108

In Winforms you can say

if ( DesignMode )
{
  // Do something that only happens on Design mode
}

is there something like this in WPF?

1

5 Answers 5

163

Indeed there is:

System.ComponentModel.DesignerProperties.GetIsInDesignMode

Example:

using System.ComponentModel;
using System.Windows;
using System.Windows.Controls;

public class MyUserControl : UserControl
{
    public MyUserControl()
    {
        if (DesignerProperties.GetIsInDesignMode(this))
        {
            // Design-mode specific functionality
        }
    }
}
3
48

In some cases I need to know, whether a call to my non-UI class is initiated by the designer (like if I create a DataContext class from XAML). Then the approach from this MSDN article is helpful:

// Check for design mode. 
if ((bool)(DesignerProperties.IsInDesignModeProperty.GetMetadata(typeof(DependencyObject)).DefaultValue)) 
{
    //in Design mode
}
1
23

For any WPF Controls hosted in WinForms, DesignerProperties.GetIsInDesignMode(this) does not work.

So, I created a bug in Microsoft Connect and added a workaround:

public static bool IsInDesignMode()
{
    if ( System.Reflection.Assembly.GetExecutingAssembly().Location.Contains( "VisualStudio" ) )
    {
        return true;
    }
    return false;
}
1
  • Shouldn't it be GetEntryAssembly() instead of GetExecutingAssembly()? The latter should be returning the assembly where this property is defined
    – fjch1997
    Commented Jun 17, 2017 at 18:40
10

Late answer, I know - but for anyone else who wants to use this in a DataTrigger, or anywhere in XAML in general:

xmlns:componentModel="clr-namespace:System.ComponentModel;assembly=PresentationFramework"

<Style.Triggers>
    <DataTrigger Binding="{Binding RelativeSource={RelativeSource Self}, 
                 Path=(componentModel:DesignerProperties.IsInDesignMode)}" 
                 Value="True">
        <Setter Property="Visibility" Value="Visible"/>
    </DataTrigger>
</Style.Triggers>
-1

Use this one:

if (Windows.ApplicationModel.DesignMode.DesignModeEnabled)
{
    //design only code here
}

(Async and File operations wont work here)

Also, to instantiate a design-time object in XAML (d is the special designer namespace)

<Grid d:DataContext="{d:DesignInstance Type=local:MyViewModel, IsDesignTimeCreatable=True}">
...
</Grid>
1
  • 1
    That class (Windows.ApplicationModel) is for Store apps, included in the Windows Runtime API. This is not an out-of-the-box WPF solution if you're just working on a regular Windows desktop application.
    – qJake
    Commented Feb 25, 2016 at 15:56

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