WPF 中是否有 DesignMode 属性？
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Is there a DesignMode property in WPF?
提问by Russ
In Winforms you can say
在 Winforms 中，您可以说
if ( DesignMode )
{
// Do something that only happens on Design mode
}
is there something like this in WPF?
WPF中有这样的东西吗？
回答by Enrico Campidoglio
Indeed there is:
确实有：
System.ComponentModel.DesignerProperties.GetIsInDesignMode
System.ComponentModel.DesignerProperties.GetIsInDesignMode
Example:
例子：
using System.ComponentModel;
using System.Windows;
using System.Windows.Controls;
public class MyUserControl : UserControl
{
public MyUserControl()
{
if (DesignerProperties.GetIsInDesignMode(this))
{
// Design-mode specific functionality
}
}
}
回答by Max Galkin
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 articleis helpful:
在某些情况下，我需要知道对我的非 UI 类的调用是否由设计器发起（就像我从 XAML 创建 DataContext 类一样）。那么这篇 MSDN 文章中的方法很有帮助：
// Check for design mode.
if ((bool)(DesignerProperties.IsInDesignModeProperty.GetMetadata(typeof(DependencyObject)).DefaultValue))
{
//in Design mode
}
回答by serhio
For any WPF Controls hosted in WinForms, DesignerProperties.GetIsInDesignMode(this)does not work.
对于WinForms 中托管的任何 WPF 控件，DesignerProperties.GetIsInDesignMode(this)都不起作用。
So, I created a bug in Microsoft Connectand added a workaround:
因此，我在 Microsoft Connect 中创建了一个错误并添加了一个解决方法：
public static bool IsInDesignMode()
{
if ( System.Reflection.Assembly.GetExecutingAssembly().Location.Contains( "VisualStudio" ) )
{
return true;
}
return false;
}
回答by Manfred Radlwimmer
Late answer, I know - but for anyone else who wants to use this in a DataTrigger, or anywhere in XAML in general:
迟到的答案，我知道 - 但对于任何想要DataTrigger在 .
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>
回答by Jeson Martajaya
Use this one:
使用这个：
if (Windows.ApplicationModel.DesignMode.DesignModeEnabled)
{
//design only code here
}
(Async and File operations wont work here)
（异步和文件操作在这里不起作用）
Also, to instantiate a design-time object in XAML (d is the special designer namespace)
此外，在 XAML 中实例化设计时对象（d 是特殊的设计器命名空间）
<Grid d:DataContext="{d:DesignInstance Type=local:MyViewModel, IsDesignTimeCreatable=True}">
...
</Grid>
