JPgrassi 提供的答案是您将如何拥有 MultiPart 数据。我认为需要添加的东西很少，所以我想写我自己的答案。
MultiPart 表单数据，顾名思义，不是单一类型的数据，而是指定表单将作为 MultiPart MIME 消息发送，因此您无法使用预定义的格式化程序来读取所有内容。您需要使用 ReadAsync 函数读取字节流并获取不同类型的数据，识别它们并反序列化它们。
有两种读取内容的方法。第一种是读取并将所有内容保存在内存中，第二种方法是使用提供程序将所有文件内容流式传输到一些随机名称的文件（带有 GUID）并以本地路径的形式提供句柄来访问文件（提供的示例jpgrassi 正在做第二个）。
第一种方法：将所有内容保存在内存中
//Async because this is asynchronous process and would read stream data in a buffer.
//If you don't make this async, you would be only reading a few KBs (buffer size)
//and you wont be able to know why it is not working
public async Task<HttpResponseMessage> Post()
{
if (!request.Content.IsMimeMultipartContent()) return null;
Dictionary<string, object> extractedMediaContents = new Dictionary<string, object>();
//Here I am going with assumption that I am sending data in two parts,
//JSON object, which will come to me as string and a file. You need to customize this in the way you want it to.
extractedMediaContents.Add(BASE64_FILE_CONTENTS, null);
extractedMediaContents.Add(SERIALIZED_JSON_CONTENTS, null);
request.Content.ReadAsMultipartAsync()
.ContinueWith(multiPart =>
{
if (multiPart.IsFaulted || multiPart.IsCanceled)
{
Request.CreateErrorResponse(HttpStatusCode.InternalServerError, multiPart.Exception);
}
foreach (var part in multiPart.Result.Contents)
{
using (var stream = part.ReadAsStreamAsync())
{
stream.Wait();
Stream requestStream = stream.Result;
using (var memoryStream = new MemoryStream())
{
requestStream.CopyTo(memoryStream);
//filename attribute is identifier for file vs other contents.
if (part.Headers.ToString().IndexOf("filename") > -1)
{
extractedMediaContents[BASE64_FILE_CONTENTS] = memoryStream.ToArray();
}
else
{
string jsonString = System.Text.Encoding.ASCII.GetString(memoryStream.ToArray());
//If you need just string, this is enough, otherwise you need to de-serialize based on the content type.
//Each content is identified by name in content headers.
extractedMediaContents[SERIALIZED_JSON_CONTENTS] = jsonString;
}
}
}
}
}).Wait();
//extractedMediaContents; This now has the contents of Request in-memory.
}
第二种方法：使用提供程序（由 jpgrassi 提供）
注意，这只是文件名。如果你想处理文件或存储在不同的位置，你需要再次流式读取文件。
public async Task<HttpResponseMessage> Post()
{
HttpResponseMessage response;
//Check if request is MultiPart
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
//This specifies local path on server where file will be created
string root = HttpContext.Current.Server.MapPath("~/App_Data");
var provider = new MultipartFormDataStreamProvider(root);
//This write the file in your App_Data with a random name
await Request.Content.ReadAsMultipartAsync(provider);
foreach (MultipartFileData file in provider.FileData)
{
//Here you can get the full file path on the server
//and other data regarding the file
//Point to note, this is only filename. If you want to keep / process file, you need to stream read the file again.
tempFileName = file.LocalFileName;
}
// You values are inside FormData. You can access them in this way
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in provider.FormData.GetValues(key))
{
Trace.WriteLine(string.Format("{0}: {1}", key, val));
}
}
//Or directly (not safe)
string name = provider.FormData.GetValues("name").FirstOrDefault();
response = Request.CreateResponse(HttpStatusCode.Ok);
return response;
}