While all examples and sources I found is to set the resource in XAML statically, I would only know in run time the name of the XML file to be connected with XMLDataProvider. Is there a way to set either in code behind or in XAML?
<Window.Resources>
<XmlDataProvider x:Key="XMLFoo" Source="Foo.xml" XPath="Foo"/>
</Window.Resources>
It could be Foo.xml, or could be Goo.xml.
Yes you can change that while runtime. Unfortunately you cannot bind it, so you have to do stuff in Code-Behind.
Here's a simple example:
(this.Resources["XMLFoo"] as XmlDataProvider).Source = new Uri("Goo.xml");
Cheers
If you are trying to have just one instance of XamlDataProvider and would like your source to change dynamically, I don't think it is possible in pure XAML as you cannot bind to Source property, since that is not a DependencyProperty.
From code-behind, you can get the instance your provider and change it's source.
var provider = (XmlDataProvider) Resources.FindName("XMLFoo");
provider.Source = new Uri("bar.xml", UriKind.Relative);
Alternatively, you can use MVVM and expose your XmlDataProvider as a property on the ViewModel and bind it to your View, you can then change the Source and refresh data from the ViewModel itself.
Related
I'm trying to set a new DataTemplate as a new Window resource in my MainWindow derived from the System.Windows.Window class. The code for the XAML is quite simple and looks like this:
<Window.Resources>
<DataTemplate DataType="{x:Type model:MyViewModel}">
<view:MyView />
</DataTemplate>
</Window.Resources>
What I exactly do here?
I try to show my data (MyViewModel) in or as a specific view (MyView). So far I do understand. Otherwise I wouldn't see the form itself, but the view model as a string with my.namespace.MyViewModel in the window.
But programmatically I do not understand, how to achieve the same. I know, that I have to add a new DataTemplate to the resources of my window. For this I have to "tell" the DataTemplate, which view to use (for the representation) and which data I want to represent, right?
So it must be something with:
DataTemplate template = new DataTemplate();
template.DataType = typeof(MyViewModel);
// Something, something ...
this.Resources.add(...);
Is this the right way to go? Or am I completely wrong?
I searched the web for solutions and also my WPF book, but there are only XAML implementations.
Why do I do that?
I have a headered content control which loads view models dynamically. The problem here is, that the user controls are sometimes dynamic and in case of data presentation I need to assign a specific (dynamic created) view to the data. So I try to load the current static user controls also in the way shown above.
Is there a way to go?
Or is there a better way to achieve the same results?
Whenever I bind text property to view model property
<TextBlock Text="{Binding SomeExampleText}"/>
on the designer I see nothing in the place where my text will appear in runtime. When I use x:Bind:
<TextBlock Text="{x:Bind ViewModel.SomeExampleText}"/>
on the designer I see "ViewModel.SomeExampleText", sometimes it does not display full length because of lacking space (if the binding path is too long).
Is there any way to have custom text displayed in designer just for preview instead of binding path or nothing as ashown above?
There are ways to create view models specifically for design time. The simplest approach is probably this:
<TextBlock Text="{x:Bind ViewModel.SomeExampleText, FallbackValue='Hello!'}"/>
That one shows the string "Hello" in the designer, both with Binding and x:Bind.
For Binding you can set the design-time data context something like this:
<Page
...
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:viewModels="using:MyNameSpace.ViewModels"
d:DataContext="{d:DesignInstance Type=viewModels:DesignTimeViewModel, IsDesignTimeCreatable=True}"
mc:Ignorable="d">
The DesignTimeViewModel doesn't need any spcecific relation to your run-time view model; it only needs to have suitable properties with the same names. If you are binding to collections, this is probably your best bet.
Create a specific view model for the Design view in order to dial in the view's layout and style with realistic data that won't enter the run-time environment.
Since x:Bind looks to the code-behind for a strongly typed data source, you'll need to mimic that data binding path in the Design view. Here's one way:
Apply the d:DataContext attribute to your view set the Type property to your view. When "IsDesignTimeCreatable" is true, it'll create a new instance of your code-behind.
d:DataContext="{d:DesignInstance Type=local:MainPage,IsDesignTimeCreatable=True}"
Your code-behind likely has a ViewModel property that can be set to a design-time state with fake data or a run-time state with real data.
This blog post shows an example: http://fast417.blogspot.com/2016/06/uwp-design-preview-with-xbind.html
I have a UserControl named "WorkspaceView" and its only purpose is to show other views as tabs. Call these views ViewA, ViewB etc. Which of these views to present should be determined on runtime, so I figured I needed a control that can present ... well ..stuff.
ContentControl to the rescue. Except ... I can't make it work. I'm trying to new up a usercontrol of type ViewA in the code behind and assign it to my MyContent, which is the ContentControl. I've tried:
public WorkspaceView()
{
InitializeComponent();
DataContext = new View(A); //Hoping that the DataContext will propagate down
}
Second attempt was
public WorkspaceView()
{
InitializeComponent();
var binding = new Binding {Source = new ViewA()};
MyContent.SetBinding(ContentControl.ContentProperty, binding);
}
In both cases, I see an empty box, but since I've hard wired a TextBlock into ViewA, I'd expect it to show me that text. What am I doing wrong?
Despite knowing that MVVM is the preferred way to develop WPF applications, I'd prefer to see how I can do this with code behind files. Later on, I will redo the application with MVVM, but first I need to get some basic understanding of WPF.
In response to the suggestions so far, I've tried
MyContent.Content = new ViewA();
but still I the text that is in ViewA does not appear. I've also at the bottom of this post included a screenshot of what the application renders.
WorkspaceView
Resource file
What is rendered
Have you tried simply doing this?
MyContent.Content = new ViewA();
EDIT
Try simplifying your code a bit and working from there. For instance:
public WorkspaceView()
{
InitializeComponent();
// Something better than UserControl should be used here
ObservableCollection<UserControl> views = new ObservableCollection<UserControl>();
views.Add(new ViewA());
views.Add(new ViewB());
DataContext = views;
}
<Border ..>
<TabControl x:Name="TabControl"
..
ItemsSource="{Binding}" />
</Border>
This code sets a WorkspaceView.DataContext to a collection of UserControls. When you specify {Binding} whithin WorkspaceView's XAML you are refering to the whole DataContext object (i.e. your collection.) This way you are setting the TabControl.ItemsSource to your collection of views.
Now you could create DataTemplates targeting the type of each view to control how each control is displayed in its tab within the TabControl.
This is a question that extends from the originally posted here:
Link to loading-xaml through runtime
I'm working on a WPF MVVM application that loads XAML content dynamically from an external source, very similar as the answer in the post above.
Here is what I got so far:
My View declares an instance of the ViewModel as a resource and creates an instance of that ViewModel
In my ViewModel constructor I'm loading a XamlString property coming from an external source (file or db..)
In my view I have a button that user clicks after ViewModel finishes loading and in the click-event code-behind I'm deserializing the dynamically loaded XAML and add it to my grid.
My question is, how can I eliminate code-behind and automate the logic so the View can render the new xaml section dynamically right after the ViewModel is done getting the XAML content and initializing the string property?
Should I use some kind of Messaging Bus so the ViewModel notifies once the property has been set so the View can add the new content?
What troubles me is the fact that ViewModels do have a reference to Views and should not be in charge of generating UI elements.
Thanks in advance!
Edit:
Just to clarify: in my particular case I am not trying to bind a Business Object or Collection (Model) to a UI element (e.g. Grid) which obviously could be accomplished through templates and binding. My ViewModel is retrieving a whole XAML Form from an external source and setting it as a string property available to the View. My question is: Who should be in charge of deserializing this XAML string property into a UI element and add it programmatically to the my grid once my Xaml string property in the VM is set?
This sounds to me more of like a View responsibility, not ViewModel. But the pattern as i understand it enforces to replace any code-behind logic with V-VM bindings.
I have a working solution now and I'd like to share it. Unfortunately I did not get rid of code-behind completely but it works as I expect it to. Here is how it works(simplified):
I have my simplified ViewModel:
public class MyViewModel : ViewModelBase
{
//This property implements INPC and triggers notification on Set
public string XamlViewData {get;set;}
public ViewModel()
{
GetXamlFormData();
}
//Gets the XAML Form from an external source (e.g. Database, File System)
public void GetXamlFormData()
{
//Set the Xaml String property
XamlViewData = //Logic to get XAML string from external source
}
}
Now my View:
<UserControl.Resources>
<ViewModel:MyViewModel x:Key="Model"></ViewModel:MyViewModel>
</UserControl.Resources>
<Grid DataContext="{StaticResource Model}">
<Grid.RowDefinitions>
<RowDefinition Height="Auto"/>
<RowDefinition/>
</Grid.RowDefinitions>
<StackPanel>
<!-- This is the Grid used as a Place Holder to populate the dynamic content!-->
<Grid x:Name="content" Grid.Row="1" Margin="2"/>
<!-- Then create a Hidden TextBlock bound to my XamlString property. Right after binding happens I will trigger an event handled in the code-behind -->
<TextBlock Name="tb_XamlString" Text="{Binding Path=XamlViewData, Mode=TwoWay, UpdateSourceTrigger=LostFocus, NotifyOnValidationError=True, ValidatesOnDataErrors=True, ValidatesOnExceptions=True}" Visibility="Hidden" Loaded="tb_XamlString_Loaded" />
</StackPanel>
</Grid>
Basically I created a hidden TextBlock bound to my XAML String property in the ViewModel and I hooked its Loaded event to an event handler in the code behind of the View:
private void tb_XamlString_Loaded(object sender, RoutedEventArgs routedEventArgs)
{
//First get the ViewModel from DataContext
MyViewModel vm = content.DataContext as MyViewModel;
FrameworkElement rootObject = XamlReader.Parse(vm.XamlViewData) as FrameworkElement;
//Add the XAML portion to the Grid content to render the XAML form dynamically!
content.Children.Add(rootObject);
}
This may not be the most elegant but gets the job done. Like some people say, in MVVM there are some cases like this where little code-behind code is needed. It doesn't hurt and also part of this solution still uses the V-VM Binding principles when using the VM to retrieve and populate the XamlString property and exposing it to the View. If we would like to Unit Test the XAML parsing and loading functionality we could delegate it to a separate class.
I hope someone finds this useful!
I'm having trouble understanding what you're saying, so my answer will be based on my interpretation. You should consider posting a sample (simplified) of what you're trying to do.
1) I think you're misunderstanding what MVVM does. MVVM is mostly a binding-based pattern. Your view model should be exposing properties containing business objects and your view should just be binding to those properties. If I am misunderstanding you, and that's what you are doing, then your problem is that your view needs to be aware of when the properties get updated (after you deserialize your xaml, etc). There are two ways to do this: INotifyPropertyChanged interface on your viewmodel, or make your view model inherit from DependencyObject, and make the properties dependency properties. I won't go into details here, because this is a large subject that you should research on Google before making a decision.
2) Generally speaking, you shouldn't use click events inside your view if you're using MVVM. Instead, create properties on the view model of type ICommand (and create ICommand implementations to match, or use an implementation of DelegateCommand (google it) which will allow you to use delegates to implement the interface. The idea is, your view binds to the property and executes the handler directly inside the viewmodel.
3) If you want to push information from the viewmodel to the view, then you should create an event on the viewmodel and subscribe to it in the view, but this is a last resort, only to be used in cases like displaying a new window, etc. Generally, you should be using binding.
4) To be more specific about what you're doing, you should be binding your Grid's ItemsSource property to some property on the view model. Note, the property on the view model should be of type ObservableCollection<T> if you want to be able to add items and get instant updates.
Hope this helps.
Is it possible to create data-template for the list box in WP7 using C# code instead of XAML??
You can't instantiate a DataTemplate in code in the same way that you can for regular controls, but you can use the XamlReader.Load() method to create a DataTemplate from a XAML string:
string xaml = #"<DataTemplate
xmlns=""http://schemas.microsoft.com/winfx/2006/xaml/presentation""
xmlns:x=""http://schemas.microsoft.com/winfx/2006/xaml"">
<!-- Template content goes here. -->
</DataTemplate>";
var dt = (DataTemplate)XamlReader.Load(xaml);
Be sure to add any additional namespaces that you might need.
The answer to this question also shows that you can create bindings in the DataTemplate in the same way: Creating a Silverlight DataTemplate in code.