C# 如何接受文件 POST
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10320232/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How To Accept a File POST
提问by Phil
I'm using asp.net mvc 4 webapi beta to build a rest service. I need to be able to accept POSTed images/files from client applications. Is this possible using the webapi? Below is how action I am currently using. Does anyone know of an example how this should work?
我正在使用 asp.net mvc 4 webapi beta 来构建休息服务。我需要能够接受来自客户端应用程序的 POSTed 图像/文件。这可以使用 webapi 吗？以下是我目前正在使用的操作。有谁知道这应该如何工作的例子？
[HttpPost]
public string ProfileImagePost(HttpPostedFile profileImage)
{
string[] extensions = { ".jpg", ".jpeg", ".gif", ".bmp", ".png" };
if (!extensions.Any(x => x.Equals(Path.GetExtension(profileImage.FileName.ToLower()), StringComparison.OrdinalIgnoreCase)))
{
throw new HttpResponseException("Invalid file type.", HttpStatusCode.BadRequest);
}
// Other code goes here
return "/path/to/image.png";
}
采纳答案by Mike Wasson
see http://www.asp.net/web-api/overview/formats-and-model-binding/html-forms-and-multipart-mime#multipartmime, although I think the article makes it seem a bit more complicated than it really is.
参见http://www.asp.net/web-api/overview/formats-and-model-binding/html-forms-and-multipart-mime#multipartmime，虽然我认为这篇文章使它看起来比确实如此。
Basically,
基本上，
public Task<HttpResponseMessage> PostFile()
{
HttpRequestMessage request = this.Request;
if (!request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = System.Web.HttpContext.Current.Server.MapPath("~/App_Data/uploads");
var provider = new MultipartFormDataStreamProvider(root);
var task = request.Content.ReadAsMultipartAsync(provider).
ContinueWith<HttpResponseMessage>(o =>
{
string file1 = provider.BodyPartFileNames.First().Value;
// this is the file name on the server where the file was saved
return new HttpResponseMessage()
{
Content = new StringContent("File uploaded.")
};
}
);
return task;
}
回答by Remy
I had a similar problem for the preview Web API. Did not port that part to the new MVC 4 Web API yet, but maybe this helps:
我在预览 Web API 上遇到了类似的问题。还没有将该部分移植到新的 MVC 4 Web API，但这可能会有所帮助：
REST file upload with HttpRequestMessage or Stream?
使用 HttpRequestMessage 或 Stream 上传 REST 文件？
Please let me know, can sit down tomorrow and try to implement it again.
请让我知道，明天可以坐下来尝试再次实施。
回答by Steve Stokes
I used Mike Wasson's answer before I updated all the NuGets in my webapi mvc4 project. Once I did, I had to re-write the file upload action:
在更新 webapi mvc4 项目中的所有 NuGet 之前，我使用了 Mike Wasson 的回答。完成后，我必须重新编写文件上传操作：
public Task<HttpResponseMessage> Upload(int id)
{
HttpRequestMessage request = this.Request;
if (!request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.UnsupportedMediaType));
}
string root = System.Web.HttpContext.Current.Server.MapPath("~/App_Data/uploads");
var provider = new MultipartFormDataStreamProvider(root);
var task = request.Content.ReadAsMultipartAsync(provider).
ContinueWith<HttpResponseMessage>(o =>
{
FileInfo finfo = new FileInfo(provider.FileData.First().LocalFileName);
string guid = Guid.NewGuid().ToString();
File.Move(finfo.FullName, Path.Combine(root, guid + "_" + provider.FileData.First().Headers.ContentDisposition.FileName.Replace("\"", "")));
return new HttpResponseMessage()
{
Content = new StringContent("File uploaded.")
};
}
);
return task;
}
Apparently BodyPartFileNames is no longer available within the MultipartFormDataStreamProvider.
显然 BodyPartFileNames 在 MultipartFormDataStreamProvider 中不再可用。
回答by Daniel Melo
Toward this same directions, I'm posting a client and server snipets that send Excel Files using WebApi, c# 4:
朝着同样的方向，我发布了一个使用 WebApi，c# 4 发送 Excel 文件的客户端和服务器片段：
public static void SetFile(String serviceUrl, byte[] fileArray, String fileName)
{
try
{
using (var client = new HttpClient())
{
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
using (var content = new MultipartFormDataContent())
{
var fileContent = new ByteArrayContent(fileArray);//(System.IO.File.ReadAllBytes(fileName));
fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
FileName = fileName
};
content.Add(fileContent);
var result = client.PostAsync(serviceUrl, content).Result;
}
}
}
catch (Exception e)
{
//Log the exception
}
}
And the server webapi controller:
和服务器 webapi 控制器：
public Task<IEnumerable<string>> Post()
{
if (Request.Content.IsMimeMultipartContent())
{
string fullPath = HttpContext.Current.Server.MapPath("~/uploads");
MyMultipartFormDataStreamProvider streamProvider = new MyMultipartFormDataStreamProvider(fullPath);
var task = Request.Content.ReadAsMultipartAsync(streamProvider).ContinueWith(t =>
{
if (t.IsFaulted || t.IsCanceled)
throw new HttpResponseException(HttpStatusCode.InternalServerError);
var fileInfo = streamProvider.FileData.Select(i =>
{
var info = new FileInfo(i.LocalFileName);
return "File uploaded as " + info.FullName + " (" + info.Length + ")";
});
return fileInfo;
});
return task;
}
else
{
throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable, "Invalid Request!"));
}
}
And the Custom MyMultipartFormDataStreamProvider, needed to customize the Filename:
PS: I took this code from another post http://www.codeguru.com/csharp/.net/uploading-files-asynchronously-using-asp.net-web-api.htm
和自定义 MyMultipartFormDataStreamProvider，需要自定义文件名：
PS：我从另一篇文章http://www.codeguru.com/csharp/.net/uploading-files-asynchronously-using-asp.net-web-api 中获取此代码.htm
public class MyMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
public MyMultipartFormDataStreamProvider(string path)
: base(path)
{
}
public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
{
string fileName;
if (!string.IsNullOrWhiteSpace(headers.ContentDisposition.FileName))
{
fileName = headers.ContentDisposition.FileName;
}
else
{
fileName = Guid.NewGuid().ToString() + ".data";
}
return fileName.Replace("\"", string.Empty);
}
}
回答by Gleno
I'm surprised that a lot of you seem to want to save files on the server. Solution to keep everything in memory is as follows:
我很惊讶你们中的很多人似乎想在服务器上保存文件。将所有内容保留在内存中的解决方案如下：
[HttpPost("api/upload")]
public async Task<IHttpActionResult> Upload()
{
if (!Request.Content.IsMimeMultipartContent())
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
var provider = new MultipartMemoryStreamProvider();
await Request.Content.ReadAsMultipartAsync(provider);
foreach (var file in provider.Contents)
{
var filename = file.Headers.ContentDisposition.FileName.Trim('\"');
var buffer = await file.ReadAsByteArrayAsync();
//Do whatever you want with filename and its binary data.
}
return Ok();
}
回答by Brent Matzelle
See the code below, adapted from this article, which demonstrates the simplest example code I could find. It includes both file and memory(faster) uploads.
请参阅下面的代码，改编自本文，它演示了我能找到的最简单的示例代码。它包括文件和内存（更快）上传。
public HttpResponseMessage Post()
{
var httpRequest = HttpContext.Current.Request;
if (httpRequest.Files.Count < 1)
{
return Request.CreateResponse(HttpStatusCode.BadRequest);
}
foreach(string file in httpRequest.Files)
{
var postedFile = httpRequest.Files[file];
var filePath = HttpContext.Current.Server.MapPath("~/" + postedFile.FileName);
postedFile.SaveAs(filePath);
// NOTE: To store in memory use postedFile.InputStream
}
return Request.CreateResponse(HttpStatusCode.Created);
}
回答by user3722373
[HttpPost]
public JsonResult PostImage(HttpPostedFileBase file)
{
try
{
if (file != null && file.ContentLength > 0 && file.ContentLength<=10485760)
{
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/") + "HisloImages" + "\", fileName);
file.SaveAs(path);
#region MyRegion
////save imag in Db
//using (MemoryStream ms = new MemoryStream())
//{
// file.InputStream.CopyTo(ms);
// byte[] array = ms.GetBuffer();
//}
#endregion
return Json(JsonResponseFactory.SuccessResponse("Status:0 ,Message: OK"), JsonRequestBehavior.AllowGet);
}
else
{
return Json(JsonResponseFactory.ErrorResponse("Status:1 , Message: Upload Again and File Size Should be Less Than 10MB"), JsonRequestBehavior.AllowGet);
}
}
catch (Exception ex)
{
return Json(JsonResponseFactory.ErrorResponse(ex.Message), JsonRequestBehavior.AllowGet);
}
}
回答by James Lawruk
Here is a quick and dirty solution which takes uploaded file contents from the HTTP body and writes it to a file. I included a "bare bones" HTML/JS snippet for the file upload.
这是一个快速而肮脏的解决方案，它从 HTTP 正文中获取上传的文件内容并将其写入文件。我为文件上传包含了一个“裸机”HTML/JS 片段。
Web API Method:
Web API 方法：
[Route("api/myfileupload")]
[HttpPost]
public string MyFileUpload()
{
var request = HttpContext.Current.Request;
var filePath = "C:\temp\" + request.Headers["filename"];
using (var fs = new System.IO.FileStream(filePath, System.IO.FileMode.Create))
{
request.InputStream.CopyTo(fs);
}
return "uploaded";
}
HTML File Upload:
HTML文件上传：
<form>
<input type="file" id="myfile"/>
<input type="button" onclick="uploadFile();" value="Upload" />
</form>
<script type="text/javascript">
function uploadFile() {
var xhr = new XMLHttpRequest();
var file = document.getElementById('myfile').files[0];
xhr.open("POST", "api/myfileupload");
xhr.setRequestHeader("filename", file.name);
xhr.send(file);
}
</script>
回答by Matt Frear
The ASP.NET Core way is now here:
ASP.NET Core 方式现在在这里：
[HttpPost("UploadFiles")]
public async Task<IActionResult> Post(List<IFormFile> files)
{
long size = files.Sum(f => f.Length);
// full path to file in temp location
var filePath = Path.GetTempFileName();
foreach (var formFile in files)
{
if (formFile.Length > 0)
{
using (var stream = new FileStream(filePath, FileMode.Create))
{
await formFile.CopyToAsync(stream);
}
}
}
// process uploaded files
// Don't rely on or trust the FileName property without validation.
return Ok(new { count = files.Count, size, filePath});
}
回答by Filix Mogilevsky
Here are two ways to accept a file. One using in memory provider MultipartMemoryStreamProviderand one using MultipartFormDataStreamProviderwhich saves to a disk. Note, this is only for one file upload at a time. You can certainty extend this to save multiple-files. The second approach can support large files. I've tested files over 200MB and it works fine. Using in memory approach does not require you to save to disk, but will throw out of memory exception if you exceed a certain limit.
这里有两种接受文件的方法。一种在内存提供程序MultipartMemoryStreamProvider 中使用，另一种使用MultipartFormDataStreamProvider保存到磁盘。请注意，这仅适用于一次上传一个文件。您可以肯定地扩展它以保存多个文件。第二种方法可以支持大文件。我已经测试了超过 200MB 的文件，它运行良好。使用 in memory 方法不需要您保存到磁盘，但如果超过某个限制，则会抛出 out of memory 异常。
private async Task<Stream> ReadStream()
{
Stream stream = null;
var provider = new MultipartMemoryStreamProvider();
await Request.Content.ReadAsMultipartAsync(provider);
foreach (var file in provider.Contents)
{
var buffer = await file.ReadAsByteArrayAsync();
stream = new MemoryStream(buffer);
}
return stream;
}
private async Task<Stream> ReadLargeStream()
{
Stream stream = null;
string root = Path.GetTempPath();
var provider = new MultipartFormDataStreamProvider(root);
await Request.Content.ReadAsMultipartAsync(provider);
foreach (var file in provider.FileData)
{
var path = file.LocalFileName;
byte[] content = File.ReadAllBytes(path);
File.Delete(path);
stream = new MemoryStream(content);
}
return stream;
}
