WPF: Composite Application with Page Navigation - wpf

I am currently writing an application to which the composite methodology fits like a glove.... almost!
I also need a way to navigate between views, including maintaining a journal for navigation backward and forward.
What is the best way to combine these two methodologies, on one hand the single Window based CAG shell with its UserControl derived views, and on the other hand the convenient NavigationWindow shell with its Page derived views and journal?
Thanks!

You can display anything in a NavigationWindow, not just Pages. A simple way to make it work is to define in the NavigationWindow's resources a DataTemplate for each ViewModel you want to display. Bind the Content property of the NavigationWindow to a property of your main ViewModel, and you're done : changing that property will update the NavigationWindow content, and the appropriate DataTemplate will be picked automatically
UPDATE
I just looked at the code of a project of mine where I used a NavigationWindow. Actually I was mistaken, it doesn't work by binding the Content (or maybe it works, but that's not what I did). Instead I created a INavigationService interface, implemented by my App class, which handles the navigation by calling the NavigationWindow.Navigate method. That way, the navigation history is maintained by the NavigationWindow.
Here's an extract from my project
MainWindow.xaml :
<NavigationWindow x:Class="MyApp.MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:vm="clr-namespace:MyApp.ViewModel"
xmlns:view="clr-namespace:MyApp.View"
Title="{Binding Content.DisplayName, RelativeSource={RelativeSource Self}, FallbackValue=The Title}"
Height="600" Width="800">
<NavigationWindow.Resources>
<DataTemplate DataType="{x:Type vm:HomeViewModel}">
<view:HomeView />
</DataTemplate>
<DataTemplate DataType="{x:Type vm:CustomerViewModel}">
<view:CustomerView />
</DataTemplate>
</NavigationWindow.Resources>
</NavigationWindow>
App.xaml.cs :
...
private void Application_Startup(object sender, StartupEventArgs e)
{
LoadConfig();
MyApp.MainWindow window = new MainWindow();
INavigationService navigationService = this;
HomeViewModel viewModel = new HomeViewModel(navigationService);
this.MainWindow = window;
window.Navigate(viewModel);
window.Show();
}
When I need to navigate to another view, I just call the Navigate method with the ViewModel as a parameter, and WPF automatically picks the appropriate DataTemplate from the resources.

Related

Create multi panel application in WPF

I'm developing WPF MVVM application and I want to create a Window with many panels that changes when user choose another panel from navigation.
I've read this article but it's not working due to Can't put a Page in a Style error. I can't find any answer about how to create a WPF application that navigate through different panels in one single window, how I can achieve what I want using MVVM pattern?
You can place various panels in a Grid, sharing the same space (overlapping) and change Visibility to make "Visible" only the one you want shown.
I've used this thecnique same time ago, and is compatible with MVVM too.
I have created an ContentPresenter and bind it to the MainWindow ViewModel and set the DataTemplate for each ViewModels.
<ContentPresenter Name="WindowContent" Content="{Binding CurrentPageViewModel}"/>
<Window.Resources>
<DataTemplate DataType="{x:Type viewModels:MainViewModel}">
<views:MainView />
</DataTemplate>
</Window.Resources>
And so when the binded property is changed, ContentPresenter display proper ViewModel and due to DataTemplate, the actual View.
public IPageViewModel CurrentPageViewModel
{
get
{
return _currentPageViewModel;
}
set
{
if (_currentPageViewModel != value)
{
_currentPageViewModel = value;
PropertyChanged?.Invoke(this, new PropertyChangedEventArgs("CurrentPageViewModel"));
}
}
}
private IPageViewModel _currentPageViewModel;
Every ViewModel implements simple IPageViewModel interface so only ViewModels could be set as content of ContentPresenter.

WPF mvvm DataTemplates change View from UserControl

I am new to WPF and the MVVM pattern. I am trying to build a 'step by step' or 'wizard' like application.
So the user should first login then select some data and finally the selected data should be stored somewhere. (This flow never changes!)
However I decided to use DataTemplates and different ViewModels for each Template and a MainViewModel for the Window which populates the Templates. (Think this should be ok regarding to different Posts here)
But now my problems are starting. I know how I can change the current view from the MainViewModel BUT I want to change the current View from the inner ViewModel. And I want to be able to collect data from one inner ViewModel to another and I have no clue how this can work.
MainViewModel:
public class MainWindowViewModel : ViewModelBase
{
public ViewModelBase CurrentView {get; set;}
public MainWindowViewModel()
{
CurrentView = new ViewModelA;
}
}
XAML:
<Window x:Class="PUSEImporter.MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:VM="clr-namespace:PUSEImporter.ViewModel"
xmlns:V="clr-namespace:PUSEImporter.View">
<Window.DataContext>
<VM:MainWindowViewModel />
</Window.DataContext>
<Window.Resources>
<DataTemplate DataType="{x:Type VM:ViewModelA}">
<V:Detail />
</DataTemplate>
<DataTemplate DataType="{x:Type VM:ViewModelB}">
<V:Overview />
</DataTemplate>
</Window.Resources>
<ContentPresenter Content="{Binding CurrentView}"/>
</Window>
So think about a button in ViewModelA (or the View from ViewModelA) and now I want to switch to ViewModelB when someone clicks on the button. And not enough the data which is collected by ViewModelA should also be available in ViewModelB.
Is this possible and the preferred way of using this techniques or do I misunderstand some concepts?
(And if this is true how should i handle things like that?)
Thanks in advance!
There are many ways to achieve what you want. In MVVM, there's one view model to each view... therefore, if your main view has a child view, then your main view model should have a property of the type of another view model. In this instance, you can add a delegate to the child view model and register a handler for it in the main view model.
This will enable you to effectively pass a signal to the main view model from the child view model, which you can react to in the parent view model in any way you want to. Rather than explain the whole story again here, please see my answers from the Passing parameters between viewmodels and How to call functions in a main view model from other view models? posts here on Stack Overflow for more information on how to achieve this.

Silverlight Switching views in a view using MVVM Light framework

I have searched and tried for days and finally must ask the question here.
I have a Silverlight 5 application, Using MVVM Light, where I want to be able to dynamically switch views in the main view.
For the sake of simplicity, lets say I have 2 buttons.
Button1 will switch to TestView1.
Button2 will switch to TestView2.
<Button Content="TestView1" Grid.Column="1" Command="{Binding CallTestView1Command}" HorizontalAlignment="Left" Margin="185,17,0,0" VerticalAlignment="Top" Width="75"/>
<Button Content="TestView2" Grid.Column="1" Command="{Binding CallTestView2Command}" HorizontalAlignment="Left" Margin="280,17,0,0" VerticalAlignment="Top" Width="75"/>
The way I have done it is by binding a relaycommand to the button and then instanciating a new viewmodel of the corresponding view.
ie:
private RelayCommand _callTestView1Command;
public RelayCommand CallTestView1Command
{
get
{
return _callTestView1Command ??
(_callTestView1Command = new RelayCommand(() =>
{
CurrentView = ViewModelLocator.NinjectKernel.Get<TestViewModel1>();
}));
}
}
The CurrentViewmodel is then set to the new viewmodel.
In the MainView I have bound the CurrentView to a ContentControl:
<Border x:Name="displayedView" Grid.Row="2">
<ContentControl Content="{Binding CurrentView}" />
</Border>
This will actually work to some extend, since the CurrentView will change but instead of actually showing the content of the view it simply shows the Namespace of the ViewModel that is instanciated.
So far I have primarily used the knowledge taken from these sources:
http://rachel53461.wordpress.com/2011/05/28/switching-between-viewsusercontrols-using-mvvm/
Loading Views into ContentControl and changing their properties by clicking buttons
but they do not solve my problem, or I do not quite understand how to actually show the views.:-(
So does anyone have a good explanation on how to switch the views correct in Silverlight 5 using MVVM Light from GalaSoft.
Thanks
The part you are missing is the DataTemplates that tell WPF how to render your ViewModels
<Window.Resources>
<DataTemplate TargetType="{x:Type local:TestViewModel1}">
<local:TestView1 />
</DataTemplate>
<DataTemplate TargetType="{x:Type local:TestViewModel2}">
<local:TestView2 />
</DataTemplate>
</Window.Resources>
When you insert an object in the Visual Tree, such as placing a ViewModel object in ContentControl.Content, it will get drawn by default using a TextBlock bound to the .ToString() of the object, which is why you are only seeing the namespace.classname of the ViewModel in your ContentControl
By defining an implicit DataTemplate in your Resources somewhere (that's a DataTemplate with only a TargetType defined - no x:Key), you are telling WPF to draw the specified object using the specified DataTemplate anytime it tries to draw that object, instead of using the default TextBlock bound to the .ToString() of the object.
It should be noted that implicit DataTemplates are not supported in earlier versions of Silverlight, however they are supported in 5.0+. For earlier versions of Silverlight, I usually use a DataTemplateSelector instead.
Id first suggest that you do not display your views via a ContentControl but look into using the navigation Frame in the silverlight toolkit. Also, we dont want our ViewModel creating Views... that'd not be so good. We don't mind, however, if our ViewModel does business logic and DETERMINES which view to show. Get the toolkit here: http://silverlight.codeplex.com/
Now setup your XAML as so in your main page:
<Border x:Name="displayedView" Grid.Row="2">
<navigation:Frame x:Name="ContentFrame" />
</Border>
Since you are using MVVM Light, we will use messaging. Your View model will get the command to change views, determine which view to change, then send a message to the main page to instruct it to change views.
Setup a listener in your main page for a navigate request as so:
public MainPage()
{
InitializeComponent();
Messenger.Default.Register<Uri>(this, "NavigationRequest", (uri) => ContentFrame.Navigate(uri));
}
Next, setup your command in your view model.
private RelayCommand _callTestView1Command;
public RelayCommand CallTestView1Command
{
get
{
return _callTestView1Command ??
(_callTestView1Command = new RelayCommand(() =>
{
Messenger.Default.Send<Uri>(new Uri("/Views/.../Page.xaml", UriKind.Relative), "NavigationRequest");
}));
}
}
These are the basics that work for me. You can expand on this and get real "architecty". For example, you can create a base class for you view models that sends the navigation requests, create a helper class that generates URIs (so they are not hard coded everywhere in your app, etc etc. Good luck!
So i actually solved this problem, in a way where there is no need to create datatemplates in the MainView, which i did not like. imo the MainView should know nothing about the views it is displaying, when we are talking about switching the views.
Prerequisite: You must use MVVM Light from GalaSoft for this solution
This is my test solution:
Two buttons are added to my MainView, Each button will open a new view. The clickevent are bound to Commands.
<Button Content="TestView1" Grid.Column="1" Command="{Binding CallTestView1Command}" HorizontalAlignment="Left" Margin="185,17,0,0" VerticalAlignment="Top" Width="75"/>
<Button Content="TestView2" Grid.Column="1" Command="{Binding CallTestView2Command}" HorizontalAlignment="Left" Margin="280,17,0,0" VerticalAlignment="Top" Width="75"/>
In the MainView i have a Border that should contain the views than can switch.
Since all views inherit from UserControl i bind the content to the property CurrentView of the MainViewModel
<Border x:Name="displayedView" Grid.Row="2">
<UserControl Content="{Binding CurrentView}" />
</Border>
In the MainViewModel i have the property CurrentView.
public const string CurrentViewPropertyName = "CurrentView";
private UserControl _currentView;
/// <summary>
/// Sets and gets the "CurrentView property.
/// Changes to that property's value raise the PropertyChanged event.
/// </summary>
public UserControl CurrentView
{
get
{
return _currentView;
}
set
{
if (_currentView == value)
{
return;
}
RaisePropertyChanging(CurrentViewPropertyName);
_currentView = value;
RaisePropertyChanged(CurrentViewPropertyName);
}
}
When a button is clicked the corresponding Command is called in the MainViewModel:
private RelayCommand _callTestView1Command;
public RelayCommand CallTestView1Command
{
get
{
return _callTestView1Command ??
(_callTestView1Command = new RelayCommand(() =>
{
CurrentView = new TestView1();
}));
}
}
private RelayCommand _callTestView2Command;
public RelayCommand CallTestView2Command
{
get
{
return _callTestView2Command ??
(_callTestView2Command = new RelayCommand(() =>
{
CurrentView = new TestView2();
}));
}
}
As seen each command will set CurrentView to a new view, and the views will switch in the MainView, because CurrentView will raise a ProperTyChanged Event.
This will actually work to some extend, since the CurrentView will
change but instead of actually showing the content of the view it
simply shows the Namespace of the ViewModel that is instanciated.
Because you are changing the CurrentView property to a viewmodel instance and bind that as the Content. This is wrong as the Content should be a view and you should set the DataContext of that view to a viewmodel.
The simplest thing you can do here is to create a View instance inside the command and set the viewmodel as its DataContext and then you can set the view to the CurrentView property. Of course this would violate the MVVM pattern so you should move this responsibility to a separate component. Instead of writing your own navigating logic I suggest you to pick up an existing solution as this kind of task is not as straightforward as it seems.
I suggest to use the Prism library

how can I set the xaml content in viewmodel

I've had a hyperlink button where i set in the button click in code behind the content to a new view if the login success.
private void OkButtonClick(object sender, RoutedEventArgs e)
{
LoginOperation loginOp = FLS.Utilities.RIAWebContext.Current.Authentication.Login(
new LoginParameters(usernameTextBox.Text, passwordTextBox.Text));
loginOp.Completed += (s2, e2) =>
{
if (loginOp.HasError)
{
errorTextBlock.Text = loginOp.Error.Message;
loginOp.MarkErrorAsHandled();
return;
}
else if (!loginOp.LoginSuccess)
{
errorTextBlock.Text = "Login failed.";
return;
}
else
{
errorTextBlock.Text = string.Empty;
Content = new WelcomeView();
}
};
}
I've now moved the code behind for MVVM in a view model and use a delegateCommand on the hyperlink button.
<UserControl ... >
<Grid ... >
...
<HyperlinkButton Content="Login" Height="23" HorizontalAlignment="Left" Margin="313,265,0,0" Name="loginButton" Command="{Binding Path=LoginCommand}" VerticalAlignment="Top" Width="75"/>
...
</Grid>
</UserControl>
But I don't know, how I make the Content = new WelcomeView(); from the code behind in the viewmodel?
A good design pattern would be to have two different Data Template, one to present the data before the login, and the second Data Template to be used after the login.
There are several ways to achieve that. The one that I typically use simply put the ViewModel (directly of using binding) the only children of Window.
In your ViewModel implement a content selector class. This is a class derived from DataTemplateSelector and uses the FindResource API to get the appropriate data template.
<Window ...>
<Window.Resources>
<DataTemplate x:key="beforeLogin">
...
</DataTemplate>
<DataTemplate x:Key="afterLogin">
...
</DataTemplate>
</Window.Resources>
<Window.ContentTemplateSelector>
<code:MyTemplateSelector />
</Window.ContentTemplateSelector>
<-- Here is the content of Window. It's the view model (data). The View will be
bind by the TemplateSelector
<code:YourViewModel />
</Window>
Check out this page: http://msdn.microsoft.com/en-us/library/system.windows.controls.contentcontrol.contenttemplateselector.aspx for a related example.
There are other design pattern. Another common idiom is simply firing a "UiRequest" event, which will be picked up by the code-behind of the view. Remember that MVVM dictates the ViewModel is "view agnostic", but it really doesn't mean "no code behind". This means the VM cannot reference anything in the view. Communication this way happens view events (e.g. data binding is just a wrapper around property changed events). So have an event UiRequest in your View Model, and design a protocol. In the constructor of the View - register a handler. In the handler, change the content (people use this idiom mainly to start a popup window, but it can be used anywhere).

how to load wpf usercontrol in MVVM pattern

I'm creating a wpf user control which is in mvvm pattern.
So we have : view(with no code in codebehind file), viewmodel,model,dataaccess files.
I have MainWindow.xaml as a view file, which I need to bind with MainWindowModel.cs.
Usually, in a a wpf application we can do this with onStartUp event in App.xaml file. But in user control, as we do not have App.xaml...How do I achieve it ?
Please help :(...Thanks in Advance !!!
You can use a ContentControl, with a DataTemplate to bind the UserControl (View) to the ViewModel :
<DataTemplate DataType="{x:Type vm:MyViewModel}">
<v:MyUserControl />
</DataTemplate>
...
<ContentControl Content="{Binding Current}" />
WPF will pick the DataTemplate automatically based on the type of the Content
I know this is an old, answered question, but I have a different approach. I like to make implicit relationships in the App.xaml file:
<Application.Resources>
<DataTemplate DataType="{x:Type ViewModels:KioskViewModel}">
<Views:KioskView />
</DataTemplate>
</Application.Resources>
With this, there is no need to set a DataContext anywhere.
UPDATE >>>
In response to #Vignesh Natraj's request, here is a fuller explanation:
Once you have set up the DataTemplate in a Resources element, you can display the KioskView in this example by adding an instance of the KioskViewModel anywhere in your XAML. This could be filling the MainWindow, or just inside a particular section of the screen. You could also host multiple instances of the KioskViewModel in a ListBox and it will generate multiple KioskView instances.
You can add an instance of the KioskViewModel to your XAML in a couple of ways, depending on your requirements. One way is to declare the XML namespace for the project that contains the KioskViewModel.cs file and simply add an instance of it in a ContentControl to the page where you want your view to appear. For example, if you had a UserControl called MainView and the KioskViewModel.cs file was in a Kiosk.ViewModels namespace, you could use basic XAML like this:
<UserControl x:Class="Kiosk.Views.MainView"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:ViewModels="clr-namespace:Kiosk.ViewModels">
<UserControl.Resources>
<ViewModels:KioskViewModel x:Key="KioskViewModel" />
<DataTemplate DataType="{x:Type ViewModels:KioskViewModel}">
<Views:KioskView />
</DataTemplate>
</UserControl.Resources>
<ContentControl Content="{StaticResource KioskViewModel}" />
</UserControl>
I prefer to use the MVVM design pattern with WPF, so I would have a base view model class providing useful functionality such as implementing the essential INotifyPropertyChanged interface. I then have a property called ViewModel in the main (top level) view model of type BaseViewModel. This provides me with a nice way to change the ViewModel property to any view model that has derived from BaseViewModel and therefore to be able to change the associated view from the view model.
For example, in the MainViewModel.cs class that is bound to MainView there is a field and relating property:
private BaseViewModel viewModel = new KioskViewModel();
public BaseViewModel ViewModel
{
get { return viewModel; }
set { viewModel = value; NotifyPropertyChanged("ViewModel"); }
}
As you can see, it starts off as a KioskViewModel instance, but can be changed to any other view at any time in response to user interaction. For this setup, the XAML is very similar, but instead of declaring an instance of the view model in the Resources element, we bind to the property in the MainViewModel:
<UserControl x:Class="Kiosk.Views.MainView"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:ViewModels="clr-namespace:Kiosk.ViewModels">
<ContentControl Content="{Binding ViewModel}" />
</UserControl>
Note that for this example, we would need to declare two (or more to make this approach useful) DataTemplates in the App.xaml file:
<Application.Resources>
<DataTemplate DataType="{x:Type ViewModels:MainViewModel}">
<Views:MainView />
</DataTemplate>
<DataTemplate DataType="{x:Type ViewModels:KioskViewModel}">
<Views:KioskView />
</DataTemplate>
</Application.Resources>
I've been using MVVM Light Toolkit which has a ViewModelLocator class that you can put properties to the viewmodels in. You then create a reference to the ViewModelLocator in your Mainwindow.xaml like so:
<vm:ViewModelLocator x:Key="Locator" d:IsDataSource="True"/>
In the grid panel, or whatever you're using, you can then set the datacontext like this:
<Grid DataContext="{Binding MainWindowViewModel, Source={StaticResource Locator}}">
...
</Grid>
You could also go with MEFedMVVM which potentially adds a bit more flexibility in terms of being able to swap different viewModel implementations into the view.
The flexibility in both of these libraries is that you don't have to use their ViewModel base classes if you don't want to - the ViewModelLocator and the MEFedMVVM can work with any class.
There are endless ways to do it, wich all fall in one of the two categories:"view first" or "model first".
In a "view first" mode the view (e.g. your mainwindow) is created first and then (e.g. in the codebehind) the View instantiates the ViewModel and sets it as its datacontext):
private void WindowLoaded(object sender, EventArgs args)
{
this.DataContext = ViewModelService.GetViewModelX();
}
In a "model first" mode the ViewModel is there first and then instanciated the View.
// method of the viewmodel
public void LoadView()
{
// in this example the view abstracted using an interface
this.View = ViewService.GetViewX();
this.View.SetDataContext(this);
this.View.Show();
}
The examples given here are just one way of many. You could look at the various MVVM frameworks and see how they do it.
We can use ObjectDataProvider to call a method inside an object ..as follows :
<ObjectDataProvider ObjectType="{x:Type local:TemperatureScale}"
MethodName="ConvertTemp"
x:Key="convertTemp">
Is there anyway to do the same using DataTemplate
You can probably look at MSDN. I find it as a good resource, though it doesn't explain how to use usercontrols,you will find your way out.

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