Question

拥有此代码会给我一个警告：Because this call is not awaited, execution of the current method continues before the call is completed. Consider applying the 'await' operator to the result of the call.

    public async Task<ActionResult> Details(Guid id)
    {
        var calendarEvent = await service.FindByIdAsync(id);
        if (calendarEvent == null) return RedirectToAction<CalendarController>(c => c.Index());
        var model = new CalendarEventPresentation(calendarEvent);
        ViewData.Model = model;
        return View();
    }

    public async Task<RedirectResult> Create(CalendarEventBindingModel binding)
    {
        var model = service.Create(binding);
        await context.SaveChangesAsync();
        return this.RedirectToAction<CalendarController>(c => c.Details(model.CalendarEventID));
    }

如果我确实添加了await运算符：

Error CS4034 The 'await' operator can only be used within an async lambda expression. Consider marking this lambda expression with the 'async' modifier.

如果我像这样添加async修饰符：

return this.RedirectToAction<CalendarController>(async c => await c.Details(model.CalendarEventID));

错误为Error CS1989 Async lambda expressions cannot be converted to expression trees

那么如何在异步控制器中使用强类型RedirectToAction（我使用的是MVC Futures）？

Answer 1

我已经弄清楚了！

我有一个基本控制器，还有一个用于异步重定向的扩展方法

protected ActionResult RedirectToAsyncAction<TController>(Expression<Func<TController, Task<ActionResult>>> action)
        where TController : Controller
    {
        Expression<Action<TController>> convertedFuncToAction = Expression.Lambda<Action<TController>>(action.Body, action.Parameters.First());
        return ControllerExtensions.RedirectToAction(this, convertedFuncToAction);
    }

这将防止警告。然后，您只需从Controller调用RedirectToAsyncAction。

public ActionResult MyAction()
    {
       // Your code
        return RedirectToAsyncAction<MyController>(c => c.MyAsyncAction(params,..)); // no warnings here
    }

使用异步控制器的强类型RedirectToAction（Futures）

1 个答案: