I got a TreeView containing different objects from different classes. Now I want to build a propertypanel, which shows up different content, depenting on what object/class is selected in the TreeView. What is the best way to build such panel? Differnt panels and collapsing panels depending on the selection(Whould make implementing the ObserverPattern this easier for me?)? Or an other approach?
I would bind the property panel (which could be just a ContentControl) to the SelectedItem in the TreeView:
<ContentPanel Content="{Binding SelectedItem, ElementName=_treeView}"/>
Then I would use DataTemplates to show the correct panel for each class of item you have:
<DataTemplate DataType="{x:Type local:SomeClass}">
<Label>This is displayed for SomeClass</Label>
</DataTemplate>
<DataTemplate DataType="{x:Type local:SomeOtherClass}">
<Label>This is displayed for SomeOtherClass</Label>
</DataTemplate>
Obviously your DataTemplates can be as complex as needed to display the various classes present in the TreeView.
Do you mean a property grid?
Related
I'm about to create a dynamic WPF UI form from DataTable data. The screens would be fairly complex. They would contain textboxes, groupboxes, checkboxes, buttons, datagrids etc. Some of them visible, some hooked up event handlers and thing like that.
What approach of creating those dynamic screens would you choose considering performance impact and complexity requirements to write and maintain source code. Please note that this code will run a LOT so it must be efficient and blazing fast. I'm considering these options:
Create Controls in code, assemble them to a tree and use the tree (Grid control) as a root element for a WPF form.
1.a) Create a XAML via XAMLReader from that screen object tree and Load it via XAMLReader inside WPF Form. Creating XAML would seem redundant to me since I can use the built tree as a Content for WPF form directly.
Use XMLDocument class to create tags, obejcts and their atributes. Create a XAMLlike that and then load that XAML in WPF form.
Thanks,
Michal
Consider displaying your form in a listview and creating a DataTemplate for each of your form fields textboxes, groupboxes, checkboxes, buttons, datagrids etc.
<ListBox ItemsSource="{Binding DataFormFields}"
<DataTemplate DataType="YourTextClass">
<StackPanel>
<TextBlock Text="{Binding LabelText}" />
<TextBox Text="{Binding ValueText}" />
</StackPanel>
</DataTemplate>
<DataTemplate DataType="YourCheckClass">
<StackPanel>
<CheckBox Content="{Binding LabelText}"
IsChecked="{Binding Checked}"/>
</StackPanel>
</DataTemplate>
For more on DataTemplates see
https://learn.microsoft.com/en-us/dotnet/framework/wpf/data/data-templating-overview
Each data template should be associated with a one of your form fields classes, using the DataType attribute, this will cause the listbox to automatically use the correct DataTemplate.
For more details:
https://learn.microsoft.com/en-us/dotnet/api/system.windows.datatemplate.datatype?view=netframework-4.7.2
I explain my issue as I'm quite new to UI design :
I have a main View which displays a TreeView on its left part. When an element is selected I'd like to show a description of the Item on the right on the same window. The design of this description depends on the nature of the Item.
So I created a View per Item Type corresponding to the different possible design.
Now When I click on the TreeView I have no idea how to show the corresponding view on the right of the same window. (I'm not asking about catching the event, just displaying a view within another view, like if I dynamically plotted a control).
Is it possible ? If not what kind of approach would you suggest ?
Many Thanks.
This seems like a great candidate for a Data Template.
Basically, create a content presenter and bind its content property to the TreeView's SelectedItem property. Now, create data templates for each of your types (using the DataType property) in the ContentTemplate property.
Now, the correct data template will be chosen with the correct data whenever you select something in your tree view.
As far as a separate dispatcher goes, I'm not sure, but I'm also not sure what scenario would require one.
More information can be found in this SO question.
Sample:
<ContentPresenter Content="{Binding Path=SelectedItem, ElementName=TreeView}">
<ContentPresenter.ContentTemplate>
<DataTemplate DataType="{x:Type Type1}">
<!-- Bunch of stuff-->
</DataTemplate>
<DataTemplate DataType="{x:Type Type2}">
<!-- Bunch of stuff-->
</DataTemplate>
</ContentPresenter.ContentTemplate>
</ContentPresenter>
I have a listbox in WPF that will contain a list of ResultsViewModel items, however the actual runtime type of these objects could be
CalculateResultsViewModel,
ScenarioResultsViewModel,
GraphResultsviewModel etc etc,
all of which extend the base abstract class ResultsViewModel.
Each of these view models should be rendered differently in the ListBox so needs a different DataTemplate. I can do that just with XAML easy enough. The difficulty is that when the viewmodels are either "processing" or when they have failed", I need them to display a DataTemplate for "processing" or "errored" which I can only so far do with Triggers. That however then means I can't use the DataTemplateSelector or a basic XAML style.
The only solution I can think of (not clean I know) is to set the DataTemplate programmatically in the SetResult() method of each viewmodel class, which is what gets called when the processing completes either successfully or with an error. In that DependencyProperty I can look at the return code and then programatically set the DataTemplate depending on the sucess/failure result. The only problem is I cannot figure out how to
Obtain a DataTemplate resource from a ResourceDictionary just using c# code - bearing in mind Im calling all of this from the viewmodel class, not the window code-behind .xaml.cs file so it doesn't have access to the properties of Window
having only a handle to the viewmodel class, somehow obtain a reference to the ListBoxItem that contains it and then programmatically set the DataTemplate on this container.
Can anyone point me in the right direction?
you can take the magic with implicit datatemplates
<ListBox ItemSource={Binding YourResults}>
<ListBox.Resources>
<DataTemplate DataType={x:Type CalculateResultsViewModel}>
<Grid></Grid>
</DataTemplate>
<DataTemplate DataType={x:Type ScenarioResultsViewModel}>
<Grid></Grid>
</DataTemplate>
<DataTemplate DataType={x:Type GraphResultsviewModel }>
<Grid></Grid>
</DataTemplate>
</ListBox.Resources>
</ListBox>
for "processing" or "errored" viewmodels you can specify a adorner overlay in all yout datatemplates (ok but you must use the triggers)
hope this helps
Let me explain you my situation.
I have a base class called Shape, and several concrete classes like Triangle, Square, etc.
I have several data templates.
I'm building just one object. So I wouldn't use an ItemControl control, I would like to use a normal panel like the grid, and show the respective data template (in DataContext has the concrete item)..
The only way to do this is using an ItemsControl? Or there's another way.. because I'm just using one item and not a collection and display the correct template.
DataTemplates are used in much more than just ItemsControls
They are used to tell WPF how to draw any object in the Visual Tree. For example, if you stick a User class object in the VisualTree, a DataTemplate can be used to tell WPF how to draw that User object
They are most frequently used in controls with an ItemsSource or Content properties, because those are the most common way of inserting data objects into the VisualTree.
In your specific case where you only want to insert one data item into the VisualTree, I would suggest a ContentControl
<ContentControl Content="{Binding MyDataObject}" />
To tell WPF how to draw MyDataObject you can either use the ContentTemplate property and set it to a DataTemplate
<ContentControl Content="{Binding MyDataObject}"
ContentTemplate="{StaticResource MyDataTemplate}" />
or define an implicit DataTemplate that tells WPF to draw any object of a specific type using a specific template.
<DataTemplate DataType="{x:Type local:MyDataObject}">
<!-- Tell WPF how to draw MyDataObject here -->
</DataTemplate>
If you want to display a single item with a data template that is selected based on the item's type, you should use ContentControl or any of its derived classes.
I have a MainWindow in which i want to add my others Views.
Users can open Mutliple number of different Views to the MainWindow.
So for adding all those Views(UserControls), which is the best control to use.
Current I am using Canvas, but its not supporting MVVM.
So how can i add multiple control.
This is what i have done till now
Note:
The control should host multiple UserControl by the same time i should be able to drag One UserControl here and there in that control and then on a click on UserControl should bring it to Front(Focused), which i did in Canvas using ZIndex.
Proper MVVM solution migh be to use ItemsControl class and bind collection of view-models as an ItemsSource.
In DataTemplate of that ItemsControl, I would specify proper view for child view-models (sort of tool-window in your case).
<ItemsControl.ItemTemplate>
<DataTemplate>
<views:ToolWindow DataContext={Binding} />
</DataTemplate>
</ItemsControl.ItemTemplate>
To achieve window-like behavior as on your picture I would specify custom panel based on Canvas which would allow drag and drop behavior.
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<panels:MyCustomMdiPanel />
</ItemsPanelTemplate>
<ItemsControl.ItemsPanel>
I suppose that you have already working canvas solution.