Have you ever tried to use your Code-Behind extensively in WPF rather than using XAML to design your objects? Certainly, there is no way you would like to do this when you have an option to avoid, but sometimes it is almost necessary to do such code.
You may argue, even you are smart enough to deal with such a situation yourself using DataTemplates and Resources and make your application truly data driven. Yes it is possible, MVVM addresses such situations, but as I must say it is not always a silver bullet to address situations like this, it is more likely you might device your work more in code. Yes, even if you ask me, I am also less comfortable in code to do the same design as I am in XAML.
Today, one of my pals asked me the code which would add some Random Ellipse in a black Canvas and these points will also move in the screen using a DoubleAnimation
. As far as my initial thought is concerned, the code seemed good to me. The code looks like:
public void CreateRandomStars()
{
int Width = (int)MyCanvas.Width;
int Height = (int)MyCanvas.Height;
Random Rnx = new Random(0);
for (int i = 0; i < 300; i++)
{
double Lx = (double)Rnx.Next(Width);
double Ty = (double)Rnx.Next(Height);
Ellipse es = new Ellipse();
es.Name = "es_" + i.ToString();
es.Width = 2;
es.Height = 2;
es.Fill = Brushes.WhiteSmoke;
es.SetValue(Canvas.LeftProperty, Lx);
es.SetValue(Canvas.TopProperty, Ty);
MyCanvas.Children.Add(es);
DoubleAnimation dAnim = new DoubleAnimation();
dAnim.From = Ty;
dAnim.To = Height;
dAnim.Duration = new Duration(new TimeSpan(0, 0, 5));
Storyboard.SetTargetName(dAnim, es.Name);
Storyboard.SetTargetProperty(dAnim, new PropertyPath(Canvas.TopProperty));
MyStoryBoard.Children.Add(dAnim);
}
}
It adds the controls in a black canvas:
<Canvas Name="MyCanvas" Background="Black" Height="310" Width="500">
<Canvas.Triggers>
<EventTrigger RoutedEvent="Canvas.Loaded">
<BeginStoryboard>
<Storyboard RepeatBehavior="Forever" Name="MyStoryBoard">
</Storyboard>
</BeginStoryboard>
</EventTrigger>
</Canvas.Triggers>
</Canvas>
Well, you might think the code will run fine, but it actually won't.
Every control in WPF needs to be registered with its Scope after the object is created so that any other object within the same scope can have access to the object. In the code above, the registration of code is not done before it is actually accessed by the StoryBoard
.
MyCanvas.Children.Add(es);
this.RegisterName(es.Name, es);
After I add the line RegisterName
for the control, the code worked fine. Hence you should remember that you should use RegisterName
for a control before you address the name as Target
of another control, especially StoryBoard
s.
Why Do You Need RegisterName ?
Each WPF control is actually using Namescopes, to uniquely identify one element inside a scope. Each scope generally defines a Table
which gets an entry when a control Name
is registered, and deleted when it is Unregistered. Any control in WPF which allows child controls needs to implement from INameScope
interface to make a common scope for the control such that you can register the same name twice in two scopes but you cannot inside the same scope. Hence WPF allows you to
The concept is much like OOPS where objects can be of the same name in the same scope. But how does WPF achieve this.
Each FrameworkElement
derives from INameScope
which maintains a HashTable
to hold all the registered names, such that when you invoke container.RegisterName()
, the name will be automatically checked into the HashTable
and assigned. Animation extensively uses the INameScope
interface to determine the object at runtime which is set to its Target
. Hence to point to the actual object, you need to Register the control which needed to be accessed by StoryBoard
s.
If you look into container.RegisterName
module, it looks like:
Here a HybridDictionary
is used, which eventually maintains a HashTable
. When one name is registered, the object gets added against the Name
into the HashTable
and eventually when the object is accessed from the StoryBoard
s, it eventually points to the actual object from the Table
. The same is true for any INameScope controls.
Conclusion
That's it. I hope this post will help you.
Thank you.