将文件发布到 ApiController

Question

我是 ApiControllers 的新手，我正在测试将文件从客户端发送到 api 并从此处保存。下面的代码确实有效，文件从客户端“复制”到服务器，但响应消息是“204 No Content” - 我应该注意这里的任何事情吗？

ApiController 代码是这样的：

    [HttpPost]      
    [Route("api/UploadFile")]     
    public async Task UploadFile()
    {
        string fileName = "myfilename.txt";
        Stream requestStream = await Request.Content.ReadAsStreamAsync();        

        using (FileStream fileStream = File.Create(@"C:\myDropFolder\" + fileName))
        {
            await requestStream.CopyToAsync(fileStream);
        }         
    }

对客户端的调用如下所示，从控制台应用程序进行测试：

 class Program
{
    static void Main(string[] args)
    {
        try
        {
            HttpWebRequest request = (HttpWebRequest)HttpWebRequest.Create("http://localhost:1741/api/UploadFile/");
            request.Method = WebRequestMethods.Http.Post;

            byte[] fileToSend = File.ReadAllBytes(@"C:\myDropFolder\tester.txt");
            request.ContentLength = fileToSend.Length;

            using (Stream requestStream = request.GetRequestStream())
            {
                // Send the file as body request. 
                requestStream.Write(fileToSend, 0, fileToSend.Length);
                requestStream.Close();
            }

            using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
                Console.WriteLine("HTTP/{0} {1} {2}", response.ProtocolVersion, (int)response.StatusCode, response.StatusDescription);
                Console.ReadLine();            

        }
        catch (Exception)
        {
            throw;
        }
    }
}

Answer 1

Web API 对于 POST 的默认响应状态代码是 204，对于 GET 是 200 (Action Results in Web API 2)。除非你返回IHttpActionResult，然后显式返回200：

[HttpPost]      
[Route("api/UploadFile")]     
public async Task<IHttpActionResult> UploadFile()
{
    string fileName = "myfilename.txt";
    Stream requestStream = await Request.Content.ReadAsStreamAsync();        

    using (FileStream fileStream = File.Create(@"C:\myDropFolder\" + fileName))
    {
        await requestStream.CopyToAsync(fileStream);
    }
    return Ok(); 
}