How to know whether a control is at design-time or not?

Question

I have a class (control), implementing ICustomTypeDescriptor, which is used both at design-time and run-time by PropertyGrid for customization. I need to expose different properties at design-time (standard controls properties like width, height and so on) and at run-time, when PropertyGrid is used in my program to change other properties of that control.

My code is like:

class MyControl : UserControl, ICustomTypeDescriptor
{
    //Some code..

    public PropertyDescriptorCollection GetProperties(Attribute[] attributes)
    {
        return GetProperties();
    }

    public PropertyDescriptorCollection GetProperties()
    {
        //I need to do something like this:
        if (designTime)
        { //Expose standart controls properties
            return TypeDescriptor.GetProperties(this, true);
        }
        else
        {
            //Forming a custom property descriptor collection
            PropertyDescriptorCollection pdc = new PropertyDescriptorCollection(null);
            //Some code..
            return pdc;
        }
    }
}

Is there an analog for a design-time flag in C#? Is it maybe better to use conditional compilation?

Possible duplicate of *[How to tell if .NET code is being run by Visual Studio designer](http://stackoverflow.com/questions/73515/how-to-tell-if-net-code-is-being-run-by-visual-studio-designer)*. — Peter Mortensen, Feb 20 '14 at 23:43

score 15 · Accepted Answer · edited Feb 20 '14 at 23:39

15

Check if DesignMode is true or false. It's a property that belongs to the control base class.

edited Feb 20 '14 at 23:39

Peter Mortensen

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answered Nov 16 '11 at 07:07

DeveloperX

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2

Actually, it belongs to the `System.ComponentModel.Component` base class. – tafa Nov 16 '11 at 07:14
`'Component.DesignMode' is inaccessible due to its protection level` How do I get around this error ? – GuidoG Sep 02 '21 at 12:37
@GuidoG Don't check the componet from the outside. The control has to check itself. – The incredible Jan Sep 20 '22 at 08:43

Juri · Answer 2 · 2011-11-16T07:19:25.057

9

The flag should be DesignMode. Hence your code should look as follows

public PropertyDescriptorCollection GetProperties()
{
   //I need to do something like this:
   if (this.DesignMode)
   { //Expose standart controls properties
       return TypeDescriptor.GetProperties(this, true);
   }
   else
   {   //Forming a custom property descriptor collection
       PropertyDescriptorCollection pdc = new PropertyDescriptorCollection(null);
       //Some code..
       return pdc;      
   }
}

Here is the according MSDN doc.

edited Nov 16 '11 at 07:19

answered Nov 16 '11 at 07:05

Juri

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I really hate redundant "this". :) – The incredible Jan Sep 20 '22 at 08:49

score 3 · Answer 3 · edited Feb 20 '14 at 23:44

3

Use the DesignMode property of the base. This will tell you about the mode.

edited Feb 20 '14 at 23:44

Peter Mortensen

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answered Nov 16 '11 at 07:08

Kangkan

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How to know whether a control is at design-time or not?

3 Answers3