Question

我正在使用asp.net mvc 4 webapi beta来构建休息服务。我需要能够从客户端应用程序接受POSTed图像/文件。这可能使用webapi吗？以下是我目前使用的操作方式。有谁知道这应该如何运作的例子？

[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";
}

Answer 1

我很惊讶很多人似乎想在服务器上保存文件。将所有内容保存在内存中的解决方案如下：

[HttpPost, Route("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();
}

Answer 2

请参阅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看起来比实际情况要复杂一些。

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; 
}

Answer 3

请参阅下面的代码，该代码改编自this article，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);
}

Answer 4

在我更新webapi mvc4项目中的所有NuGets之前，我使用了Mike Wasson的答案。完成后，我不得不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;
    }

显然，MultipartFormDataStreamProvider中不再提供BodyPartFileNames。

Answer 5

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。我包括了一个＆＃34;裸骨＆＃34;用于文件上传的HTML / JS代码段。

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>

Answer 6

ASP.NET Core方式现在是here：

[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});
}

Answer 7

朝着同样的方向，我发布了一个使用WebApi发送Excel文件的客户端和服务器snipets，c＃4：

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
    }
}

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!"));
    }
}

自定义MyMultipartFormDataStreamProvider，需要自定义文件名：

PS：我从另一篇帖子http://www.codeguru.com/csharp/.net/uploading-files-asynchronously-using-asp.net-web-api.htm

中获取了此代码

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);
    }
}

Answer 8

[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);

    }
}

Answer 9

以下是两种接受文件的方法。一个使用内存提供程序 MultipartMemoryStreamProvider ，另一个使用 MultipartFormDataStreamProvider 保存到磁盘。请注意，这仅适用于一次上传一个文件。您可以确定地扩展它以保存多个文件。第二种方法可以支持大文件。我测试了超过200MB的文件，它工作正常。在内存中使用方法不需要您保存到磁盘，但如果超过某个限制，则会抛出内存不足。

        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;
    }

Answer 10

我遇到了类似的预览Web API问题。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?

请让我知道，明天可以坐下来尝试再次实施。

Answer 11

即使对于.Net Core，此问题也有很多很好的答案。我使用了两个框架，提供的代码示例运行正常。所以我不再重复。就我而言，重要的是如何使用 Swagger 这样的文件上传操作：

这是我的回顾：

ASP .Net WebAPI 2

要上传文件，请使用： MultipartFormDataStreamStreamProvider 在此处查看答案
如何use it with Swagger

.NET Core

要上传文件，请使用： IFormFile 在此处查看答案或MS documentation
如何use it with Swagger

Answer 12

API控制器：

[HttpPost]
public HttpResponseMessage Post()
    {
           var httpRequest = System.Web.HttpContext.Current.Request;

            if (System.Web.HttpContext.Current.Request.Files.Count < 1)
            {
                //TODO
            }
            else
            {

                try
                { 
                    foreach (string file in httpRequest.Files)
                    { 
                        var postedFile = httpRequest.Files[file];
                        BinaryReader binReader = new BinaryReader(postedFile.InputStream);
                        byte[] byteArray = binReader.ReadBytes(postedFile.ContentLength);

                    }

                }
                catch (System.Exception e)
                {
                   //TODO
                }
            }
 return Request.CreateResponse(HttpStatusCode.Created);
}

Answer 13

补充Matt Frear的答案-这是ASP NET Core的替代方案，可直接从Stream读取文件，而无需从磁盘保存和读取文件：

public ActionResult OnPostUpload(List<IFormFile> files)
    {
        try
        {
            var file = files.FirstOrDefault();
            var inputstream = file.OpenReadStream();

            XSSFWorkbook workbook = new XSSFWorkbook(stream);

            var FIRST_ROW_NUMBER = {{firstRowWithValue}};

            ISheet sheet = workbook.GetSheetAt(0);
            // Example: var firstCellRow = (int)sheet.GetRow(0).GetCell(0).NumericCellValue;

            for (int rowIdx = 2; rowIdx <= sheet.LastRowNum; rowIdx++)
               {
                  IRow currentRow = sheet.GetRow(rowIdx);

                  if (currentRow == null || currentRow.Cells == null || currentRow.Cells.Count() < FIRST_ROW_NUMBER) break;

                  var df = new DataFormatter();                

                  for (int cellNumber = {{firstCellWithValue}}; cellNumber < {{lastCellWithValue}}; cellNumber++)
                      {
                         //business logic & saving data to DB                        
                      }               
                }
        }
        catch(Exception ex)
        {
            throw new FileFormatException($"Error on file processing - {ex.Message}");
        }
    }

如何接受文件POST

13 个答案: