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How to switch between ViewModels in multi-window WPF application?

I am trying to create a WPF Model-View-ViewModel that has separate LoginWindow, RegistrationWindow and MainDashboardWindow.
I have already read Rachel Lim's solution on navigating with MVVM using ApplicationViewModel and ApplicationView that contains other views but I have trouble understanding something.
I was planning on using LoginViewModel and RegistrationViewmodel (none of them will inherit from BaseViewModel) by setting the DataContext of LoginWindow and RegistrationWindow and then after logging in to application start using a MainDashboardViewModel to switch between viewmodel's on MainDashboardWindow.
Is this the correct way to go?
If this is not the correct way to go, how can I implement an application-wide viewmodel that can be switched between windows (only one window will be opened at a time)?

To implement MVVM properly, you need to understand what is View, Model and ViewModel.
View is the UI that WON'T HAVE ANY CODE. The DataContext in View is the ViewModel class, one View should be associated with one ViewModel. You should set the DataContext by XAML code.
Model is the class contains the data and some methods to process for that data. I highly recommend you using Repository Pattern when design Model for you program.
ViewModel is where you put the business method in it. All ViewModel must implement from BaseViewModel and DON'T KNOW ANYTHING ABOUT VIEW.
In your case, you need to create 3 different Views, 3 ViewModels for 3 Views.
Let's start with your Login windows. First you need to specify what ViewModel for Login windows by indicate it on Login view.
<Window x:Class="SampleApplication.LoginWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:local="clr-namespace:SampleApplication.ViewModels"
Title="Login" Height="350" Width="525">
<Window.DataContext>
<local:LoginViewModel/>
</Window.DataContext>
You have 2 buttons in Login windows. one is Login and on is Register. Each button must be binded with an RelayCommand in ViewModel. By this way you can properly implement MVVM. To navigate between windows you must close you current windows and open a new one. To close windows, see this, to open windows, just create an object of your view and call ViewObject.Show();

Use SendKeys in a UserControl

I want to use the C# System.Windows.Forms.SendKeys.SendWait() Method to send Keystrokes from an OnScreenKeyboard to a Textbox. Since I may use this OnScreenKeyboard at other places too I created a UserControl with View (for the Design of the Keyboard) and Viewmodel (basically for calling the Sendkeys.SendWait() Method) in an extra project.
Within this project I created a MainView where I included the UserControl via a ContentControl as you can see in the Code below. CurrentPage basically refers to the Viewmodel of the Keyboard.
<Window.Resources>
<DataTemplate DataType="{x:Type viewModel:KeyboardViewmodel}">
<view:KeyboardView/>
</DataTemplate>
</Window.Resources>
<Grid>
<Border Background="White">
<HeaderedContentControl Content="{Binding Path=CurrentPage}"/>
</Border>
<TextBox Width="120"/>
</Grid>
I now have the OnScreenKeyboard and a Textbox in my Window. Clicking into the Textbox and pressing buttons of my OnScreenKeyboard will result in text appearing in my Textbox. All Controls within the KeyboardView are set to avoid getting focus. This is necessary to maintain focus on the Textbox.
The Buttons in the KeyboardView all bind to the Command ClickCommandin my KeyboardViewmodel. Here is the code of the KeyboardViewmodel:
public class KeyboardViewmodel : BaseModel
{
public BaseCommand ClickCommand { get; set; }
public KeyboardViewmodel()
{
ClickCommand = new BaseCommand(PressAndRelease);
}
public void PressAndRelease(object key)
{
if (((string)key).Length <= 1)
SendKeys.SendWait((string)key);
else
SendKeys.SendWait("{" + (string)key + "}");
}
}
Now I did create a NuGet Package with these Files and imported them to the project where I want to use my OnScreenKeyboard.
I did do basically the same as when I tested the OnScreenKeyboard before.
But let me explain the structure of the project a little more:
I have a MainView + MainViewmodel. The MainViewmodel manages the navigation between available pages. To show these pages I have - as in the short example before - a ContentControl whose content is bound to a CurrentPage Property. The MainViewis a normal Window, all other Views are UserControls.
In one of these pages I need an OnScreenKeyboard (DetailsView + DetailsViewmodel). So it seemed logical to me to use another ContentControl within the DetailsView:
<Border Grid.Column="0" Grid.Row="4" Grid.ColumnSpan="3" Height="Auto" Width="Auto">
<ContentControl Content="{Binding Path=OnScreenKeyboard}"/>
</Border>
I create the KeyboardViewmodel in the constructor of the DetailsViewmodel. The constructor of the DetailsViewmodel is called in the MainViewmodel at startup.
So now everything works out fine so far, the OnScreenKeyboard is shown on the correct page in the correct place. If I click a button of the OnScreenKeyboard the proper bound command is called and the SendKeys.SendWait() Method is called.
But no text appears in the TextBox. I have a very bad understanding of the SendKeys.SendAwait() Method. Also, the MSDN Documentation seems to be not very exhaustive on this topic.
It states: "Sends the given keys to the active application, and then waits for the messages to be processed."
Now. The Active / Focused Application is my Application. So my guess is that the KeyStrokes should be processed by my Textbox.
My Questions:
Any guesses how to debug the 'SenWait()' Method further e.g. track where the strokes are really sent to or something like that?
Is this the correct way for sending KeyStrokes to an active Application? It seems like SendKeys comes from Windows Forms, I use WPF.
Should I just pass my Textbox as reference to the OnScreenKeyboard and write directly to the referenced Textbox? This would make me much less flexible in regards of reusability.
Update:
As pointed out in the comments this could probably be a duplicate question.
I am well aware of the various different solutions and have already considerd them:
http://wpfkb.codeplex.com/
http://www.codeproject.com/Articles/32568/A-Touch-Screen-Keyboard-Control-in-WPF
http://www.codeproject.com/Articles/145579/A-Software-Virtual-Keyboard-for-Your-WPF-Apps
But as one may understand these projects are looking all way too powerfull for my simple needs.
Here a screenshot to provide a better understanding of my needs:
It is really as simple as that. 4 rows of buttons that will never change, no other controls / functionality than sending the CommandParameter of the pressed button to the Textbox / Active Form.
Researching on that specific problem hasn't shown any problems like that. In most other SO Questions the problem is to send Data to another Window, not to send Data WITHIN the current Window.
So I don't consider this question as duplicate.

Visual Studio 2013 WPF design-time data isn't displayed

i have some UserControls that are shown fine in designer, but i can't make any changes to the design-time example content from the constructor. It seems like it is not executed at all.
XAML:
<UserControl x:Class="Example.Test"
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"
mc:Ignorable="d"
d:DesignHeight="300" d:DesignWidth="300">
<Grid>
<TextBlock Name="testx" Foreground="White" HorizontalAlignment="Center" VerticalAlignment="Center"></TextBlock>
</Grid>
Code:
using System.ComponentModel;
using System.Windows.Controls;
namespace Example
{
/// <summary>
/// Interaction logic for Test.xaml
/// </summary>
public partial class Test : UserControl
{
public Test()
{
InitializeComponent();
if (DesignerProperties.GetIsInDesignMode(this))
testx.Text = " IN DESIGN!";
}
}
}
I've tried many options, but still can't get it how to display design-time data in WPF designer :( Different context binding also shows nothing...
PS: Tried clean VS2012 and VS2013 projects on Win8. NOTHING WORKS! :( I don't know what to do, haven't found anything similar on the net... Is it sufficient to just add design check in constructor and set existent control text? It should work, right?

K, the short answer is: You're on the right path.
The long one is: It's a bit more complicated than that.
Your example will "kinda" work, as in, if you'll put an else testx.Text = RUNTIME; after your if, like that:
if (DesignerProperties.GetIsInDesignMode(this))
testx.Text= " IN DESIGN!";
else
testx.Text= " Runtime";
you'll see what you want on runtime, but you're design time will stay empty.
For the Design time, you also need to set the context if I'm not mistaken.
If you're using any of the MVVM framework out there, you kinda get this functionality for "free". As in, you'll have a "in design time" property and you can set whatever data you want for the design. The catch is that you need to have an empty constructor if my memory serves me right.
You'll also use bindings, and not set the text property directly.
I remember that the default WPF and binding for design time was lacking a bit last time I tried to do something like that in "vanilla" wpf (as in, no MVVM, no bindings), but I believe that with a bit of a hack it's achievable. Again, can't remember it from the top of my head.

MVVM Light - Multiple ViewModels (and connecting them up)

I am trying to learn the MVVM pattern (C#), having come from a Windows Forms background. I am using the MVVM Light toolkit, and so far I think it is brilliant.
I have made several small applications, however one thing I am struggling with is introducing a second view.
I want to (for example), have a button on my MainViewModel, which via a RelayCommand, opens up a new Window - let's say an "About" window. I have done hours of research on the web for this however it seems I can't get my AboutViewModel to communicate with/show my AboutView.
I have placed a receiving messenger in the code-behind constructor of the AboutView.xaml - however I can't get it to receive any messages from the AboutViewModel, and thus can't make it 'Show()'.
If anyone has an example of an Mvvm Light WPF app using multiple views that would be great :)

There are two ways I can think to do this easily
The first would be to use a Popup instead of a new Window. For example, I often put properties in my ViewModel for PopupContent and IsPopupVisible, and set those values anytime I want to display my Popup control. For example, a ShowAboutPopup relay command might run something like this:
void ShowAboutPopup()
{
PopupContent = new AboutViewModel();
IsPopupVisible = true;
}
You can display it using a Popup object, or a custom UserControl. I prefer to use my own custom Popup UserControl, which will usually end up looking like this:
<Window>
<Canvas x:Name="RootPanel">
<SomePanel>
<!-- Regular content goes here -->
</SomePanel>
<local:PopupPanel Content="{Binding PopupContent}"
local:PopupPanel.IsPopupVisible="{Binding IsPopupVisible}"
local:PopupPanel.PopupParent="{Binding ElementName=RootPanel}" />
</Canvas>
</Window>
The PopupContent property is a ViewModel (such as an AboutViewModel), and DataTemplates are used to tell WPF to draw specific ViewModels with specific Views
<Window.Resources>
<DataTemplate DataType="{x:Type local:AboutViewModel}">
<local:AboutView />
</DataTemplate>
</Window.Resources>
The other method is to have some kind of ApplicationViewModel that runs on startup, and is responsible for the overall application state, which includes which window(s) are open.
Typically I prefer to have a single ApplicationView that contains a ContentControl to display the current page
<Window>
<ContentControl Content="{Binding CurrentViewModel}" />
</Window>
however it can also be used to manage multiple windows. If you do use it to manage multiple Window objects, be warned that this will not be a pure ViewModel because it will need to access some View-specific objects, and referencing UI objects it not something a ViewModel should do. For example, it may subscribe to receive ShowWindow messages, and upon receiving those messages it would create the specified View and show it, and possibly hide the current window as well.
Personally, I try to avoid multiple windows as much as possible. My usual method is to have a single View that contains consistent application objects for any page, and a ContentControl containing dynamic content that changes. I have an example using this navigation style on my blog if you're interested

As i can see you want a navigation in your MVVM app?
Word goes to the creator of MVVM Light - Laurent Bugnion - with his post about using Navigation Service for switching Views. It's actually about Windows Phone & Silverlight but same should apply to WPF.
Also this answer in related question uses this approach.

How to manipulate a window object from another class in WPF

I'm new in WPF and C#. I know a lot of VB.NET and I'm used to the way when I call a form object like textboxes, etc. I'm calling it from another form. Now, I'm using WPF, I'm confused. Because I have a Main Window. And I want to add and item to a listbox in the Main Window from a Class. In VB.Net , its just like this.
IN FORM2
Form1.Textbox.Text = "";
Wherein I can't do it in WPF. Can someone please Help me. Thanks!

WPF windows defined in XAML have their controls publicly accessible from other classes and forms, unless you specifically mark them with the x:FieldModifier attribute as private.
Therefore, if you make an instance of your main window accessible in another class, be it a Window or anything else, you'll be able to populate controls from within this second class.
A particular scenario is when you want to update the contents of a control in your main window from a child window that you have opened on top of it. Is such a case, you may set the child window's Owner property to the current, main window, in order to access it while the child is visible. For instance, let's say you have defined these two windows:
// MainWindow
<Window x:Class="TestApplication.MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="MainWindow" Height="350" Width="525">
<Grid>
<ListBox Name="mainListBox" Height="250" HorizontalAlignment="Stretch" VerticalAlignment="Top"/>
<Button Content="Open Another Window" HorizontalAlignment="Center" VerticalAlignment="Bottom" Margin="20" Click="OpenAnotherWindow_Click"/>
</Grid>
</Window>
and
// AnotherWindow
<Window x:Class="TestApplication.AnotherWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="AnotherWindow" Height="300" Width="300">
<Grid>
<Button Content="Add New Item to Main Window" HorizontalAlignment="Center" VerticalAlignment="Center" Click="AddNewItem_Click"/>
</Grid>
</Window>
each in its own XAML file.
In MainWindow's code behind, inside the button click handler, you show an instance of AnotherWindow as a dialog and set its Owner property to MainWindow's instance:
private void OpenAnotherWindow_Click(object sender, RoutedEventArgs e)
{
AnotherWindow anotherWindow = new AnotherWindow();
anotherWindow.Owner = this;
anotherWindow.ShowDialog();
}
Now, you can access the MainWindow's instance from AnotherWindow's Owner property, in order to add a new item to the ListBox control defined on it, in the button click handler in AnotherWindow's code behind:
private void AddNewItem_Click(object sender, RoutedEventArgs e)
{
MainWindow mainWindow = Owner as MainWindow;
mainWindow.mainListBox.Items.Add(new Random().Next(1000).ToString());
}
It simply adds a new random number to the ListBox, in order to show how the code accesses and modifies the control's data in MainWindow.

Pure WPF solution, but also may be easiest in your case, is using a Data Binding in WPF.
Every form's control is binded to some data on ModelView (pure MVVM approach) or to data (more or less like yuo can do it in WindowsForms). So the "only" thing you have to do is to read/write data binded to controls on UI of that form.
For example, you have TextBox on Windows and want to read a data from it.
This TextBox is binded to some string property of the class that is responsible for holding the data for the controls on that form (just an example, in real world could be 1000 other solutions, based on developer decisions). So what you need, is not to say: "window give textbox" and after read TextBox's content, but simply read binded string property.
Sure it's very simply description of a stuff. But just to give you a hint. Follow databinding link provided above to learn more about this stuff. Do not afraid of a lot of stuff there, it's after all is not a complicated idea and also pretty intuitive. To make that stuff to work in simply case you will not need to make huge efforts by me. The stuff becomes really complex when you end up into real world applications.

This will get all active windows:
foreach (Window item in Application.Current.Windows)
{
}