Inherit from Wizard control

[Martin]

Hi,

I want to inherit from the Wizard control in ASP.Net 2.0, to add some
properties, events and modify/replace the default template for navigation.

I've started with a very simple class:
public class SequencedWizard:Wizard
{
public SequencedWizard()
{
//
// TODO: Add constructor logic here
//
}
protected override void OnInit(EventArgs e)
{
base.OnInit(e);
// modify default templates here
}
}

When I use this control in an aspx page using
<%@ Register Assembly="WebControlsLib" Namespace="WebControlsLib"
TagPrefix="cc1" %>

and

<cc1:SequencedWizard ID="SequencedWizard1"
runat="server"></cc1:SequencedWizard>

I get the message "Unknown server tag 'cc1:SequencedWizard'

How should I trouble shoot this?
I'm not expecting to write any Designer controls - hoping I get the Wizard
designer controls "for free". Is that right?

Are there any articles describing how to extend the Wizard control like
this?

Thanks
Martin

[Martin]

Durrr

Mix up over namespaces


"Martin" <x@y.z> wrote in message
news:uCwBoW8iGHA.4344@TK2MSFTNGP05.phx.gbl...

> Hi,
>
> I want to inherit from the Wizard control in ASP.Net 2.0, to add some
> properties, events and modify/replace the default template for navigation.
>
> I've started with a very simple class:
> public class SequencedWizard:Wizard
> {
> public SequencedWizard()
> {
> //
> // TODO: Add constructor logic here
> //
> }
> protected override void OnInit(EventArgs e)
> {
> base.OnInit(e);
> // modify default templates here
> }
> }
>
> When I use this control in an aspx page using
> <%@ Register Assembly="WebControlsLib" Namespace="WebControlsLib"
> TagPrefix="cc1" %>
>
> and
>
> <cc1:SequencedWizard ID="SequencedWizard1"
> runat="server"></cc1:SequencedWizard>
>
> I get the message "Unknown server tag 'cc1:SequencedWizard'
>
> How should I trouble shoot this?
> I'm not expecting to write any Designer controls - hoping I get the Wizard
> designer controls "for free". Is that right?
>
> Are there any articles describing how to extend the Wizard control like
> this?
>
> Thanks
> Martin
>
>
>
>