来自HttpRequestMessage内容的文件名

时间:2013-02-18 13:52:15

标签: asp.net-mvc rest post file-upload asp.net-web-api

我实现了POST Rest服务以将文件上传到我的服务器。我现在的问题是我想按类型限制上传的文件。比方说,我只想允许上传.pdf文件。

我试图做的是

            Task<Stream> task = this.Request.Content.ReadAsStreamAsync();
            task.Wait();
            FileStream requestStream = (FileStream)task.Result;

但遗憾的是,无法将Stream强制转换为FileStream并通过requestStream.Name访问该类型。

是否有一种简单的方法(除了将流写入磁盘并检查类型)以获取文件类型?

1 个答案:

答案 0 :(得分:8)

如果您将文件上传到Web API并且想要访问文件数据(Content-Disposition),则应将该文件作为MIME多部分(multipart/form-data)上传。

Here我展示了一些关于如何从HTML表单,Javascript和.NET上传的例子。

然后你可以这样做,这个例子只检查pdf / doc文件:

public async Task<HttpResponseMessage> Post()
    {
        if (!Request.Content.IsMimeMultipartContent())
        {
            throw new HttpResponseException(Request.CreateResponse(HttpStatusCode.NotAcceptable,
                                                                   "This request is not properly formatted - not multipart."));
        }

        var provider = new RestrictiveMultipartMemoryStreamProvider();

        //READ CONTENTS OF REQUEST TO MEMORY WITHOUT FLUSHING TO DISK
        await Request.Content.ReadAsMultipartAsync(provider);

        foreach (HttpContent ctnt in provider.Contents)
        {
            //now read individual part into STREAM
            var stream = await ctnt.ReadAsStreamAsync();

            if (stream.Length != 0)
            {
                using (var ms = new MemoryStream())
                {
                    //do something with the file memorystream
                }
            }
        }
        return Request.CreateResponse(HttpStatusCode.OK);
    }
}

public class RestrictiveMultipartMemoryStreamProvider : MultipartMemoryStreamProvider
{
    public override Stream GetStream(HttpContent parent, HttpContentHeaders headers)
    {
        var extensions = new[] {"pdf", "doc"};
        var filename = headers.ContentDisposition.FileName.Replace("\"", string.Empty);

        if (filename.IndexOf('.') < 0)
            return Stream.Null;

        var extension = filename.Split('.').Last();

        return extensions.Any(i => i.Equals(extension, StringComparison.InvariantCultureIgnoreCase))
                   ? base.GetStream(parent, headers)
                   : Stream.Null;

    }
}