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05.17.2008 at 02:51PM PDT, ID: 23411148
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8.0

How to make a disabled button gray during designtime

Asked by arildj78 in .NET, Microsoft Visual C#.Net

Tags: ,

I'm building a custom usercontrol based off System.Windows.Forms.Button. It will be a flat button with a solid background color, single pixel frame around the edge and a single forecolor.

Now I would like the forecolor to change to gray (both for design time and runtime) when the enabled property is set to false.

The code below works during runtime.
I can get it to work during designtime by replacing Enabled with Enabled2 (and skipping the new keyword). Downside with this is that the user of my control will see both Enabled and Enabled2 in the property window.

        private bool _Enabled = true;
        public new bool Enabled
        {
            get { return this._Enabled; }
            set
            {
       //       Console.Beep()
                _Enabled = value;
                base.Enabled = _Enabled;
                RefreshPEPbutton();        // This checks the _Enabled and changes
                                                         // forecolor if control is disabled
            }
        }

I've tried to add Console.Beep() as a debugging tool. If the property is named Enabled2, I get a beep when changing it. If the property is named Enabled, nothing happens during design time.

Where did I go wrong?
Arild
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[+][-]05.18.2008 at 01:22AM PDT, ID: 21591939

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[+][-]05.20.2008 at 05:51AM PDT, ID: 21605624

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About this solution

Zones: .NET, Microsoft Visual C#.Net
Tags: C# .NET v2.0, Visual Studio 2005
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Solution Provided By: dweppenaar
Participating Experts: 2
Solution Grade: A
 
 
[+][-]07.20.2008 at 05:37AM PDT, ID: 22045613

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[+][-]07.24.2008 at 06:01PM PDT, ID: 22085148

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