In a CM WPF application I have following items:
<Window> <!-- Shell Window -->
<ContentControl x:Name="MainRegion">
<ContentControl x:Name="SearchRegion">
<Window>
Views:
MainRegionView
SearchRegionView
View Models:
MainRegionViewModel //implements Conductor<Screen>
SearchRegionViewModel //implements Conductor<Screen>
I neet to switch the view of SearchRegionViewModel from MainRegionViewModel.
Following code is executed inside MainRegionViewModel.
ViewModelTest test = new ViewModelTest();
//Calling the method from object reference
searchRegionViewModel.ActivateItem(test);
Problem:
the view is not loaded in SearchRegionView.
How can i load the view to SearchRegion?
Your MainRegionViewModel & SearchRegionView need to implement Screen and your ShellViewModel is the one that need to implement the Conductor. After that, you need to bind the ActiveItem in the XAML (ShellView.xaml).
<ContentControl x:Name="ActiveItem"></ContentControl>
Inside your ShellViewModel use Activate(YourViewModel).
Here is an Example
If you want to change the ActiveItem inside your MainRegionViewModel, You need to look at the EventAggregator to publish a message to your ShellViewModel to Activate the desire ViewModel
Related
I am quite new to WPF development, and currently I am trying to use the MVVM on my application development. I have read a lot about MVVM navigation and switching views, but I can't find a solution for my current situation. Let's explain what it is:
First of all, I have my main View element, a Dockpanel, with some fixed areas, and a main "dynamic" area where the content should change, depending on actions:
<DockPanel>
<Label Content="Top Fixed element"/>
<StackPanel Orientation="Vertical" Height="auto" Width="150" DockPanel.Dock="Left">
<Label Content="SomeOptions"/>
<!-- some more elements -->
</StackPanel>
<Label DockPanel.Dock="Bottom" Content="Foot"/>
<ContentControl Content="{Binding CurrentMainViewElementViewModel}"/>
</DockPanel>
I have defined some DataTemplates that I would like to load in this ContentControl, here there is one of the Data Templates as example:
<Window.Resources>
<DataTemplate DataType="{x:Type ViewModel:FileLoaderVM}">
<View:FileLoaderView/>
</DataTemplate>
</Window.Resources>
This FileLoader (View and View Model are implemented, using the RelayCommand and the INotifyPropertyChanged) opens a dialog box after clicking a button, where after selecting a file it is opened and parsed, and show all the found elements inside a ListView with multiple selection(in this case, persons with their data).
What I want to do now is to load another user control in this ContentControl, when I click a button. This button is defined in my view model like this:
public ICommand LoadPersons
{
get { return new RelayCommand(param => this.loadSelectedPersons(), param => (SelectedPersons!=null && SelectedPersons.Any()));}
}
My question comes at this point, how can I modify the content of the ContentControl, loading another User Control instead of the current one directly from my view model (in this "this.loadSelectedPersons()")?
If this is not possible, how should I approach to solve this problem?
Next to this action, I want to show all the previously selected elements and manipulate in different possible ways (inserting in a DB, saving in another file and so on), and I have already for that the appropriate User Control, that I would like to show in my main view element in the ContentControl section, keeping the other elements as they are originally.
lets see if i get you right.
you have a mainviewmodel with a property (CurrentMainViewElementViewModel) bound to the ContentControl. your MainViewmodel set the FileLoaderVM to this Property. now you wanna show a "new/other" Viewmodel when a File is seleted in your FileLoaderVM?
why dont you simply expose a event from your FileLoaderVM and subscribe to this event in your MainViewModel? if you do so your MainViewModel can then set the "new/other" Viewmodel to the ContentControl
To change content of ContentControl you do not load another user control, but change value of CurrentMainViewElementViewModel (to which ContentControl.Content is bound) to a new ViewModel, which will load another UserControl (defined in DataTemplate same way as FileLoaderVM is).
This looks like a job for main ViewModel (where CurrentMainViewElementViewModel is located).
Easiest solution is to provide a method in that ViewModel
public Switch()
{
CurrentMainViewElementViewModel = SomeViewModel;
}
and call this method from FileLoaderVM.
This question is in reference to this link Opening new window in MVVM WPF.
I want to open new window in some service.
Following is my code
This is a window service which I am calling from ViewModel
public class WindowService : IWindowService
{
public void ShowWindow(object viewModel)
{
var win = new Window {Content = viewModel};
win.Show();
}
}
Following is my App.xaml code
<DataTemplate DataType="{x:Type viewModel:MainViewModel}" >
<viewModel:ChildWindow />
</DataTemplate>
Now this works fine for all windows which have different ViewModel.
But when I want to open another window which is using same view model but their view is not same,I can not define data template of same type in App.xaml.
How do I open multiple new window which have same ViewModel?
Should I create different ViewModel for each window?
I have shown an alternative way of opening and closing windows here.
You could add a DependencyProperty e.g. DataContext to the OpenCloseWindowBehavior, use it to pass the ViewModel, and hook up the window's DataContext to this ViewModel in the code of the behavior. Tell me if you need more help.
<local:OpenCloseWindowBehavior WindowType="local:YellowWindow" Open="{Binding YellowOpen, Mode=TwoWay}" DataContext="{Binding SomeViewModel}" />
I using from mvvm in my application. I want know how to define my user control in mvvm pattern.
Must I define it by using from mvvm, or I can define it generally?
Let's just call the control that embeds the user control MainWindow, and the user control UserControl. Since you are in MVVM pattern, you have at least one View Model for the outer view - I usually use the name MainVm.
You have two choices for the user control: They can share the same View Model, or you could have a sub view model, just for the UserControl, i.e. UserVm.
For your first choice, you do nothing. You define UserControl (Visual Studio 'add new item' -> User Control is a pretty good start). Then, you simply embed it in Main Window.
<Window
x:Class="SO.MainWindow"
...
xmlns:src="clr-namespace:SO"
...
>
...
<src:UserControl />
...
</Window>
UserControl will inherit the same DataContext from MainWindow, and do all the {Binding} as you would do in the MainWindow.
If you want to have a sub view model (UserVm) - it would typically be a public property of the MainVm (say, userVm). In that case, you'll set the DataContext of the UserControl when you reference it.
<src:UserControl DataContext="{Binding Path=userVm}" />
Another popular paradigm would be to declare the DataTemplate instead of the UserControl. If you do that, you just need to put the UserVm (either instantiate it in the XAML, or through binding):
<Window x:Class="MainWindow" ...>
<Window.Resources>
<DataTemplate x:Key="UserDt"> <!-- or user TargetType instead of x:Key -->
...
</DataTemplate>
</Window.Resources>
...
<!-- You can put in a ContentControl like here: -->
<ContentControl Content="{Binding Path=userVm}"
ContentTemplate="{StaticResource UserDt}" />
<!-- or, if you defined TargetType for the DT, you can simply instantiate
the sub VM here. I don't like this apporach but it exists. -->
<src:UserVm />
</Window>
I think that depends on the user control. The user control can be just a view, in which case you would compose a larger control or page which has this user control as part of the whole. The larger control or page would provide the view and the view model parts for this view.
Or you could create a self contained user control which has all of mvvm and use events to interact with the larger user control that it is a part of.
I suspect you'll get better reuse and modularisation with the second approach.
In short: it depends.
My Main window defines the markup for the application, for this specific scenario lets say I have a grid with 2 columns.
First column will have navigation links, and second column will display the different views.
There are 2 views (and 2 viewmodels) defined in mainwindow xaml:
<Window.Resources>
<DataTemplate DataType="{x:Type vm:Window1ViewModel}">
<vw:Window1View/>
</DataTemplate>
<DataTemplate DataType="{x:Type vm:Window2ViewModel}">
<vw:Window2View/>
</DataTemplate>
</Window.Resources>
And in second grid column that displays the views i got :
<ContentControl Content="{Binding Path=ViewModel}" HorizontalAlignment="Left">
</ContentControl>
Where ViewModel is a property that I set accordingly to a view(viewmodel) that i want to display.
Like :
ViewModel = new Window1ViewModel();
(datacontext of the mainwindowview is set to MainWindowViewModel)
So there is no problem to switch between views from the MainWindowViewModel.
My problem is how to switch within Window1ViewModel into Window1ViewMode2?
The various ViewModels don't "know" about other ViewModels.
Only MainWindowViewModel knos about others...
How can I solve this?
Maybe I should define a custom Event (with parameter), MainWindowViewModel will subscribe and other viewmodels will trigger it and then MainWindowViewModel will switch to the needed view?
the solution you describe is one possibility. One other I can think of is using some kind of Navigation-Service (static class or interface you pass to all your child-Viewmodels) that do this kind of work.
If your MainWindowViewModel creates all the others I would stick to the interface solution. You can for example let the MainWindowVM implement such a interface and inject it into all the child-vm on creation. This is much the same as your event-approach but instead of the childs-providing and the main having to subscribe you have the main give something ... IMHO the better approach.
Ok, may be I understood your point. You want that controller actually be modelview which notifies to mainmodelview about the fact that it have to be swapped with someone else.
Considering that we are talking about WPF, create DependecyProperty on mainmodelview , and set it from childview, which in code behind will trigger modelviews swap.
I have a WPF Window which contains few UserControls, those controls contain another. And now, what is the most principal way how to create ViewModel for this Window and where to bind it.
I do expect that one firstly needs to create ViewModel for each of sub-controls.
There are a few ways to do this.
Inject the VM
I would recommend this method.
If your window is created in the App class like
var window = new MyWindow();
window.Show();
I would assign the VM before showing the window:
var window = new MyWindow();
window.DataContext = GetDataContextForWindow();
window.Show();
If one of your controls needs an own view model assign the VM wile creating the control instance.
DataBind
If you want to set the VM of a control you can bind the DataContext property to an VM instance provided by the surrounding VM.
<Controls:MyControl DataContext={Binding MyControlsVm} />
Code Behind
You may set the VM using the init method in code behind like
public MyWindow()
{
InitializeComponent();
DataContext = CreateViewModel;
}
You may use a trick if you don't want to create a VM for your main page:
public MyWindow()
{
InitializeComponent();
DataContext = this;
}
and just use the code behind class as VM.
I see the view as a visual representation of the ViewModel so I like WPF picking the view based on the instance of the ViewModel it wants to render.
I call this the View Locator pattern, I use this method to instantiate my view because I have found it to be very simple to implement.
It basically puts an entry in the ResourceDictionary of your app that tells WPF to use an IValueConverter to look up and instantiate the View when it comes across a ViewModel.
So a working example would be:
In your app.xaml:
<Application x:Class="MyApp.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
StartupUri="MainWindow.xaml" >
<Application.Resources>
<ResourceDictionary Source="Resources.xaml"/>
</Application.Resources>
</Application>
In resources.xaml:
<DataTemplate DataType="{x:Type vm:ViewModelBase}">
<ContentControl Content="{Binding Converter={StaticResource ViewModelConverter}}"/>
</DataTemplate>
Set the DataContext of your startup Window Control e.g.
public MainWindow : Window
{
InitializeComponent();
DataContext = new MainViewModel();
}
And you're pretty much done. So if you have a MainViewModel like so:
public class MainViewModel : ViewModelBase
{
public ChildViewModel1 Child1 {get;set;}
public ChildViewModel2 Child2 {get;set;}
}
and you have a UserControl that resolves to your MainViewModel like so:
<UserControl x:Class="MainView">
<StackPanel>
<ContentPresenter Content="{Binding Child1}"/>
<ContentPresenter Content="{Binding Child2}"/>
</StackPanel>
</UserControl>
So your ViewModelConverter will return an instance of the appropriate View without any extra effort on your part.
On the child controls issue, why wouldn't one of the properties of the root view model be an instance of the child view model that you would pass onto the child control? The other option would be a converter that converts the non-view model based property into an instance of the child view model (like an adapter pattern).
You might be interested in the sample applications of the WPF Application Framework (WAF). They show how composite Views and ViewModels can be instantiated and how they interact which each other.