WPF 中是否有 DesignMode 属性？

在 Winform 中你可以说

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

WPF 里有这种东西吗？

wpf
.net-3.5

51513 次浏览

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最佳答案

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
}
}
}

小开

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
}

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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;
}

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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>

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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>