Question about ViewModel Management (DesignTime Vs Run Time)

Question

I have a fairly basic WPF UI whereby user requests cause a new tab to open in my TabControl. The TabControl is bound to an ObservableCollection<ViewModelBase>

I add ViewModel instances to this collection, and the corresponding tab's content is displayed based on templates like this:

    <DataTemplate DataType="{x:Type viewModels:UserUploadsViewModel}">
        <userControls:UserUploads />
    </DataTemplate>

Now let's say that inside of the UserUploads control I'd like to wire up a ViewModel in XAML to help with the designing, like this:

<UserControl x:Class=".....UserUploads"
   .....
    DataContext="{Binding Source={StaticResource ViewModelLocater},
                  Path=UserAdministrationViewModel}">

This property will return a ViewModel with live services at runtime, and a ViewModel with mock data at design time.

Question: Will this XAML interfere with what I'm doing in binding a TabItems content to a ViewModel instance, and relying on the dataTemplate above to render the right View? If so, is there a way to get both of these concepts to work together?

score 31 · Accepted Answer · edited Jul 01 '21 at 15:08

31

There is an easier way to do this. Have a DesignTimeUserAdministrationViewModel and populate it with static data in the constructor and refer that in UserControl as:

<UserControl d:DataContext="{d:DesignInstance designTimeVMs:DesignTimeUserAdministrationViewModel, IsDesignTimeCreatable=True}">

This way you have a design time test data bound to d:DataContext and runtime live data bound to the actual DataContext. More details here.

edited Jul 01 '21 at 15:08

StayOnTarget

11,743
10
52
81

answered Jul 14 '11 at 16:51

anivas

6,437
6
37
45

That looks great. So if it's design time, that (design time view model) object will be wired up, but at runtime it'll do nothing? – Adam Rackis Jul 14 '11 at 18:08
1

Yes, it's added for blendability. – anivas Jul 14 '11 at 18:17
Fantastic! Using `IsDesignTimeCreatable=True` will populate the view with data. If set to `False`, it will only provide the "shape" of the data, which gives full data binding support in the designer, instead of binding blindly. Also, you can use a .xaml file to fully specify example data with `{d:DesignData}`. See [Walkthrough: Using a DesignInstance to Bind to Data in the Designer](http://msdn.microsoft.com/en-us/library/dd490796(VS.100).aspx) and [Walkthrough: Using Sample Data in the WPF Designer](http://msdn.microsoft.com/en-us/library/ee823176(v=vs.100).aspx) – cod3monk3y Jul 21 '14 at 20:58
This solution is super! – peter70 Oct 06 '14 at 12:36
5

Your link is dead - you deleted your WordPress site – CAD bloke Apr 16 '16 at 23:09
The link to the answer has gone stale, but the MSDN page for a walkthrough of setting up design instance data context for editing XAML markup is explained here: https://learn.microsoft.com/en-us/previous-versions/visualstudio/visual-studio-2010/dd490796(v=vs.100) – Tore Aurstad Jun 27 '19 at 18:22

score 1 · Answer 2 · answered Apr 14 '14 at 21:33

The built in MS stuff is not bad, but another more elegant and structurally sound alternative which I am busy incorporating in my project is: http://msdn.microsoft.com/en-us/magazine/dn169081.aspx

Basically, you use the MVVM Light toolkit with the SimpleIoc container it comes with and end up with the ability to serve up data for the following three scenarios:

Design time, Run time, and Test time.

Better still, the whole point of MVVM Light is to have your stuff be directly editable in Blend and there is a whole series of videos and blogs and sample apps describing it all. I wish I had found these earlier in my WPF explorations.

score 1 · Answer 3 · answered Jul 14 '11 at 16:49

Yes I think it will interfere with your current setup

The ViewModelLocator is a static class that returns a dummy object at design time, and a static ViewModel at runtime. This means that

The ViewModelLocator, not your ParentViewModel, contains your TabViewModels
You cannot have multiple instances of the same Tab (ViewModel) open at once
You cannot manage Open/Closed tabs unless you reference the UserControl, which is a violation of the MVVM principle where the ViewModel doesn't know of the View
You can't instantiate new copies of the TabViewModel with parameterized constructors. For example, OpenTabs.Add(new CustomerViewModel(CustomerId));

Perhaps an alternative could be a Converter? One that returns a static object if in design time, or the bound object during runtime? I've never tested such a thing but in theory it should work :)

Question about ViewModel Management (DesignTime Vs Run Time)

3 Answers3

Linked