[ACCEPTED]-How to tell if .NET code is being run by Visual Studio designer-gui-designer

To find out if you're in "design mode":

• Windows Forms components (and controls) have a DesignMode property.
• Windows Presentation Foundation controls should use the IsInDesignMode attached property.

0

if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)
{
  // Design time logic
}

0

The Control.DesignMode property is probably 12 what you're looking for. It tells you if 11 the control's parent is open in the designer.

In 10 most cases it works great, but there are 9 instances where it doesn't work as expected. First, it 8 doesn't work in the controls constructor. Second, DesignMode 7 is false for "grandchild" controls. For 6 example, DesignMode on controls hosted in 5 a UserControl will return false when the 4 UserControl is hosted in a parent.

There 3 is a pretty easy workaround. It goes something 2 like this:

public bool HostedDesignMode
{
  get 
  {
     Control parent = Parent;
     while (parent!=null)
     {
        if(parent.DesignMode) return true;
        parent = parent.Parent;
     }
     return DesignMode;
  }
}

I haven't tested that code, but 1 it should work.

The most reliable approach is:

public bool isInDesignMode
{
    get
    {
        System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
        bool res = process.ProcessName == "devenv";
        process.Dispose();
        return res;
    }
}

0

The most reliable way to do this is to ignore 3 the DesignMode property and use your own 2 flag that gets set on application startup.

Class:

public static class Foo
{
    public static bool IsApplicationRunning { get; set; }
}

Program.cs:

[STAThread]
static void Main()
{
     Foo.IsApplicationRunning = true;
     // ... code goes here ...
}

Then 1 just check the flag whever you need it.

if(Foo.IsApplicationRunning)
{
    // Do runtime stuff
}
else
{
    // Do design time stuff
}

I had the same problem in Visual Studio 4 Express 2013. I tried many of the solutions 3 suggested here but the one that worked for 2 me was an answer to a different thread, which I will repeat here in case 1 the link is ever broken:

protected static bool IsInDesigner
{
    get { return (Assembly.GetEntryAssembly() == null); }
}

The devenv approach stopped working in VS2012 2 as the designer now has its own process. Here 1 is the solution I am currently using (the 'devenv' part is left there for legacy, but without VS2010 I am not able to test that though).

private static readonly string[] _designerProcessNames = new[] { "xdesproc", "devenv" };

private static bool? _runningFromVisualStudioDesigner = null;
public static bool RunningFromVisualStudioDesigner
{
  get
  {
    if (!_runningFromVisualStudioDesigner.HasValue)
    {
      using (System.Diagnostics.Process currentProcess = System.Diagnostics.Process.GetCurrentProcess())
      {
        _runningFromVisualStudioDesigner = _designerProcessNames.Contains(currentProcess.ProcessName.ToLower().Trim());
      }
    }

    return _runningFromVisualStudioDesigner.Value;
  }
}

/// <summary>
/// Are we in design mode?
/// </summary>
/// <returns>True if in design mode</returns>
private bool IsDesignMode() {
    // Ugly hack, but it works in every version
    return 0 == String.CompareOrdinal(
        "devenv.exe", 0,
        Application.ExecutablePath, Application.ExecutablePath.Length - 10, 10);
}

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System.Diagnostics.Debugger.IsAttached

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It's hack-ish, but if you're using VB.NET and 4 when you're running from within Visual Studio 3 My.Application.Deployment.CurrentDeployment will be Nothing, because you haven't deployed 2 it yet. I'm not sure how to check the equivalent 1 value in C#.

using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
{
    bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv";
    return inDesigner;
}

I tried the above code (added a using statement) and 9 this would fail on some occasions for me. Testing 8 in the constructor of a usercontrol placed 7 directly in a form with the designer loading 6 at startup. But would work in other places.

What 5 worked for me, in all locations is:

private bool isDesignMode()
{
    bool bProcCheck = false;
    using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
    {
        bProcCheck = process.ProcessName.ToLower().Trim() == "devenv";
    }

    bool bModeCheck = (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime);

    return bProcCheck || DesignMode || bModeCheck;
}

Maybe 4 a bit overkill, but it works, so is good 3 enough for me.

The success in the example 2 noted above is the bModeCheck, so probably 1 the DesignMode is surplus.

I'm not sure if running in debug mode counts 3 as real, but an easy way is to include an 2 if statement in your code that checkes for 1 System.Diagnostics.Debugger.IsAttached.

You check the DesignMode property of your control:

if (!DesignMode)
{
//Do production runtime stuff
}

Note 3 that this won't work in your constructor 2 because the components haven't been initialized 1 yet.

We use the following code in UserControls 5 and it does the work. Using only DesignMode 4 will not work in your app that uses your 3 custom user controls as pointed out by other 2 members.

    public bool IsDesignerHosted
    {
        get { return IsControlDesignerHosted(this); }
    }

    public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
    {
        if (ctrl != null)
        {
            if (ctrl.Site != null)
            {
                if (ctrl.Site.DesignMode == true)
                    return true;
                else
                {
                    if (IsControlDesignerHosted(ctrl.Parent))
                        return true;
                    else
                        return false;
                }
            }
            else
            {
                if (IsControlDesignerHosted(ctrl.Parent))
                    return true;
                else
                    return false;
            }
        }
        else
            return false;
    }

Basically the logic above boils 1 down to:

    public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
    {
        if (ctrl == null) return false;
        if (ctrl.Site != null && ctrl.Site.DesignMode) return true;
        return IsControlDesignerHosted(ctrl.Parent);
    }

When running a project, its name is appended 2 with ".vshost".

So, I use this:

    public bool IsInDesignMode
    {
        get
        {
            Process p = Process.GetCurrentProcess();
            bool result = false;

            if (p.ProcessName.ToLower().Trim().IndexOf("vshost") != -1)
                result = true;
            p.Dispose();

            return result;
        }
    }

It works for 1 me.

If you created a property that you don't 11 need at all at design time, you can use 10 the DesignerSerializationVisibility attribute and set it to Hidden. For 9 example:

protected virtual DataGridView GetGrid()
{
    throw new NotImplementedException("frmBase.GetGrid()");
}

[DesignerSerializationVisibility(DesignerSerializationVisibility.Hidden)]
public int ColumnCount { get { return GetGrid().Columns.Count; } set { /*Some code*/ } }

It stopped my Visual Studio crashing 8 every time I made a change to the form with 7 NotImplementedException() and tried to save. Instead, Visual Studio 6 knows that I don't want to serialize this 5 property, so it can skip it. It only displays 4 some weird string in the properties box 3 of the form, but it seems to be safe to 2 ignore.

Please note that this change does 1 not take effect until you rebuild.

If you are in a form or control you can 1 use the DesignMode property:

if (DesignMode)
{
        DesignMode Only stuff
}

System.ComponentModel.Component.DesignMode == true

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I found the DesignMode property to be buggy, at 3 least in previous versions of Visual Studio. Hence, I 2 made my own using the following logic:

Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";

Kind 1 of a hack, I know, but it works well.

To solve the problem, you can also code 1 as below:

private bool IsUnderDevelopment
{
    get
    {
        System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
        if (process.ProcessName.EndsWith(".vshost")) return true;
        else return false;
    }

}

Here's another one:

        //Caters only to thing done while only in design mode
        if (App.Current.MainWindow == null){ // in design mode  }

        //Avoids design mode problems
        if (App.Current.MainWindow != null) { //applicaiton is running }

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After testing most of the answers here, unfortunately 13 nothing worked for me (VS2015). So I added 12 a little twist to JohnV's answer, which didn't work out 11 of the box, since DesignMode is a protected 10 Property in the Control class.

First I made 9 an extension method which returns the DesignMode's 8 Property value via Reflection:

public static Boolean GetDesignMode(this Control control)
{
    BindingFlags bindFlags = BindingFlags.Instance | BindingFlags.NonPublic | BindingFlags.Static;
    PropertyInfo prop = control.GetType().GetProperty("DesignMode", bindFlags);
    return (Boolean)prop.GetValue(control, null);
}

and then I 7 made a function like JohnV:

public bool HostedDesignMode
{
    get
    {
        Control parent = Parent;
        while (parent != null)
        {
            if (parent.GetDesignMode()) return true;
            parent = parent.Parent;
        }
        return DesignMode;
    }
}

This is the only 6 method that worked for me, avoiding all 5 the ProcessName mess, and while reflection 4 should not be used lightly, in this case 3 it did all the difference! ;)

EDIT:

You can also 2 make the second function an extension method 1 like this:

public static Boolean IsInDesignMode(this Control control)
{
    Control parent = control.Parent;
    while (parent != null)
    {
        if (parent.GetDesignMode())
        {
            return true;
        }
        parent = parent.Parent;
    }
    return control.GetDesignMode();
}

For WPF (hopefully this is useful for those 3 WPF people stumbling upon this question):

if (System.ComponentModel.DesignerProperties.GetIsInDesignMode(new DependencyObject()))
{
}

GetIsInDesignMode requires 2 a DependencyObject. If you don't have one, just 1 create one.