无法将类对象添加到实体框架模型

Question

我正在创建一个发布 API 来将数据存储在数据库中。我创建了一个类并将类成员数据添加到模型中，但出现错误

给出错误 - 无法从“System.Web.Mvc.JsonRequestBehavior”转换为“Newtonsoft.Json.JsonSerializerSettings”

public ActionResult AddTemprature([FromBody]tempretureViewModel paramTemplate)
{
    try
    {
        DateTime currentTime = DateTime.Now;
        using (smartpondEntities DB = new smartpondEntities())
        {
            WebApi.Temperature temptable = new WebApi.Temperature();
            // and you need paramTemplate props to template object
            DB.Temperatures.Add(temptable);
            DB.SaveChanges();
            return Json(true, JsonRequestBehavior.AllowGet);
        }
    }
    catch (Exception Ex)
    {

    }
    return Json(false, JsonRequestBehavior.AllowGet);
}

public  class tempretureViewModel
{

    public decimal? temp1 { get; set; }
    public decimal? PHValue { get; set; }
    public decimal? DOValue { get; set; }
    public DateTime? currentTime { get; set; }
}

WebApiConfig

 public static void Register(HttpConfiguration config)
    {


        // Web API configuration and services

        // Web API routes
        config.MapHttpAttributeRoutes();

        config.Routes.MapHttpRoute(
            name: "DefaultApi",
            routeTemplate: "api/{controller}/{id}",
            defaults: new { id = RouteParameter.Optional }
        );
        GlobalConfiguration.Configuration.Formatters.Clear();
        GlobalConfiguration.Configuration.Formatters.Add(new System.Net.Http.Formatting.JsonMediaTypeFormatter());
    }

Answer 1

它与 asp.net、json 实体框架、模型视图控制器、asp.net-web-api2 无关。它是关于 c# 的。 WebApi.Controllers.tempretureViewModel'和'WebApi.Temperature'”不同的类型。

public ActionResult AddTemprature([FromBody]tempretureViewModel paramTemplate)
        {
            try
            {
                currentTime = DateTime.Now;
                using (smartpondEntities DB = new smartpondEntities())
                {
                    WebApi.Temperature temptable = new WebApi.Temperature();
                    // and you need to set paramTemplate props to template object

                    DB.Temperatures.Add(temptable);
                    DB.SaveChanges();
                    return Json(true, JsonRequestBehavior.AllowGet);
                }
            }
            catch (Exception Ex)
            {

            }
            return Json(false, JsonRequestBehavior.AllowGet);
        }

更新：WebApiConfig.cs

GlobalConfiguration.Configuration.Formatters.Clear();
GlobalConfiguration.Configuration.Formatters.Add(new System.Net.Http.Formatting.JsonMediaTypeFormatter());

Answer 2

使用类上面的命名空间而不是使用类 我用粗体标记了它。 WebApi.Temperature temptable = new WebApi.Temperature(); 添加命名空间

public ActionResult AddTemprature([FromBody]tempretureViewModel paramTemplate)
{
    try
    {
        DateTime currentTime = DateTime.Now;
        using (smartpondEntities DB = new smartpondEntities())
        {
            **WebApi.Temperature temptable = new WebApi.Temperature();**
            // and you need paramTemplate props to template object
            DB.Temperatures.Add(temptable);
            DB.SaveChanges();
            return Json(true, JsonRequestBehavior.AllowGet);
        }
    }
    catch (Exception Ex)
    {

    }
    return Json(false, JsonRequestBehavior.AllowGet);
}

public  class tempretureViewModel
{

    public decimal? temp1 { get; set; }
    public decimal? PHValue { get; set; }
    public decimal? DOValue { get; set; }
    public DateTime? currentTime { get; set; }
}