c# - 如何接受文件 POST

Question

我正在使用 asp.net mvc 4 webapi beta 来构建一个休息服务。我需要能够接受来自客户端应用程序的 POST 图像/文件。这可以使用webapi吗？以下是我目前正在使用的操作。有谁知道这个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";
}

score 398 · Accepted Answer

我很惊讶你们中的很多人似乎都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("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();
}

score 177 · Accepted Answer

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使它看起来比确实如此。

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; 
}

score 118 · Accepted Answer

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);
}

score 27 · Accepted Answer

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});
}

score 16 · Accepted Answer

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“基本”HTML/JS 片段。

网络 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>

score 13 · Accepted Answer

在我更新我的 webapi mvc4 项目中的所有 NuGet 之前，我使用了 Mike Wasson 的答案。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;
    }

显然 BodyPartFileNames 在 MultipartFormDataStreamProvider 中不再可用。

score 10 · Accepted Answer

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
    }
}

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!"));
    }
}

还有自定义 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中获取了这段代码.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);
    }
}

score 6 · Accepted Answer

[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);

    }
}

score 4 · Accepted Answer

这里有两种接受文件的方法。一种使用内存提供程序MultipartMemoryStreamProvider，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，它工作正常。使用内存中的方法不需要你保存到磁盘，但是如果超过一定的限制就会抛出内存不足的异常。

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;
}

score 2 · Accepted Answer

即使对于 .Net Core，这个问题也有很多很好的答案。我正在使用两个框架，提供的代码示例工作正常。所以我不会重复它。就我而言，重要的是如何使用Swagger的文件上传操作，如下所示：

这是我的回顾：

ASP .Net WebAPI 2

上传文件使用：MultipartFormDataStreamProvider在这里查看答案
如何与 Swagger 一起使用

.NET 核心

要上传文件使用：IFormFile请参阅此处的答案或MS 文档
如何与 Swagger 一起使用

score 1 · Accepted Answer

对于预览版 Web API，我遇到了类似的问题。尚未将该部分移植到新的 MVC 4 Web API，但这可能会有所帮助：

使用 HttpRequestMessage 或 Stream 上传 REST 文件？

请让我知道，明天可以坐下来尝试再次实施。

score 1 · Accepted Answer

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);
}

score 0 · Accepted Answer

补充 Matt Frear 的回答 - 这将是一个 ASP NET Core 替代方案，用于直接从 Stream 读取文件，而不是从磁盘保存和读取它：

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}");
        }
    }

c# - 如何接受文件 POST

13 回答 13

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