form);
response.EnsureSuccessStatusCode();
httpClient.Dispose();
string sd = response.Content.ReadAsStringAsync().Result;
【讨论】：
谢谢，我不想依赖 HttpClient，如果本机支持的话。 HttpClient 可通过 WP8 中的 NuGet 获得 MultipartFormDataContent 仅在 WP8.1 中可用 MultipartFormDataContent 在 .NET 4.5（不仅仅是 WP）中可用 您可以通过流添加文件，而不是将整个文件内容作为字节[]保存在内存中。var fileStream = new FileStream(filePath, FileMode.Open); form.Add(new StreamContent(fileStream), "profile_pic");
永远不要调用新的 HttpClient。哟可以破坏你的软件aspnetmonsters.com/2016/08/2016-08-27-httpclientwrong【参考方案3】：
我知道这是一个旧线程，但我一直在努力解决这个问题，我想分享我的解决方案。
此解决方案适用于来自System.Net.Http 的HttpClient 和MultipartFormDataContent。您可以使用.NET Core 1.0 或更高版本或.NET Framework 4.5 或更高版本发布它。
简单概括一下，它是一种异步方法，它接收您要执行 POST 的 URL、用于发送字符串的键/值集合和用于发送文件的键/值集合作为参数。
private static async Task<HttpResponseMessage> Post(string url, NameValueCollection strings, NameValueCollection files)
var formContent = new MultipartFormDataContent(/* If you need a boundary, you can define it here */);
// Strings
foreach (string key in strings.Keys)
string inputName = key;
string content = strings[key];
formContent.Add(new StringContent(content), inputName);
// Files
foreach (string key in files.Keys)
string inputName = key;
string fullPathToFile = files[key];
FileStream fileStream = File.OpenRead(fullPathToFile);
var streamContent = new StreamContent(fileStream);
var fileContent = new ByteArrayContent(streamContent.ReadAsByteArrayAsync().Result);
formContent.Add(fileContent, inputName, Path.GetFileName(fullPathToFile));
var myHttpClient = new HttpClient();
var response = await myHttpClient.PostAsync(url, formContent);
//string stringContent = await response.Content.ReadAsStringAsync(); // If you need to read the content
return response;
您可以像这样准备您的 POST（您可以根据需要添加很多字符串和文件）：
string url = @"http://yoursite.com/upload.php"
NameValueCollection strings = new NameValueCollection();
strings.Add("stringInputName1", "The content for input 1");
strings.Add("stringInputNameN", "The content for input N");
NameValueCollection files = new NameValueCollection();
files.Add("fileInputName1", @"FullPathToFile1"); // Path + filename
files.Add("fileInputNameN", @"FullPathToFileN");
最后，像这样调用方法：
var result = Post(url, strings, files).GetAwaiter().GetResult();
如果你愿意，你可以检查你的状态码，并显示原因如下：
if (result.StatusCode == HttpStatusCode.OK)
// Logic if all was OK
else
// You can show a message like this:
Console.WriteLine(string.Format("Error. StatusCode: 0 | ReasonPhrase: 1", result.StatusCode, result.ReasonPhrase));
如果有人需要，这里我举一个小例子来说明如何使用 PHP 接收存储文件（在我们的 .Net 应用程序的另一端）：
<?php
if (isset($_FILES['fileInputName1']) && $_FILES['fileInputName1']['error'] === UPLOAD_ERR_OK)
$fileTmpPath = $_FILES['fileInputName1']['tmp_name'];
$fileName = $_FILES['fileInputName1']['name'];
move_uploaded_file($fileTmpPath, '/the/final/path/you/want/' . $fileName);
我希望你觉得它有用，我很注意你的问题。
【讨论】：
Async().Result 不是异步的
@OwnageIsMagic 我无法理解您指的是代码的哪一部分。让我们检查一下这里发生的事情的顺序： 1) HttpClient() 对象中的 PostAsync(...)，它是异步的。 2) 由于 1)，我的 Post 方法返回一个异步任务。 3) 由于 2)，我将 Post(...) 方法与 GetAwaiter().GetResult() 挂钩，以等待异步任务结束。我错过了什么？干杯
我指的是streamContent.ReadAsByteArrayAsync().Result。但是GetAwaiter().GetResult() 也不是异步的【参考方案4】：
这是带有基本身份验证 C# 的多部分数据发布
public string UploadFilesToRemoteUrl(string url)
try
Dictionary<string, object> formFields = new Dictionary<string, object>();
formFields.Add("requestid", "\"id\":\"idvalue\"");
string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
request.ContentType = "multipart/form-data; boundary=" + boundary;
// basic authentication.
var username = "userid";
var password = "password";
string credidentials = username + ":" + password;
var authorization = Convert.ToBase64String(Encoding.Default.GetBytes(credidentials));
request.Headers["Authorization"] = "Basic " + authorization;
request.Method = "POST";
request.KeepAlive = true;
Stream memStream = new System.IO.MemoryStream();
WriteFormData(formFields, memStream, boundary);
FileInfo fileToUpload = new FileInfo(@"filelocation with name");
string fileFormKey = "file";
if (fileToUpload != null)
WritefileToUpload(fileToUpload, memStream, boundary, fileFormKey);
request.ContentLength = memStream.Length;
using (Stream requestStream = request.GetRequestStream())
memStream.Position = 0;
byte[] tempBuffer = new byte[memStream.Length];
memStream.Read(tempBuffer, 0, tempBuffer.Length);
memStream.Close();
requestStream.Write(tempBuffer, 0, tempBuffer.Length);
using (var response = request.GetResponse())
Stream responseSReam = response.GetResponseStream();
StreamReader streamReader = new StreamReader(responseSReam);
return streamReader.ReadToEnd();
catch (WebException ex)
using (WebResponse response = ex.Response)
HttpWebResponse httpResponse = (HttpWebResponse)response;
using (var streamReader = new StreamReader(response.GetResponseStream()))
return streamReader.ReadToEnd();
// write form id.
public static void WriteFormData(Dictionary<string, object> dictionary, Stream stream, string mimeBoundary)
string formdataTemplate = "\r\n--" + mimeBoundary +
"\r\nContent-Disposition: form-data; name=\"0\";\r\n\r\n1";
if (dictionary != null)
foreach (string key in dictionary.Keys)
string formitem = string.Format(formdataTemplate, key, dictionary[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
stream.Write(formitembytes, 0, formitembytes.Length);
// write file.
public static void WritefileToUpload(FileInfo file, Stream stream, string mimeBoundary, string formkey)
var boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "\r\n");
var endBoundaryBytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + mimeBoundary + "--");
string headerTemplate = "Content-Disposition: form-data; name=\"0\"; filename=\"1\"\r\n" +
"Content-Type: application/octet-stream\r\n\r\n";
stream.Write(boundarybytes, 0, boundarybytes.Length);
var header = string.Format(headerTemplate, formkey, file.Name);
var headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
stream.Write(headerbytes, 0, headerbytes.Length);
using (var fileStream = new FileStream(file.FullName, FileMode.Open, FileAccess.Read))
var buffer = new byte[1024];
var bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
stream.Write(buffer, 0, bytesRead);
stream.Write(endBoundaryBytes, 0, endBoundaryBytes.Length);
【讨论】：
【参考方案5】：以下是在将文件作为多格式数据发送时对我有用的方法：
public T HttpPostMultiPartFileStream<T>(string requestURL, string filePath, string fileName)
string content = null;
using (MultipartFormDataContent form = new MultipartFormDataContent())
StreamContent streamContent;
using (var fileStream = new FileStream(filePath, FileMode.Open))
streamContent = new StreamContent(fileStream);
streamContent.Headers.Add("Content-Type", "application/octet-stream");
streamContent.Headers.Add("Content-Disposition", string.Format("form-data; name=\"file\"; filename=\"0\"", fileName));
form.Add(streamContent, "file", fileName);
using (HttpClient client = GetAuthenticatedHttpClient())
HttpResponseMessage response = client.PostAsync(requestURL, form).GetAwaiter().GetResult();
content = response.Content.ReadAsStringAsync().GetAwaiter().GetResult();
try
return JsonConvert.DeserializeObject<T>(content);
catch (Exception ex)
// Log the exception
return default(T);
上面使用的GetAuthenticatedHttpClient可以是：
private HttpClient GetAuthenticatedHttpClient()
HttpClient httpClient = new HttpClient();
httpClient.BaseAddress = new Uri(<yourBaseURL>));
httpClient.DefaultRequestHeaders.Add("Token, <yourToken>);
return httpClient;
【讨论】：
感谢MultipartFormDataContent 的想法，这正是我想要的。【参考方案6】：
置顶到@loop 答案。
我们得到以下 Asp.Net MVC 错误， 无法连接到远程服务器
修复： 在 Web.Confing 中添加以下代码后，问题已为我们解决
<system.net>
<defaultProxy useDefaultCredentials="true" >
</defaultProxy>
</system.net>
【讨论】：
【参考方案7】：大家好，经过一天的网络搜索，我终于用下面的源代码解决了问题 希望能帮到你
public UploadResult UploadFile(string fileAddress)
HttpClient client = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();
HttpContent content = new StringContent("fileToUpload");
form.Add(content, "fileToUpload");
var stream = new FileStream(fileAddress, FileMode.Open);
content = new StreamContent(stream);
var fileName =
content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
Name = "name",
FileName = Path.GetFileName(fileAddress),
;
form.Add(content);
HttpResponseMessage response = null;
var url = new Uri("http://192.168.10.236:2000/api/Upload2");
response = (client.PostAsync(url, form)).Result;
【讨论】：
【参考方案8】：我还想将内容上传到服务器，这是一个 Spring 应用程序，我终于发现我需要为它设置一个内容类型才能将其解释为文件。就像这样：
...
MultipartFormDataContent form = new MultipartFormDataContent();
var fileStream = new FileStream(uniqueTempPathInProject, FileMode.Open);
var streamContent = new StreamContent(fileStream);
streamContent.Headers.ContentType=new MediaTypeHeaderValue("application/zip");
form.Add(streamContent, "file",fileName);
...
【讨论】：
这是我看到的最干净的解决方案。您可以做的唯一改进是在前三行中的每一行前面添加“使用”。【参考方案9】：对于在尝试以多部分形式上传时搜索 403 禁止问题的人，以下可能会有所帮助，因为根据服务器配置，由于 MultipartFormDataContent 标头不正确，您将获得 MULTIPART_STRICT_ERROR "!@eq 0" 的情况。 请注意，imagetag/filename 变量都包含引号 (\") 例如 filename="\"myfile.png\"" .
MultipartFormDataContent form = new MultipartFormDataContent();
ByteArrayContent imageContent = new ByteArrayContent(fileBytes, 0, fileBytes.Length);
imageContent.Headers.TryAddWithoutValidation("Content-Disposition", "form-data; name="+imagetag+"; filename="+filename);
imageContent.Headers.TryAddWithoutValidation("Content-Type", "image / png");
form.Add(imageContent, imagetag, filename);
【讨论】：
【参考方案10】：以下代码读取文件，将其转换为字节数组，然后向服务器发出请求。
public void PostImage()
HttpClient httpClient = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();
byte[] imagebytearraystring = ImageFileToByteArray(@"C:\Users\Downloads\icon.png");
form.Add(new ByteArrayContent(imagebytearraystring, 0, imagebytearraystring.Count()), "profile_pic", "hello1.jpg");
HttpResponseMessage response = httpClient.PostAsync("your url", form).Result;
httpClient.Dispose();
string sd = response.Content.ReadAsStringAsync().Result;
private byte[] ImageFileToByteArray(string fullFilePath)
FileStream fs = File.OpenRead(fullFilePath);
byte[] bytes = new byte[fs.Length];
fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
fs.Close();
return bytes;
【讨论】：
【参考方案11】：这是我的最终工作代码。我的网络服务需要一个文件（POST 参数名称为“file”）和一个字符串值（POST 参数名称为“userid”）。
/// <summary>
/// Occurs when upload backup application bar button is clicked. Author : Farhan Ghumra
/// </summary>
private async void btnUploadBackup_Click(object sender, EventArgs e)
var dbFile = await ApplicationData.Current.LocalFolder.GetFileAsync(Util.DBNAME);
var fileBytes = await GetBytesAsync(dbFile);
var Params = new Dictionary<string, string> "userid", "9" ;
UploadFilesToServer(new Uri(Util.UPLOAD_BACKUP), Params, Path.GetFileName(dbFile.Path), "application/octet-stream", fileBytes);
/// <summary>
/// Creates HTTP POST request & uploads database to server. Author : Farhan Ghumra
/// </summary>
private void UploadFilesToServer(Uri uri, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(uri);
httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
httpWebRequest.Method = "POST";
httpWebRequest.BeginGetRequestStream((result) =>
try
HttpWebRequest request = (HttpWebRequest)result.AsyncState;
using (Stream requestStream = request.EndGetRequestStream(result))
WriteMultipartForm(requestStream, boundary, data, fileName, fileContentType, fileData);
request.BeginGetResponse(a =>
try
var response = request.EndGetResponse(a);
var responseStream = response.GetResponseStream();
using (var sr = new StreamReader(responseStream))
using (StreamReader streamReader = new StreamReader(response.GetResponseStream()))
string responseString = streamReader.ReadToEnd();
//responseString is depend upon your web service.
if (responseString == "Success")
MessageBox.Show("Backup stored successfully on server.");
else
MessageBox.Show("Error occurred while uploading backup on server.");
catch (Exception)
, null);
catch (Exception)
, httpWebRequest);
/// <summary>
/// Writes multi part HTTP POST request. Author : Farhan Ghumra
/// </summary>
private void WriteMultipartForm(Stream s, string boundary, Dictionary<string, string> data, string fileName, string fileContentType, byte[] fileData)
/// The first boundary
byte[] boundarybytes = Encoding.UTF8.GetBytes("--" + boundary + "\r\n");
/// the last boundary.
byte[] trailer = Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
/// the form data, properly formatted
string formdataTemplate = "Content-Dis-data; name=\"0\"\r\n\r\n1";
/// the form-data file upload, properly formatted
string fileheaderTemplate = "Content-Dis-data; name=\"0\"; filename=\"1\";\r\nContent-Type: 2\r\n\r\n";
/// Added to track if we need a CRLF or not.
bool bNeedsCRLF = false;
if (data != null)
foreach (string key in data.Keys)
/// if we need to drop a CRLF, do that.
if (bNeedsCRLF)
WriteToStream(s, "\r\n");
/// Write the boundary.
WriteToStream(s, boundarybytes);
/// Write the key.
WriteToStream(s, string.Format(formdataTemplate, key, data[key]));
bNeedsCRLF = true;
/// If we don't have keys, we don't need a crlf.
if (bNeedsCRLF)
WriteToStream(s, "\r\n");
WriteToStream(s, boundarybytes);
WriteToStream(s, string.Format(fileheaderTemplate, "file", fileName, fileContentType));
/// Write the file data to the stream.
WriteToStream(s, fileData);
WriteToStream(s, trailer);
/// <summary>
/// Writes string to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, string txt)
byte[] bytes = Encoding.UTF8.GetBytes(txt);
s.Write(bytes, 0, bytes.Length);
/// <summary>
/// Writes byte array to stream. Author : Farhan Ghumra
/// </summary>
private void WriteToStream(Stream s, byte[] bytes)
s.Write(bytes, 0, bytes.Length);
/// <summary>
/// Returns byte array from StorageFile. Author : Farhan Ghumra
/// </summary>
private async Task<byte[]> GetBytesAsync(StorageFile file)
byte[] fileBytes = null;
using (var stream = await file.OpenReadAsync())
fileBytes = new byte[stream.Size];
using (var reader = new DataReader(stream))
await reader.LoadAsync((uint)stream.Size);
reader.ReadBytes(fileBytes);
return fileBytes;
非常感谢Darin Rousseau 的帮助。
【讨论】：
【参考方案12】：它适用于window phone 8.1。你可以试试这个。
Dictionary<string, object> _headerContents = new Dictionary<string, object>();
const String _lineEnd = "\r\n";
const String _twoHyphens = "--";
const String _boundary = "*****";
private async void UploadFile_OnTap(object sender, System.Windows.Input.GestureEventArgs e)
Uri serverUri = new Uri("http:www.myserver.com/Mp4UploadHandler", UriKind.Absolute);
string fileContentType = "multipart/form-data";
byte[] _boundarybytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _lineEnd);
byte[] _trailerbytes = Encoding.UTF8.GetBytes(_twoHyphens + _boundary + _twoHyphens + _lineEnd);
Dictionary<string, object> _headerContents = new Dictionary<string, object>();
SetEndHeaders(); // to add some extra parameter if you need
httpWebRequest = (HttpWebRequest)WebRequest.Create(serverUri);
httpWebRequest.ContentType = fileContentType + "; boundary=" + _boundary;
httpWebRequest.Method = "POST";
httpWebRequest.AllowWriteStreamBuffering = false; // get response after upload header part
var fileName = Path.GetFileName(MediaStorageFile.Path);
Stream fStream = (await MediaStorageFile.OpenAsync(Windows.Storage.FileAccessMode.Read)).AsStream(); //MediaStorageFile is a storage file from where you want to upload the file of your device
string fileheaderTemplate = "Content-Disposition: form-data; name=\"0\"" + _lineEnd + _lineEnd + "1" + _lineEnd;
long httpLength = 0;
foreach (var headerContent in _headerContents) // get the length of upload strem
httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes(string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value)).Length;
httpLength += _boundarybytes.Length + Encoding.UTF8.GetBytes("Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd).Length
+ Encoding.UTF8.GetBytes(_lineEnd).Length * 2 + _trailerbytes.Length;
httpWebRequest.ContentLength = httpLength + fStream.Length; // wait until you upload your total stream
httpWebRequest.BeginGetRequestStream((result) =>
try
HttpWebRequest request = (HttpWebRequest)result.AsyncState;
using (Stream stream = request.EndGetRequestStream(result))
foreach (var headerContent in _headerContents)
WriteToStream(stream, _boundarybytes);
WriteToStream(stream, string.Format(fileheaderTemplate, headerContent.Key, headerContent.Value));
WriteToStream(stream, _boundarybytes);
WriteToStream(stream, "Content-Disposition: form-data; name=\"uploadedFile\";filename=\"" + fileName + "\"" + _lineEnd);
WriteToStream(stream, _lineEnd);
int bytesRead = 0;
byte[] buffer = new byte[2048]; //upload 2K each time
while ((bytesRead = fStream.Read(buffer, 0, buffer.Length)) != 0)
stream.Write(buffer, 0, bytesRead);
Array.Clear(buffer, 0, 2048); // Clear the array.
WriteToStream(stream, _lineEnd);
WriteToStream(stream, _trailerbytes);
fStream.Close();
request.BeginGetResponse(a =>
//get response here
try
var response = request.EndGetResponse(a);
using (Stream streamResponse = response.GetResponseStream())
using (var memoryStream = new MemoryStream())
streamResponse.CopyTo(memoryStream);
responseBytes = memoryStream.ToArray(); // here I get byte response from server. you can change depends on server response
if (responseBytes.Length > 0 && responseBytes[0] == 1)
MessageBox.Show("Uploading Completed");
else
MessageBox.Show("Uploading failed, please try again.");
catch (Exception ex)
, null);
catch (Exception ex)
fStream.Close();
, httpWebRequest);
private static void WriteToStream(Stream s, string txt)
byte[] bytes = Encoding.UTF8.GetBytes(txt);
s.Write(bytes, 0, bytes.Length);
private static void WriteToStream(Stream s, byte[] bytes)
s.Write(bytes, 0, bytes.Length);
private void SetEndHeaders()
_headerContents.Add("sId", LocalData.currentUser.SessionId);
_headerContents.Add("uId", LocalData.currentUser.UserIdentity);
_headerContents.Add("authServer", LocalData.currentUser.AuthServerIP);
_headerContents.Add("comPort", LocalData.currentUser.ComPort);
【讨论】：
【参考方案13】：这个简单的版本也可以。
public void UploadMultipart(byte[] file, string filename, string contentType, string url)
var webClient = new WebClient();
string boundary = "------------------------" + DateTime.Now.Ticks.ToString("x");
webClient.Headers.Add("Content-Type", "multipart/form-data; boundary=" + boundary);
var fileData = webClient.Encoding.GetString(file);
var package = string.Format("--0\r\nContent-Disposition: form-data; name=\"file\"; filename=\"1\"\r\nContent-Type: 2\r\n\r\n3\r\n--0--\r\n", boundary, filename, contentType, fileData);
var nfile = webClient.Encoding.GetBytes(package);
byte[] resp = webClient.UploadData(url, "POST", nfile);
如果需要，添加任何额外的必需标头。
【讨论】：
如何在Windows phone 8.1 Rt中执行？ 谢谢你，狼！这对我很有帮助。 在与MultipartFormDataContent 搏斗失败后，这种方法效果很好。它不那么迷人，但如果您需要对 HTTP 消息进行详细控制，这是一个很好的解决方案。
数以百万计的感谢你，伙计，我花了 3 天时间解决了这个问题，而不是运气。Thiis 是完美的解决方案。
@wolf5 在上面的例子中输入什么内容我正在尝试发送 jpg 图片【参考方案14】：
我玩了一会儿，想出了一个简化的、更通用的解决方案：
private static string sendHttpRequest(string url, NameValueCollection values, NameValueCollection files = null)
string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
// The first boundary
byte[] boundaryBytes = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "\r\n");
// The last boundary
byte[] trailer = System.Text.Encoding.UTF8.GetBytes("\r\n--" + boundary + "--\r\n");
// The first time it itereates, we need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick
byte[] boundaryBytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n");
// Create the request and set parameters
HttpWebRequest request = (HttpWebRequest) WebRequest.Create(url);
request.ContentType = "multipart/form-data; boundary=" + boundary;
request.Method = "POST";
request.KeepAlive = true;
request.Credentials = System.Net.CredentialCache.DefaultCredentials;
// Get request stream
Stream requestStream = request.GetRequestStream();
foreach (string key in values.Keys)
// Write item to stream
byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"0\";\r\n\r\n1", key, values[key]));
requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
requestStream.Write(formItemBytes, 0, formItemBytes.Length);
if (files != null)
foreach(string key in files.Keys)
if(File.Exists(files[key]))
int bytesRead = 0;
byte[] buffer = new byte[2048];
byte[] formItemBytes = System.Text.Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"0\"; filename=\"1\"\r\nContent-Type: application/octet-stream\r\n\r\n", key, files[key]));
requestStream.Write(boundaryBytes, 0, boundaryBytes.Length);
requestStream.Write(formItemBytes, 0, formItemBytes.Length);
using (FileStream fileStream = new FileStream(files[key], FileMode.Open, FileAccess.Read))
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
// Write file content to stream, byte by byte
requestStream.Write(buffer, 0, bytesRead);
fileStream.Close();
// Write trailer and close stream
requestStream.Write(trailer, 0, trailer.Length);
requestStream.Close();
using (StreamReader reader = new StreamReader(request.GetResponse().GetResponseStream()))
return reader.ReadToEnd();
;
你可以这样使用它：
string fileLocation = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + Path.DirectorySeparatorChar + "somefile.jpg";
NameValueCollection values = new NameValueCollection();
NameValueCollection files = new NameValueCollection();
values.Add("firstName", "Alan");
files.Add("profilePicture", fileLocation);
sendHttpRequest("http://example.com/handler.php", values, files);
在 PHP 脚本中，您可以像这样处理数据：
echo $_POST['firstName'];
$name = $_POST['firstName'];
$image = $_FILES['profilePicture'];
$ds = DIRECTORY_SEPARATOR;
move_uploaded_file($image['tmp_name'], realpath(dirname(__FILE__)) . $ds . "uploads" . $ds . $image['name']);
【讨论】：
一次将文件一个块写入流的好工作，而不是将其全部加载到一个字节[]中并传递它。【参考方案15】：你可以使用这个类：
using System.Collections.Specialized;
class Post_File
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");
byte[] boundarybytesF = System.Text.Encoding.ASCII.GetBytes("--" + boundary + "\r\n"); // the first time it itereates, you need to make sure it doesn't put too many new paragraphs down or it completely messes up poor webbrick.
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
wr.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8";
var nvc2 = new NameValueCollection();
nvc2.Add("Accepts-Language", "en-us,en;q=0.5");
wr.Headers.Add(nvc2);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
Stream rs = wr.GetRequestStream();
bool firstLoop = true;
string formdataTemplate = "Content-Disposition: form-data; name=\"0\"\r\n\r\n1";
foreach (string key in nvc.Keys)
if (firstLoop)
rs.Write(boundarybytesF, 0, boundarybytesF.Length);
firstLoop = false;
else
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, new FileInfo(file).Name, 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);
catch (Exception ex)
if (wresp != null)
wresp.Close();
wresp = null;
finally
wr = null;
使用它：
NameValueCollection nvc = new NameValueCollection();
//nvc.Add("id", "TTR");
nvc.Add("table_name", "uploadfile");
nvc.Add("commit", "uploadfile");
Post_File.HttpUploadFile("http://example/upload_file.php", @"C:\user\yourfile.docx", "uploadfile", "application/vnd.ms-excel", nvc);
示例服务器upload_file.php:
m('File upload '.(@copy($_FILES['uploadfile']['tmp_name'],getcwd().'\\'.'/'.$_FILES['uploadfile']['name']) ? 'success' : 'failed'));
function m($msg)
echo '<div style="background:#f1f1f1;border:1px solid #ddd;padding:15px;font:14px;text-align:center;font-weight:bold;">';
echo $msg;
echo '</div>';
【讨论】：
非常好，你的代码就像一个魅力。 =) 这正是我正在寻找的普通表单字段和文件字段的解决方案。您的代码也很容易扩展以进行多文件上传。谢谢！ 如果nvc为空，我相信你在开头发送了“太多新段落”（“\r\n”）。以上是关于如何使用 HTTP POST multipart/form-data 将文件上传到服务器？的主要内容，如果未能解决你的问题，请参考以下文章
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