WPF good practice about open window and back button

Question

I'm beginner programmer in WPF MVVM C#. When user install my app he will follow a tutorial. So, for each "screen" of tutorial I created a Window. On each screen, I have a Back Button.

So, I'm really confused on the good practice about open new window and back to the previous Window. What is the best way to do this? Currently, I'm doing:

First Window (Next Step Button) - ViewModel1

<Button Command="{Binding NextWindowCommand}" >


public void NextWindow(object parameter)
{
    var newWindow = new Window2();
    newWindow.Show();
    newWindow.DataContext = new ViewModel2();
    CloseWindow();
}

Second Window (Back Button) ViewModel2

<Button Command="{Binding BackWindowCommand}" >

    public void Back(object parameter)
    {
        var backWindow = new Window1();
        backWindow.Show();
        backWindow.DataContext = new ViewModel1();
        CloseWindow();
    }

Close Window Method (My ViewModelBase):

    public bool? CloseWindowFlag
    {
        get { return _CloseWindowFlag; }
        set
        {
            _CloseWindowFlag = value;
            RaisePropertyChanged("CloseWindowFlag");
        }
    }


public virtual void CloseWindow(bool? result = true)
{
    Application.Current.Dispatcher.BeginInvoke(DispatcherPriority.Background, new Action(() =>
    {
        CloseWindowFlag = CloseWindowFlag == null 
            ? true 
            : !CloseWindowFlag;
    }));
}

StepUp StepUp · Accepted Answer · 2016-03-12T21:23:50

It is appropriate to use DataTemplates if you want to dynamically switch Views depending on the ViewModel:

<Window>
   <Window.Resources>
      <DataTemplate DataType="{x:Type ViewModelA}">
         <localControls:ViewAUserControl/>
      </DataTemplate>
      <DataTemplate DataType="{x:Type ViewModelB}">
         <localControls:ViewBUserControl/>
      </DataTemplate>
   <Window.Resources>
  <ContentPresenter Content="{Binding CurrentView}"/>
</Window>

If Window.DataContext is an instance of ViewModelA, then ViewA will be displayed and Window.DataContext is an instance of ViewModelB, then ViewB will be displayed.

The best example I've ever seen and read it is made by Rachel Lim. See the example.

And I highly recommend you pay attention to Prism framework. This tutorial is very nice.

Update:

Let me show an example where it can be seen where you should put DataTemplates:

<Window x:Class="SimpleMVVMExample.ApplicationView"
        ...The code omitted for the brevity...
        Title="Simple MVVM Example with Navigation" Height="350" Width="525">
    <Window.Resources>
       <DataTemplate DataType="{x:Type ViewModelA}">
         <localControls:ViewAUserControl/>
      </DataTemplate>
      <DataTemplate DataType="{x:Type ViewModelB}">
         <localControls:ViewBUserControl/>
      </DataTemplate>
    </Window.Resources>

    <DockPanel>
        <Border DockPanel.Dock="Left" BorderBrush="Black" BorderThickness="0,0,1,0">
            <ItemsControl ItemsSource="{Binding ListOfViewModels}">
                <ItemsControl.ItemTemplate>
                    <DataTemplate>
                        <Button Content="{Binding Name}"
                                Command="{Binding DataContext.ChangePageCommand, RelativeSource={RelativeSource AncestorType={x:Type Window}}}"
                                CommandParameter="{Binding }"
                                Margin="2,5"/>
                    </DataTemplate>
                </ItemsControl.ItemTemplate>
            </ItemsControl>
        </Border>

        <ContentControl Content="{Binding CurrentDataTemplateViewModel}" />
    </DockPanel>
</Window>

To see full work example, please, see Rachel Lim's page.

To download the example, please, click here.

WPF good practice about open window and back button

3 Answers