Is there a way to check if WPF is currently executing in design mode or not?

只愿长相守 提交于 2019-11-26 03:34:21

问题


Does anyone know of some global state variable that is available so that I can check if the code is currently executing in design mode (e.g. in Blend or Visual Studio) or not?

It would look something like this:

//pseudo code:
if (Application.Current.ExecutingStatus == ExecutingStatus.DesignMode) 
{
    ...
}

The reason I need this is: when my application is being shown in design mode in Expression Blend, I want the ViewModel to instead use a \"Design Customer class\" which has mock data in it that the designer can view in design mode.

However, when the application is actually executing, I of course want the ViewModel to use the real Customer class which returns real data.

Currently I solve this by having the designer, before he works on it, go into the ViewModel and change \"ApplicationDevelopmentMode.Executing\" to \"ApplicationDevelopmentMode.Designing\":

public CustomersViewModel()
{
    _currentApplicationDevelopmentMode = ApplicationDevelopmentMode.Designing;
}

public ObservableCollection<Customer> GetAll
{
    get
    {
        try
        {
            if (_currentApplicationDevelopmentMode == ApplicationDevelopmentMode.Developing)
            {
                return Customer.GetAll;
            }
            else
            {
                return CustomerDesign.GetAll;
            }
        }
        catch (Exception ex)
        {
            throw new Exception(ex.Message);
        }
    }
}

回答1:


I believe you are looking for GetIsInDesignMode, which takes a DependencyObject.

Ie.

// 'this' is your UI element
DesignerProperties.GetIsInDesignMode(this);

Edit: When using Silverlight / WP7, you should use IsInDesignTool since GetIsInDesignMode can sometimes return false while in Visual Studio:

DesignerProperties.IsInDesignTool

Edit: And finally, in the interest of completeness, the equivalent in WinRT / Metro / Windows Store applications is DesignModeEnabled:

Windows.ApplicationModel.DesignMode.DesignModeEnabled



回答2:


You can do something like this:

DesignerProperties.GetIsInDesignMode(new DependencyObject());



回答3:


public static bool InDesignMode()
{
    return !(Application.Current is App);
}

Works from anywhere. I use it to stop databound videos from playing in the designer.




回答4:


When Visual Studio auto generated some code for me it used

if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode(this)) 
{
    ...
}



回答5:


There are other (maybe newer) ways of specifying design-time data in WPF, as mentioned in this related answer.

Essentially, you can specify design-time data using a design-time instance of your ViewModel:

d:DataContext="{d:DesignInstance Type=v:MySampleData, IsDesignTimeCreatable=True}"

or by specifying sample data in a XAML file:

d:DataContext="{d:DesignData Source=../DesignData/SamplePage.xaml}">

You have to set the SamplePage.xaml file properties to:

BuildAction:               DesignData
Copy to Output Directory:  Do not copy
Custom Tool:               [DELETE ANYTHING HERE SO THE FIELD IS EMPTY]

I place these in my UserControl tag, like this:

<UserControl
    ...
    xmlns:d="http://schemas.microsoft.com/expression/blend/2008" 

    xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
    ...
    d:DesignWidth="640" d:DesignHeight="480"
    d:DataContext="...">

At run-time, all of the "d:" design-time tags disappear, so you'll only get your run-time data context, however you choose to set it.

Edit You may also need these lines (I'm not certain, but they seem relevant):

xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" 
mc:Ignorable="d" 



回答6:


And if you extensively use Caliburn.Micro for your large WPF / Silverlight / WP8 / WinRT application you could use handy and universal caliburn's Execute.InDesignMode static property in your view-models as well (and it works in Blend as good as in Visual Studio):

using Caliburn.Micro;

// ...

/// <summary>
/// Default view-model's ctor without parameters.
/// </summary>
public SomeViewModel()
{
    if(Execute.InDesignMode)
    {
        //Add fake data for design-time only here:

        //SomeStringItems = new List<string>
        //{
        //  "Item 1",
        //  "Item 2",
        //  "Item 3"
        //};
    }
}



回答7:


I've only tested this with Visual Studio 2013 and .NET 4.5 but it does the trick.

public static bool IsDesignerContext()
{
  var maybeExpressionUseLayoutRounding =
    Application.Current.Resources["ExpressionUseLayoutRounding"] as bool?;
  return maybeExpressionUseLayoutRounding ?? false;
}

It's possible though that some setting in Visual Studio will change this value to false, if that ever happens we can result to just checking whether this resource name exist. It was null when I ran my code outside the designer.

The upside of this approach is that it does not require explicit knowledge of the specific App class and that it can be used globally throughout your code. Specifically to populate view models with dummy data.




回答8:


Accepted answer didn't work for me (VS2019).

After inspecting what was going on, I came up with this:

    public static bool IsRunningInVisualStudioDesigner
    {
        get
        {
            // Are we looking at this dialog in the Visual Studio Designer or Blend?
            string appname = System.Reflection.Assembly.GetEntryAssembly().FullName;
            return appname.Contains("XDesProc");
        }
    }



回答9:


I have an idea for you if your class doesn't need an empty constructor.

The idea is to create an empty constructor, then mark it with ObsoleteAttribute. The designer ignores the obsolete attribute, but the compiler will raise an error if you try to use it, so there's no risk of accidentaly using it yourself.

(pardon my visual basic)

Public Class SomeClass

    <Obsolete("Constructor intended for design mode only", True)>
    Public Sub New()
        DesignMode = True
        If DesignMode Then
            Name = "Paula is Brillant"
        End If
    End Sub

    Public Property DesignMode As Boolean
    Public Property Name As String = "FileNotFound"
End Class

And the xaml:

<UserControl x:Class="TestDesignMode"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" 
             xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
             xmlns:vm="clr-namespace:AssemblyWithViewModels;assembly=AssemblyWithViewModels"
             mc:Ignorable="d" 
             >
  <UserControl.Resources>
    <vm:SomeClass x:Key="myDataContext" />
  </UserControl.Resources>
  <StackPanel>
    <TextBlock d:DataContext="{StaticResource myDataContext}" Text="{Binding DesignMode}" Margin="20"/>
    <TextBlock d:DataContext="{StaticResource myDataContext}" Text="{Binding Name}" Margin="20"/>
  </StackPanel>
</UserControl>

This won't work if you really need the empty constructor for something else.



来源:https://stackoverflow.com/questions/834283/is-there-a-way-to-check-if-wpf-is-currently-executing-in-design-mode-or-not

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