结合actionresult和jsonresult
我想做这个：
public ActionResult Details(int id)
{
Object ent = new{ prop1 = 1, prop2 = 2};
if (Request.AcceptTypes.Contains("application/json"))
return Json(ent, JsonRequestBehavior.AllowGet);
ViewData.Model = ent;
return View();
}
没有找到相关结果
已邀请:
1 个回复
姬第柔炒
[IsJsonRequest] public abstract class BaseController : Controller { public bool IsJsonRequest { get; set; } } The ActionFilterAttribute public class IsJsonRequest: ActionFilterAttribute { public override void OnActionExecuting(ActionExecutingContext filterContext) { var myController = filterContext.Controller as MyController; if (myController != null) { if (filterContext.HttpContext.Request.AcceptTypes.Contains("application/json")) { myController.IsJsonRequest = true; } else { myController.IsJsonRequest = false; } } } } public class TestController : BaseController { public ActionResult Details(int id) { if (IsJsonRequest) return Json Data else return view } }