如何从action属性返回json对象?

public override void OnActionExecuting(ActionExecutingContext filterContext)
{
   if (/* whatever */)
   {
      var result = new JsonResult();
      result.Data = /* json data */;
      filterContext.Result = result;
      return;
   }

   base.OnActionExecuting(filterContext);
   return;
}

我觉得它很有用 Json.NET 生成json输出。这有很多优点,例如json属性可以在某些条件下隐藏。

public override void OnActionExecuting(ActionExecutingContext filterContext)
{
    if (/* whatever */)
    {
        var result = new ResultModel(); // your json model
        ContentResult content = new ContentResult();
        content.ContentType = "application/json";
        content.Content = JsonConvert.SerializeObject(result);
        filterContext.Result = content;
        base.OnActionExecuting(filterContext);
    }
}