[英]JsonResult return Json in ASP.NET CORE 2.1
Controller that worked in ASP.NET Core 2.0: 在ASP.NET Core 2.0中工作的控制器:
[Produces("application/json")]
[Route("api/[controller]")]
[ApiController]
public class GraficResourcesApiController : ControllerBase
{
private readonly ApplicationDbContext _context;
public GraficResourcesApiController(ApplicationDbContext context)
{
_context = context;
}
[HttpGet]
public JsonResult GetGrafic(int ResourceId)
{
var sheduling = new List<Sheduling>();
var events = from e in _context.Grafic.Where(c=>c.ResourceId == ResourceId)
select new
{
id = e.Id,
title = e.Personals.Name,
start = e.DateStart,
end = e.DateStop,
color = e.Personals.Color,
personalId = e.PersonalId,
description = e.ClientName
};
var rows = events.ToArray();
return Json(rows);
}
}
in ASP.NET Core 2.1 在ASP.NET Core 2.1中
return Json (rows);
writes that Json does not exist in the current context. 写道Json在当前上下文中不存在。 If we remove Json leaving simply
如果我们删除Json,就简单地离开
return rows;
then writes that it was not possible to explicitly convert the type List () to JsonResult 然后写道,不可能将类型List()显式转换为JsonResult
How to convert to Json now? 如何立即转换为Json?
In asp.net-core-2.1 ControllerBase
does not have a Json(Object)
method. 在asp.net-core-2.1中,
ControllerBase
没有Json(Object)
方法。 However Controller
does. 但是,
Controller
可以。
So either refactor the current controller to be derived from Controller
因此,要么将当前控制器重构为从
Controller
派生
public class GraficResourcesApiController : Controller {
//...
}
to have access to the Controller.Json
Method or you can initialize a new JsonResult
yourself in the action 可以访问
Controller.Json
方法 ,也可以在操作中自己初始化一个新的JsonResult
return new JsonResult(rows);
which is basically what the method does internally in Controller
这基本上是该方法在
Controller
内部进行的操作
/// <summary>
/// Creates a <see cref="JsonResult"/> object that serializes the specified <paramref name="data"/> object
/// to JSON.
/// </summary>
/// <param name="data">The object to serialize.</param>
/// <returns>The created <see cref="JsonResult"/> that serializes the specified <paramref name="data"/>
/// to JSON format for the response.</returns>
[NonAction]
public virtual JsonResult Json(object data)
{
return new JsonResult(data);
}
/// <summary>
/// Creates a <see cref="JsonResult"/> object that serializes the specified <paramref name="data"/> object
/// to JSON.
/// </summary>
/// <param name="data">The object to serialize.</param>
/// <param name="serializerSettings">The <see cref="JsonSerializerSettings"/> to be used by
/// the formatter.</param>
/// <returns>The created <see cref="JsonResult"/> that serializes the specified <paramref name="data"/>
/// as JSON format for the response.</returns>
/// <remarks>Callers should cache an instance of <see cref="JsonSerializerSettings"/> to avoid
/// recreating cached data with each call.</remarks>
[NonAction]
public virtual JsonResult Json(object data, JsonSerializerSettings serializerSettings)
{
if (serializerSettings == null)
{
throw new ArgumentNullException(nameof(serializerSettings));
}
return new JsonResult(data, serializerSettings);
}
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