How can I make a Controller Action take a dynamic parameter?
I would like to be able to post any serialized object to an action method and instantiate a new object of the posted type in order to use TryUpdateModel
. They didn't teach me any of this stuff in the QBasic help file... How can I instantiate the unknown type based on the posted data?
If it would help, I could theoretically include the name of the type as a string in the posted data. I was hoping to avoid that because it seemed like I would need the full name of the type.
public void Save(object/dynamic whatever, string typename) {
//Instantiate posted type
//TryUpdateModel
context.Entry(Thing).State = EntityState.Modified;
context.SaveChanges();
}
Here is an example of a serialized object
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
From Fiddler
Thing.Id 1
Thing.Name blah
Thing.OptionID 1
Thing.ListItems.index开发者_运维知识库 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
You could write a custom model binder which uses reflection and the typeName
parameter:
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;
}
}
and then simply:
[HttpPost]
public ActionResult Save([ModelBinder(typeof(MyModelBinder))] object model)
{
context.Entry(model).State = EntityState.Modified;
context.SaveChanges();
return View();
}
精彩评论