我遇到如何制作通用控制器的问题。我有Model,id属性可以是string或int的类型。当我想编辑一个id为int的模型时,我找不到导致错误的动作,因为我实现了 公共抽象ActionResult Edit(字符串id)的覆盖方法。
存在一种传递我的路线可以承受的通用参数(字符串或整数)的方法吗?
public abstract class BaseController<T> : Controller
{
//protected UnitOfWork unitOfWork = new UnitOfWork(System.Web.HttpContext.Current.User.Identity.Name);
protected UnitOfWork unitOfWork = new UnitOfWork("");
public abstract ActionResult List();
public abstract ActionResult Index();
public abstract ActionResult Details(string id);,
public abstract ActionResult CreateByModal();
public abstract ActionResult Create();
[HttpPost]
[ValidateAntiForgeryToken]
public abstract ActionResult Create(T entity);
public abstract ActionResult Edit(string id);
[HttpPost]
[ValidateAntiForgeryToken]
public abstract ActionResult Edit(T entity);
public abstract ActionResult Delete(string id);
protected override void Dispose(bool disposing)
{
//db.Dispose();
base.Dispose(disposing);
}
}
public class ModeloController : BaseController<Modelo>
{
public override ActionResult Edit(string id)
{
//Here i have a int Id im my Model and the error in my route
Modelo modelo = unitOfWork.ModeloRepository.Find(id);
if (modelo == null)
{
return HttpNotFound();
}
return PartialView("_Edit", modelo);
}
}
public class GeneroController : BaseController<Genero>
{
//Here i have a string Id im my Model
public override ActionResult Edit(string id)
{
Genero genero = unitOfWork.GeneroRepository.Find(id);
if (genero == null)
{
return HttpNotFound();
}
return PartialView("_Edit", genero);
}
}
答案 0 :(得分:2)
一种选择是重载具有id
作为参数的每个方法:
public abstract ActionResult Details(string id);
public abstract ActionResult Details(int id);
或者为id
的类型添加第二个通用参数:
public abstract class BaseController<T, U> : Controller
{
public abstract ActionResult Details(U id);
public abstract ActionResult Edit(U id);
无论哪种方式,您都要对来电者承担更多责任,以确保使用正确的类型。
答案 1 :(得分:1)
您可以添加其他泛型类型参数来指示给定类的键类型。
public abstract class AbstractController<TEntity, TId> : Controller
{
// irrelevant stuff omitted
public ActionResult Edit(TEntity id);
}