如何使Controller Action采用动态参数？

Question

我希望能够将任何序列化对象发布到操作方法并实例化已发布类型的新对象以便使用TryUpdateModel.他们没有在QBasic帮助文件中教我任何这些东西 ......如何根据发布的数据实例化未知类型？

如果它有用,我理论上可以在发布的数据中包含类型的名称作为字符串.我希望避免这种情况,因为我似乎需要该类型的全名.

public void Save(object/dynamic whatever, string typename) {
    //Instantiate posted type
    //TryUpdateModel
    context.Entry(Thing).State = EntityState.Modified;
    context.SaveChanges();
}

以下是序列化对象的示例

Thing.Id=1&Thing.Name=blah&Thing.OptionID=1&Thing.ListItems.index=1&Thing.ListItems%5B1%5D.Id=1&Thing.ListItems%5B1%5D.Name=whatever&Thing.ListItems%5B1%5D.OptionID=2&Thing.ListItems%5B1%5D.ThingID=1&Thing.ListItems%5B1%5D.EntityState=16

来自Fiddler

Thing.Id                            1
Thing.Name                          blah
Thing.OptionID                      1
Thing.ListItems.index               1
Thing.ListItems[1].Id               1
Thing.ListItems[1].Name             whatever
Thing.ListItems[1].OptionID         2
Thing.ListItems[1].ThingID          1
Thing.ListItems[1].EntityState      16

Answer 1

您可以编写一个使用反射和typeName参数的自定义模型绑定器:

public class MyModelBinder : DefaultModelBinder
{
    protected override object CreateModel(ControllerContext controllerContext, ModelBindingContext bindingContext, Type modelType)
    {
        var typeValue = bindingContext.ValueProvider.GetValue("typename");
        if (typeValue == null)
        {
            throw new Exception("Impossible to instantiate a model. The \"typeName\" query string parameter was not provided.");
        }
        var type = Type.GetType(
            (string)typeValue.ConvertTo(typeof(string)),
            true
        );
        var model = Activator.CreateInstance(type);
        bindingContext.ModelMetadata = ModelMetadataProviders.Current.GetMetadataForType(() => model, type);
        return model;
    }
}

然后简单地说:

[HttpPost]
public ActionResult Save([ModelBinder(typeof(MyModelBinder))] object model) 
{
    context.Entry(model).State = EntityState.Modified;
    context.SaveChanges();
    return View();
}