ASP.net Cast FindControl output to a parameter

Question

I want to cast the output of FindControl method to a control of a specific type but I want the type itself being passed as an argument. I first tried:

public static List<T> GetList<T>(..., Type tp, ...)
{
...
tp castedCtrl = (panel.FindControl(ctrlPrefixName + i.ToString()) as tp);
...
}

This is my preferred result but doesn't compile at all. I also tried :

dynamic castedCtrl = Convert.ChangeType(panel.FindControl(ctrlPrefixName + i.ToString()), tp);

but this requires all my tp 's implement IConvertible interface which is not desired and required a lot of unintended work.

How can I cast the output of FindControl output to my desired type (which in action I want to pass the name of a user control class).

Does the type change frequently that you want to pass it as an argument every time? — Dania, Mar 16 '16 at 08:38

Lesmian · Accepted Answer · 2016-03-16T08:42:19.257

1

Here you go:

public static List<T> GetList<T, U>(Page page, string ctrlPrefixName) 
   where T : class 
   where U : Control
{
   int i = 1;
   U castedCtrl = page.FindControl(ctrlPrefixName + i.ToString()) as U;
   return new List<T>();
}

In order to use it:

GetList<MyType, DropDownList>(this, "MyControl_");

edited Mar 16 '16 at 08:42

answered Mar 16 '16 at 08:31

Lesmian

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Please consider T is a seperate class – Farshid Mar 16 '16 at 08:32
T in my code actually represents that ascx's data object – Farshid Mar 16 '16 at 08:33
So you can write where T : class and then you can use any class you want. But if you look for control your class will need to inherit from Control class anyway. – Lesmian Mar 16 '16 at 08:34
I want to pass type among arguments, T is used for a diffrent purpose (to pack that control's info in an object) – Farshid Mar 16 '16 at 08:34
You cannot use type as variable, but you can add a second generic type to method declaration. – Lesmian Mar 16 '16 at 08:42
Thanks @Lesiman, This is the approach I also reached and truly works – Farshid Mar 16 '16 at 08:49

ASP.net Cast FindControl output to a parameter

1 Answers1