I want to add a dynamic number of ellipses to a canvas and set the ellipse position (canvas.top, canvas.left). I tried binding to an ItemsControl but it each item (ellipse) has a container, so I cant set the ellipses position directly. I don't want an items container, I just want a canvas that contains ellipses.
Can this be done?
Try this - worked for me -- I use it to freely place textblocks on a canvas.
Re: Re: Positioning Items when Canvas is the ItemsPanel of an ItemsControl
02-26-2010 7:17 AM |
There is an alternative simpler solution that does work in silverlight 3.
<Canvas>
<ItemsControl ItemsSource={Binding MyItems}>
<ItemsControl.ItemTemplate>
<DataTemplate>
<Canvas>
<TextBlock Canvas.Left={Binding Left} Canvas.Top={Binding Top} Text={Binding Text} />
</Canvas>
</DataTemplate>
</ItemsControl.ItemTemplate>
</ItemsControl>
</Canvas>
If the MyItems is a list of items that are of a class that has Left, Top and Text public properties, this works fine. I have also tested with Line and Border to draw simple bar graph graphics in silverlight 3.
From the bottom of this post:
http://forums.silverlight.net/forums/p/29753/450510.aspx#450510
Combine it with a Silverlight DataTemplateSelector and you can change the objects you draw based on view model properties:
http://www.codeproject.com/KB/silverlight/SLTemplateSelector.aspx
Ordinarily I would say use an ItemsControl in conjunction with a Canvas:
<ItemsControl ItemsSource="{Binding Ellipses}">
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<Canvas/>
</ItemsPanelTemplate>
</ItemsControl.ItemsPanel>
<ItemsControl.ItemsContainerStyle>
<Style>
<Setter Property="Canvas.Left" Value="{Binding Left}"/>
<Setter Property="Canvas.Top" Value="{Binding Top}"/>
</Style>
</ItemsControl.ItemsContainerStyle>
</ItemsControl>
But in a display of Silverlight suckiness, the ItemContainerStyle property does not work on ItemsControl. It has support in ItemsControl, but it's not exposed by ItemsControl itself. Instead, it's up to subclasses of ItemsControl - such as ListBox - to expose it. Oh, and those subclasses have to be provided by Microsoft because the functionality is protected internal, so you can't just subclass ItemsControl and expose this stuff yourself. :S
So you could use ListBox instead, possibly by subclassing it and changing its item container to something simpler than a ListBoxItem. Or you could just use ListBox directly and fiddle around until the ListBoxItems look the way you want them to (i.e. not selected).
Related
I am trying to present the usercontrol, CharacterEditorView, from within an ItemsControl. (The use of "Canvas" in the ItemsPanelTemplate can be changed, Grid?). The CharacterEditorView is to overlay on top of a character string, so position is crucial. The ItemsControl is:
<ItemsControl Grid.Row="0" Grid.RowSpan="1" ItemsSource="{Binding CharacterPads}" >
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<Canvas />
</ItemsPanelTemplate>
</ItemsControl.ItemsPanel>
<ItemsControl.ItemContainerStyle>
<Style>
<Setter Property="Control.Margin" Value="{Binding Margin}"/>
</Style>
</ItemsControl.ItemContainerStyle>
</ItemsControl>
where the binding of CharacterPads is defined as:
private ObservableCollection<CharacterEditorViewModel> characterPads = new ObservableCollection<CharacterEditorViewModel>();
public ObservableCollection<CharacterEditorViewModel> CharacterPads
{
get
{
return characterPads;
}
}
I would like to present EACH CharacterEditorView within a specific rectangle known only at the time of its viewmodel creation. That is, each CharacterEditorView will have a different
rectangle to be presented in:
<UserControl x:Class="Nova5.UI.Views.CharacterEditorView"
.................................................................
<Grid Background="Red">
<TextBlock Text="TextBlock" Margin="{Binding ...?????}" />
</Grid>
</UserControl>
What binding is needed to position EACH CharacterEditorView at its own position?
Thanks for any ideas.
If you want to bind the TextBlock's Margin, you need some bindable property of type Thickness within the CharacterEditorViewModel and specify this as the binding target (or, for cleaner code, provide numeric properties in the ViewModel and implement a ValueConverter that creates a Thickness instance out of them).
Alternatively, since the ItemsPanelTemplate is already defined as Canvas, bind those numeric properties directly to Canvas.Left and Canvas.Top, as HighCore suggested.
In WPF using MVVM I would like to set a property in the view model to the displayed text when the mouse is clicked. That is I want the PreviewMouseDown event from the ItemsControl to set a property in the viewmodel.
In the following XAML, I am using an ItemsControl to display Strings from a FormattedText ObservableCollection. All goes well with the XAML below to display the FormattedText.
But, how can I bind a PreviewMouseDown to each of the generated items for the view model?
All my attempts to use DataTemplate within the ItemsControl ultimately lead to:
System.Windows.Data Error: 26 : ItemTemplate and ItemTemplateSelector are ignored for items already of the ItemsControl's container type;
XAML
<ItemsControl
ItemsSource="{Binding Strings}" >
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<Canvas
Background="Transparent"
Width="{x:Static h:Constants.widthCanvas}"
Height="{x:Static h:Constants.heightCanvas}"
/>
</ItemsPanelTemplate>
</ItemsControl.ItemsPanel>
</ItemsControl>
Adding
h:MouseBehaviour.PreviewMouseDownCommand="{Binding PreviewMouseDown}"
to the Canvas definition never results in the command being called and I can't add it in a DataTemplate.
Any help or better idea is appreciated.
as items in an ItemsControl are hosted in ContentPresenter so if you bind your command to the same it will be applied to the Item's in the ItemsControl
so for that purpose we can use a generic Style for ContentPresenter in the resources of ItemsControl or any parent container
eg
<Style TargetType="ContentPresenter">
<Setter Property="h:MouseBehaviour.PreviewMouseDownCommand"
Value="{Binding PreviewMouseDown}" />
</Style>
above example is based on assumption that PreviewMouseDown command is in the view model for each item, if the command is in the parent view model then you may perhaps use
<Style TargetType="ContentPresenter">
<Setter Property="h:MouseBehaviour.PreviewMouseDownCommand"
Value="{Binding PreviewMouseDown, RelativeSource={RelativeSource FindAncestor,AncestorType=ItemsControl}}" />
</Style>
Sorry that the title is a bit vague but I could'n come up with a better one.
For arguments sake let's say I'm developing a simple drawing application, where the user just clicks and drags to draw a line (I'm not really developing that, but just to keep it simple).
I have a custom shape for the line to draw. Now I want to add new lines to the view as needed, but I'd like to use an ObservableCollection property via data binding on the view model to do that. Usaully I would use an ItemsControl. But of course the ItemsControl automatically positions it's items, which is not what I want.
Does anyone have an idea how to do that? Is there a way to disable the layout functions of an ItemsControl?
You can change the ItemsPanelTemplate of an ItemsControl so it uses a Canvas instead of a StackPanel to hold its items, then use the ItemContainerStyle to bind the Canvas.Top and Canvas.Left properties to your data object to position them.
<ItemsControl ItemsSource="{Binding MyCollection}">
<!-- ItemsPanelTemplate -->
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<Canvas />
</ItemsPanelTemplate>
</ItemsControl.ItemsPanel>
<!-- ItemContainerStyle -->
<ItemsControl.ItemContainerStyle>
<Style>
<Setter Property="Canvas.Top" Value="{Binding Y}" />
<Setter Property="Canvas.Left" Value="{Binding X}" />
</Style>
</ItemsControl.ItemContainerStyle>
</ItemsControl>
I have a blog article about the ItemsControl that explains in more detail how an ItemsControl works if you're interested.
What can stop a TreeView from virtualizing if the TreeView is set up as follows?
<TreeView
ItemsSource="{Binding}"
VirtualizingStackPanel.IsVirtualizing="True">
<TreeView.ItemsPanel>
<ItemsPanelTemplate>
<VirtualizingStackPanel />
</ItemsPanelTemplate>
</TreeView.ItemsPanel>
<TreeView.ItemContainerStyle>
<Style
TargetType="{x:Type TreeViewItem}">
<Setter
Property="IsExpanded"
Value="{Binding IsExpanded, Mode=TwoWay}"/>
</Style>
</TreeView.ItemContainerStyle>
</TreeView>
I have one that is not virtualizing, when i expand the nodes (and use snoop to check) i have all of the TreeViewItems being created. I am wondering if there is some combination of containers that would prevent the TreeView from virtualizing its content. (like hosting it in a StackPanel for example)
The problem was with the styling. After some research we found that there was an unnamed style targeting the TreeView (i.e. one with DataType={x:Type TreeView} without an x:Key) and one targetting the TreeViewItem in our App.xaml (or equivalent) It was overriding the ControlTemplate for each respectively.
These styles did not have the triggers to set the ItemsPanel to a VirtualizingStackPanel and had no mention of any virtualization. When the styles are removed the TreeView works fine. Even though the local properties set the ItemsPanel and the VirtualizingStackPanel.Isvirtualizing="True" on the TreeView these properties were not being propogated to the TreeViewItems so the top level of the TreeView would virtualize whilst the sub categories would not (as their virtualization behaviour was dependant on the TreeViewItem)
Had the same problem. In my case, the initial size of the TreeView was not limited (TreeView is within a popup). Therfore, the virtualization panel initialized all controls for the first time.
Setting the MaxHeigt of the TreeView to 1000 solvend the problem.
What would be the bast way to change the orientation of the WPF treeview. I would like to work the expand-collapse-functionality to work left to right instead of top to down. I.e. when I click on on the expand button of a treenode I would its subnode to appear right of the parent and the indent should work top-down instead. Also the vertical lines that connect the node must now be horizontal.
Here is a great article by Josh Smith on CodeProject detaling exactly how to do this kind of thing.
To expand on John Smith's CodeProject article, if you want to have horizontal layout on only a particular level in the tree (instead of on all levels like his article shows), then just set the ItemsPanel property on the TreeViewItem at the level you want to have a StackPanel.
It wasn't intuitive to me at first, but you can get to this property through the ItemContainerStyle property of the HierarchicalDataTemplate for the layer above the layer you want to be horizontal.
Like this:
<ItemsPanelTemplate
x:Key="ItemsPanelForHorizontalItems">
<StackPanel
Orientation="Horizontal"/>
</ItemsPanelTemplate>
<HierarchicalDataTemplate
x:Key="DataTemplateForLayerAboveHorizontalItems"
DataType="{x:Type viewModel:ThingHavingHorizontalItems}"
ItemsSource="{Binding HorizontalItems}"
ItemTemplate="{StaticResource DataTemplateForLayerWithHorizontalItems}">
<HierarchicalDataTemplate.ItemContainerStyle>
<Style
TargetType="TreeViewItem">
<Setter
Property="ItemsPanel"
Value="{StaticResource ItemsPanelForHorizontalItems}"/>
</Style>
</HierarchicalDataTemplate.ItemContainerStyle>
<ContentControl
Content="{Binding}"
ContentTemplate="{StaticResource DataTemplateForThingHavingHorizontalItems}"/>
</HierarchicalDataTemplate>
Following this pattern will let you set horizontal layout for any individual layer within your tree, except the root layer. And if you want the root layer to be horizontal, then just set the ItemsPanel property on the TreeView to use a horizontal StackPanel.