Currently I'm doing it so:
public MainWindow()
{
InitializeComponent();
Title = Properties.Resources.WindowName;
}
How to do the same through the WPF binding?
EDIT: It still doesn't work in XAML.
Environment:VS2010, .NET 4.0, Windows 7.
Reproduction steps:
Create class library ClassLibrary1 with code:
namespace ClassLibrary1
{
static public class Class1
{
static public string Something
{
get { return "something"; }
}
}
}
Create WPF windows application in VS2010 .NET 4.0.
Edit main window's XAML:
<Window x:Class="ahtranslator.MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:ClassLibrary1="clr-namespace:ClassLibrary1;assembly=ClassLibrary1"
Title="{Binding Source={x:Static ClassLibrary1:Class1}, Path=Something}"
Height="350" Width="525" Icon="/ahtranslator;component/Icon1.ico" WindowStyle="SingleBorderWindow" ShowInTaskbar="False" DataContext="{Binding}">
...
Compilation error message:
MainWindow.xaml(7,130): error MC3029: 'ClassLibrary1:Class1' member is not valid because it does not have a qualifying type name.
Also I found this topic My.Resources in WPF XAML?.
And it seems all should work but it doesn't.
Microsoft doesn't give description for this error message. Only another topic in help forum http://social.msdn.microsoft.com/Forums/en/wpf/thread/4fe7d58d-785f-434c-bef3-31bd9e400691, which doesn't help either.
In code it would look like this i think:
Binding titleBinding = new Binding("WindowName");
titleBinding.Source = Properties.Resources;
this.SetBinding(Window.Title, titleBinding);
This only makes sense if changes may occur to the title and the binding will be notified of those changes (WindowName has to either be a Dependency Property or Resources needs to implement INotifyPropertyChanged)
If Properties is a namespace (as would be the case with the default VS-generated properties) you need to declare it somewhere using xmlns & use x:Static:
<Window
...
xmlns:prop="clr-namespace:App.Properties"
Title="{Binding Source={x:Static prop:Resources.WindowName}}">
Another note: If you use the managed resources of Visual Studio you need to make sure that the access modifier of the properties is public, default is internal which will throw an exception since binding only works for public properties.
just remove this:
... ;assembly=ClassLibrary1"
I actually have the Title in a static resource defined at the top of the application and I bind the Title and anything else I want to it
<s:String x:Key="ApplicationName">My Application</s:String>
Have you tried to change the access modifier of the resource from internal to public?
I have just had some problem with that right now.
/// <summary>
/// Looks up a localized string similar to Has been impossible to load the configuration information.
/// </summary>
internal static string ERROR_NoConfigurationLoaded {
get {
return ResourceManager.GetString("ERROR_NoConfigurationLoaded", resourceCulture);
}
}
to
/// <summary>
/// Looks up a localized string similar to Has been impossible to load the configuration information.
/// </summary>
public static string ERROR_NoConfigurationLoaded {
get {
return ResourceManager.GetString("ERROR_NoConfigurationLoaded", resourceCulture);
}
}
Related
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);
}
}
}
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
You can do something like this:
DesignerProperties.GetIsInDesignMode(new DependencyObject());
public static bool InDesignMode()
{
return !(Application.Current is App);
}
Works from anywhere. I use it to stop databound videos from playing in the designer.
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"
When Visual Studio auto generated some code for me it used
if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode(this))
{
...
}
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"
//};
}
}
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");
}
}
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.
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.
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);
}
}
}
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
You can do something like this:
DesignerProperties.GetIsInDesignMode(new DependencyObject());
public static bool InDesignMode()
{
return !(Application.Current is App);
}
Works from anywhere. I use it to stop databound videos from playing in the designer.
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"
When Visual Studio auto generated some code for me it used
if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode(this))
{
...
}
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"
//};
}
}
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");
}
}
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.
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.
In a UWP-Windows 10 C#/XAML app using Template10, I'm currently trying to have my pages/views inherit from a base class that inherits from Windows.UI.Xaml.Controls.Page.
This has been working correctly, however when I try to make the base class generic, and include a type argument in the child class and the XAML in the following way, it does not work:
namespace App.Views
{
public abstract class InfoListViewBase<M> : Page where M : InfoListViewModelBase
{
public InfoListViewBase() { }
}
public sealed partial class ModelPage : InfoListViewBase<Model>
{
public Model()
{
InitializeComponent();
NavigationCacheMode = NavigationCacheMode.Disabled;
}
}
}
<local:InfoListViewBase
x:Class="App.Views.ModelPage"
x:TypeArguments="l:Model"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:Behaviors="using:Template10.Behaviors"
xmlns:Core="using:Microsoft.Xaml.Interactions.Core"
xmlns:Interactivity="using:Microsoft.Xaml.Interactivity"
xmlns:controls="using:Template10.Controls"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:local="using:App.Views"
xmlns:l="using:Library"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" mc:Ignorable="d">
</local:InfoListViewBase>
The error I receive is:
The name "InfoListViewBase`1" does not exist in the namespace "using:App.Views".
The error shows up every time I add the line x:TypeArguments="l:Model" to the XAML.
Multiple rebuilds, cleans, etc., in visual studio have not solved the problem.
Is there something I am doing wrong in the implementation of the generic in XAML?
Unfortunately, generic parameters in XAML still aren't supported in UWP app. You cannot use x:TypeArguments in XAML currently. But you may reference this thread to try to have a workaround.
If you still want this feature ,you can also submit this feature request to UserVoice.
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);
}
}
}
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
You can do something like this:
DesignerProperties.GetIsInDesignMode(new DependencyObject());
public static bool InDesignMode()
{
return !(Application.Current is App);
}
Works from anywhere. I use it to stop databound videos from playing in the designer.
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"
When Visual Studio auto generated some code for me it used
if (!System.ComponentModel.DesignerProperties.GetIsInDesignMode(this))
{
...
}
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"
//};
}
}
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");
}
}
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.
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.
The XAML designer crashes visual studio 2010 if the view model that is set as the Data Context registers itself in a static class.
View
<Window x:Class="CTL.Editor.View.EditorWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:editor="clr-namespace:CTL.Editor.ViewModel.Editor"
Height="300" Width="300">
<Window.DataContext>
<editor:EditorWindowViewModel />
</Window.DataContext>
<Grid>
</Grid>
</Window>
ViewModel:
public EditorWindowViewModel()
{
ApplicationViewModel.EditorWindows.Add(this);
}
~EditorWindowViewModel()
{
ApplicationViewModel.EditorWindows.Remove(this);
}
Is there any way around this? Maybe a # directive?
You can use the DesignerProperties.IsInDesignMode to supress execution while in design mode. Just wrap your code in an if statement: if(!DesignerProperties.IsInDesignTool)
However it is often a good idea to find the root cause of the problem by debugging the designer exception. Here is an good article that should get you started.
For those seeking a bit more detail than Postlagerkarte's answer:
A way to use IsInDesignMode that is MVVM-friendly is shown below.
if (DesignerProperties.GetIsInDesignMode(new DependencyObject()))
{
....
}
My issue was caused by the fact that ApplicationViewModel's constructor what loading a config file and apparently Visual Studio didn't like that, didn't find the file or didn't search for the file in the right place when it was running my code.
What I ended up doing was:
public static bool DesignMode
{
get { return DesignerProperties.GetIsInDesignMode(new DependencyObject()); }
}
static ApplicationViewModel()
{
if (!DesignMode)
{
Configuration = Configuration.LoadConfigurationDocument();
}
}
Note: There is a Configuration static member on ApplicationViewModel and a Configuration class which loads the config.