如何接受文件POST

我正在使用asp.net mvc 4 webapi beta来构建一个休息服务。我需要能够接受张贴的图像/文件从客户端应用程序。这可能使用webapi吗?下面是如何行动，我目前正在使用。有人知道一个例子吗?

[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";
}

当前回答

这里有两种接受文件的方法。一个使用内存提供程序MultipartMemoryStreamProvider，另一个使用保存到磁盘的MultipartFormDataStreamProvider。注意，这一次只针对一个文件上传。您可以扩展它来保存多个文件。第二种方法可以支持大文件。我测试过超过200MB的文件，它工作得很好。使用内存方法不需要保存到磁盘，但如果超过一定的限制，将抛出内存外异常。

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;
}

2017-10-02 15:25:27

其他回答

我在预览Web API时遇到了类似的问题。还没有将这部分移植到新的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，可以坐下来，明天再试着实施它。

2012-04-25 16:58:22

请参阅http://www.asp.net/web-api/overview/formats-and-model-binding/html-forms-and-multipart-mime#multipartmime，尽管我认为这篇文章让它看起来比实际情况更复杂。

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; 
}

2012-04-26 05:39:03

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);
}

2013-12-03 16:33:26

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);
}

2018-11-12 15:41:41

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;
}

2017-10-02 15:25:27

如何接受文件POST

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