我试图返回一个状态代码304未修改的web api控制器中的GET方法。

我成功的唯一方法是这样的:

public class TryController : ApiController
{
    public User GetUser(int userId, DateTime lastModifiedAtClient)
    {
        var user = new DataEntities().Users.First(p => p.Id == userId);
        if (user.LastModified <= lastModifiedAtClient)
        {
             throw new HttpResponseException(HttpStatusCode.NotModified);
        }
        return user;
    }
}

这里的问题是,它不是一个异常,它只是没有修改,所以客户端缓存是OK的。 我还希望返回类型是一个用户(所有的web api示例显示与GET)不返回HttpResponseMessage或类似的东西。


当前回答

我讨厌撞旧文章,但这是在谷歌搜索的第一个结果,我有一个问题(即使有你们的支持)。所以这里什么都没有……

希望我的解决方案能帮助那些同样感到困惑的人。

namespace MyApplication.WebAPI.Controllers
{
    public class BaseController : ApiController
    {
        public T SendResponse<T>(T response, HttpStatusCode statusCode = HttpStatusCode.OK)
        {
            if (statusCode != HttpStatusCode.OK)
            {
                // leave it up to microsoft to make this way more complicated than it needs to be
                // seriously i used to be able to just set the status and leave it at that but nooo... now 
                // i need to throw an exception 
                var badResponse =
                    new HttpResponseMessage(statusCode)
                    {
                        Content =  new StringContent(JsonConvert.SerializeObject(response), Encoding.UTF8, "application/json")
                    };

                throw new HttpResponseException(badResponse);
            }
            return response;
        }
    }
}

然后从BaseController继承

[RoutePrefix("api/devicemanagement")]
public class DeviceManagementController : BaseController
{...

然后使用它

[HttpGet]
[Route("device/search/{property}/{value}")]
public SearchForDeviceResponse SearchForDevice(string property, string value)
{
    //todo: limit search property here?
    var response = new SearchForDeviceResponse();

    var results = _deviceManagementBusiness.SearchForDevices(property, value);

    response.Success = true;
    response.Data = results;

    var statusCode = results == null || !results.Any() ? HttpStatusCode.NoContent : HttpStatusCode.OK;

    return SendResponse(response, statusCode);
}

其他回答

我不知道答案所以问ASP。NET团队。

因此,诀窍是将签名更改为HttpResponseMessage并使用Request.CreateResponse。

[ResponseType(typeof(User))]
public HttpResponseMessage GetUser(HttpRequestMessage request, int userId, DateTime lastModifiedAtClient)
{
    var user = new DataEntities().Users.First(p => p.Id == userId);
    if (user.LastModified <= lastModifiedAtClient)
    {
         return new HttpResponseMessage(HttpStatusCode.NotModified);
    }
    return request.CreateResponse(HttpStatusCode.OK, user);
}

试试这个:

return new ContentResult() { 
    StatusCode = 404, 
    Content = "Not found" 
};

如果您希望将操作签名保留为返回User,还可以执行以下操作:

public User GetUser(int userId, DateTime lastModifiedAtClient) 

如果你想返回200以外的值,那么你在你的动作中抛出一个HttpResponseException,并传递你想要发送给客户端的HttpResponseMessage。

.net core 2.2返回304状态码。这是使用ApiController。

    [HttpGet]
    public ActionResult<YOUROBJECT> Get()
    {
        return StatusCode(304);
    }

您还可以选择返回带有响应的对象

    [HttpGet]
    public ActionResult<YOUROBJECT> Get()
    {
        return StatusCode(304, YOUROBJECT); 
    }

我知道这里有几个很好的答案,但这是我需要的,所以我想我应该添加这个代码,以防任何人需要返回任何状态代码和响应体,他们想在4.7。x与webAPI。

public class DuplicateResponseResult<TResponse> : IHttpActionResult
{
    private TResponse _response;
    private HttpStatusCode _statusCode;
    private HttpRequestMessage _httpRequestMessage;
    public DuplicateResponseResult(HttpRequestMessage httpRequestMessage, TResponse response, HttpStatusCode statusCode)
    {
        _httpRequestMessage = httpRequestMessage;
        _response = response;
        _statusCode = statusCode;
    }

    public Task<HttpResponseMessage> ExecuteAsync(CancellationToken cancellationToken)
    {
        var response = new HttpResponseMessage(_statusCode);
        return Task.FromResult(_httpRequestMessage.CreateResponse(_statusCode, _response));
    }
}