MVVM load data during or after ViewModel construction? - data-binding

My generic question is as the title states, is it best to load data during ViewModel construction or afterward through some Loaded event handling?
I'm guessing the answer is after construction via some Loaded event handling, but I'm wondering how that is most cleanly coordinated between ViewModel and View?
Here's more details about my situation and the particular problem I'm trying to solve:
I am using the MVVM Light framework as well as Unity for DI. I have some nested Views, each bound to a corresponding ViewModel. The ViewModels are bound to each View's root control DataContext via the ViewModelLocator idea that Laurent Bugnion has put into MVVM Light. This allows for finding ViewModels via a static resource and for controlling the lifetime of ViewModels via a Dependency Injection framework, in this case Unity. It also allows for Expression Blend to see everything in regard to ViewModels and how to bind them.
So anyway, I've got a parent View that has a ComboBox databound to an ObservableCollection in its ViewModel. The ComboBox's SelectedItem is also bound (two-way) to a property on the ViewModel. When the selection of the ComboBox changes, this is to trigger updates in other views and subviews. Currently I am accomplishing this via the Messaging system that is found in MVVM Light. This is all working great and as expected when you choose different items in the ComboBox.
However, the ViewModel is getting its data during construction time via a series of initializing method calls. This seems to only be a problem if I want to control what the initial SelectedItem of the ComboBox is. Using MVVM Light's messaging system, I currently have it set up where the setter of the ViewModel's SelectedItem property is the one broadcasting the update and the other interested ViewModels register for the message in their constructors. It appears I am currently trying to set the SelectedItem via the ViewModel at construction time, which hasn't allowed sub-ViewModels to be constructed and register yet.
What would be the cleanest way to coordinate the data load and initial setting of SelectedItem within the ViewModel? I really want to stick with putting as little in the View's code-behind as is reasonable. I think I just need a way for the ViewModel to know when stuff has Loaded and that it can then continue to load the data and finalize the setup phase.
Thanks in advance for your responses.

For events you should use the EventToCommand in MVVM Light Toolkit. Using this you can bind any event of any ui element to relaycommand. Check out his article on EventToCommand at
http://blog.galasoft.ch/archive/2009/11/05/mvvm-light-toolkit-v3-alpha-2-eventtocommand-behavior.aspx
Download the sample and have a look. Its great. You won't need any codebehind then. An example is as follows:
<Page x:Class="cubic.cats.Wpf.Views.SplashScreenView"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:i="clr-namespace:System.Windows.Interactivity;assembly=System.Windows.Interactivity"
xmlns:cmd="clr-namespace:GalaSoft.MvvmLight.Command;assembly=GalaSoft.MvvmLight.Extras"
mc:Ignorable="d"
d:DesignHeight="300" d:DesignWidth="300"
Title="SplashScreenPage">
<i:Interaction.Triggers>
<i:EventTrigger EventName="Loaded">
<cmd:EventToCommand Command="{Binding LoadedCommand}" />
</i:EventTrigger>
</i:Interaction.Triggers>
<Grid>
<Label Content="This is test page" />
</Grid>
</Page>
and the view mode could be like this
public class SplashScreenViewModel : ViewModelBase
{
public RelayCommand LoadedCommand
{
get;
private set;
}
/// <summary>
/// Initializes a new instance of the SplashScreenViewModel class.
/// </summary>
public SplashScreenViewModel()
{
LoadedCommand = new RelayCommand(() =>
{
string a = "put a break point here to see that it gets called after the view as been loaded";
});
}
}
if you would like the view model to have the EventArgs, you can simple set PassEventArgsToCommand to true:
<i:Interaction.Triggers>
<i:EventTrigger EventName="Loaded">
<cmd:EventToCommand PassEventArgsToCommand="True" Command="{Binding LoadedCommand}" />
</i:EventTrigger>
</i:Interaction.Triggers>
and the view model will be like
public class SplashScreenViewModel : ViewModelBase
{
public RelayCommand<MouseEventArgs> LoadedCommand
{
get;
private set;
}
/// <summary>
/// Initializes a new instance of the SplashScreenViewModel class.
/// </summary>
public SplashScreenViewModel()
{
LoadedCommand = new RelayCommand<MouseEventArgs>(e =>
{
var a = e.WhateverParameters....;
});
}
}

The following solution is similar to the one already provided and accepted, but it does not use a command in the view model to load the data, but a "normal method".
I think commands are more suited for user actions (commands can be available and not available at runtime), that is why a use a regular method call, but also by setting a an interaction trigger in the view.
I suggest this:
Create a view model class.
Instantiate the view model class within the xaml of the view by creating it inside the DataContext property.
Implement a method to load the data in your view model, e.g. LoadData.
Set up the view, so that this method is called when the view loads.
This is done by an interaction trigger in your view which is linked to the method in the view model (references to "Microsoft.Expression.Interactions" and "System.Windows.Interactivity" are needed):
View (xaml):
<Window x:Class="MyWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="Test"
xmlns:viewModel="clr-namespace:ViewModels"
xmlns:i="http://schemas.microsoft.com/expression/2010/interactivity"
xmlns:ei="http://schemas.microsoft.com/expression/2010/interactions"
>
<Window.DataContext>
<viewModel:ExampleViewModel/>
</Window.DataContext>
<i:Interaction.Triggers>
<i:EventTrigger EventName="Loaded">
<ei:CallMethodAction TargetObject="{Binding}" MethodName="LoadData"/>
</i:EventTrigger>
</i:Interaction.Triggers>
This will call the LoadData method in the ViewModel at runtime when the view is loaded. This is where you load your data.
public class ExampleViewModel
{
/// <summary>
/// Constructor.
/// </summary>
public ExampleViewModel()
{
// Do NOT do complex stuff here
}
public void LoadData()
{
// Make a call to the repository class here
// to set properties of your view model
}
If the method in the repository is an async method, you can make the LoadData method async too, but this is not needed in each case.
By the way, generally I would not load data in the constructor of the view model.
In the example above the (parameter less) constructor of the view model is called when the designer shows your view. Doing complex things here can cause errors in the designer when showing your view (for the same reason I would not make complex things in the views constructor).
In some scenarios code in the view models constructor can even cause issues at runtime, when the view models constructors executes, set properties of the view model which are bound to elements in the view, while the view object is not completely finished creating.

Ok, then. :-)
You can bind to a method in the ViewModel by using a behavior.
Here is a link that will help you with that.
http://expressionblend.codeplex.com/

I decided to just have the XAML declaratively bound to a Loaded event handler on the View's code-behind, which in turn just called a method on the ViewModel object, via the View's root element UserControl DataContext.
It was a fairly simple, straight forward, and clean solution. I guess I was hoping for a way to bind the Loaded event to the ViewModel object in the same declarative way you can with ICommands in the XAML.
I may have given Klinger the official answer credit, but he posted a comment to my question, and not an answer. So I at least gave him a one-up on his comment.

I had this same problem when dealing with messages between a parent window and a child window. Just change the order in which your view models are created in your ViewModelLocator class. Make sure that all view models which are dependent on a message are created before the view model that sends the message.
For example, in your ViewModelLocator class's constructor:
public ViewModelLocator()
{
if (s_messageReceiverVm == null)
{
s_messageReceiverVm = new MessageReceiverVM();
}
if (s_messageSenderVm == null)
{
s_messageSenderVm = new MessageSenderVM();
}
}

Related

How to change property value of a view inside CarouselView's ItemTemplate?

I have a CarouselView with ItemTemplate containing a Label and a Grid. Outside of that CarouselView, I want to make a button to modify Carousel's current item's Grid's visibility. Because it's inside ItemTemplate, I can't use x:Name to refer to that specific Grid, so how can I refer to the current item's Grid so I can change its property value? Thank you.
You will want to do that through databinding. As you already mentioned, you can't use x:Name. This is because you're inside of a template. The value in x:Name would be duplicated for each time that template is applied to a concrete item in your list, in this case a CarouselView. Moreover; if you use virtualization for that list, a template might not even exist at all at that point in time. All reasons why you can't use x:Name to reference anything inside of a template.
I don't have any info about the code you want to use this with, so I'll make something up.
If the backing collection for your CarouselView is a ObservableCollection<MyItem>, then your CarouselView might look something like this:
<!-- Databinding scope here is MyViewModel -->
<CarouselView ItemsSource="{Binding MyItemsCollection}">
<CarouselView.ItemTemplate>
<!-- Databinding scope here is MyItem -->
<DataTemplate>
<Button Text="Delete" IsVisible="{Binding CanDelete}" />
</DataTemplate>
</CarouselView.ItemTemplate>
</CarouselView>
So you will have a backing view model which has a MyItemsCollection, and your page (that holds the CarouselView) has set the BindingContext to a new instance of MyViewModel:
public class MyViewModel
{
public ObservableCollection<MyItem> MyItemsCollection { get; set; } = new ObservableCollection<MyItem>();
private void LoadData()
{
var items = _myItemService.GetItems();
foreach (var item in items)
{
MyItemsCollection.Add(item);
}
}
}
Whenever you want to influence the IsVisible you will want to set the CanDelete of the MyItem that it's about to false. Let's assume MyItem looks like this:
public class MyItem
{
public bool CanDelete { get; set; }
}
You will need to implement the INotifyPropertyChanged interface on it so that the UI will pick up on any changes that are made to property values.
Now whenever you set the CanDelete of a certain instance of MyItem to false, that will change your UI. E.g.: MyItemsCollection[3].CanDelete = false;
On my YouTube channel I added a playlist with videos about data binding which might help in cases like these.
PS. At the time of writing IsVisible is bugged in .NET MAUI

How to create shared form in ASP.NET CORE Razor Pages?

I have to create a reusable form in my page header. It should be displayed on each page. Simple input and submit button that will send a POST request.
Options that I'm aware of are partial views or view components.
I checked a documentation about view components but there is no mention that it works with forms. Only InvokeAsync method is available to initialize a view.
With partial view, it may be hard to define its page model and I don't understand where to put its POST handler.
One other option I see, somehow to place a form directly on _Layout.cshtml, but, again, it doesn't have its page model (afaik),
So what is a way to create a shared form, and where POST request should be handled?
You should use a view component, as this allows you to have a somewhat self-contained model and view interaction.
public class SharedFormViewComponent : ViewComponent
{
public Task<IViewComponentResult> InvokeAsync() =>
Task.FromResult(View(new SharedFormViewModel()));
}
Then, put your form HTML code in Views\Shared\Components\SharedForm\Default.cshtml. For your form's action, you'll need to specify a route. More on that in a sec. Then, to display your form:
#await Component.InvokeAsync("SharedForm")
Now, as you've somewhat discerned, view components can't be posted to. It's not an issue of "not supporting forms"; they're literally not part of the request pipeline, and therefore can't respond to requests such as a POST. You'll need a distinct action that will handle the POST on some controller. On your component's form tag, then:
<form asp-action="SharedFormHandlerAction" asp-controller="Foo" asp-area="" method="post">
The asp-area attribute should be provided just in case this is used in the context of different areas.
You'll also need a "return URL". This will be the URL of the current page, so that after the user successfully posts the form, they'll go back to the page they submitted it from. You can achieve that by adding a hidden input to your form:
<input type="hidden" name="returnUrl" value="#(Context.Request.Query["returnUrl"].FirstOrDefault() ?? (Context.Request.Path + Context.Request.QueryString))" />
Your handler action should then take a param like string returnUrl = null, and on success you should do:
return !string.IsNullOrEmpty(returnUrl) && Url.IsLocalUrl(returnUrl)
? Redirect(returnUrl)
: RedirectToAction("SomeDefaultAction");
Where things get a little tricky is in handling validation errors. Since you're posting to a different action, you can't return the previous view the user was on to show the validation errors within the from in the layout or whatever. Instead, you'll want a view specific to this handler action which will simply be your shared form. You can use the view for your view component here as a partial:
<partial name="~\Views\Shared\Components\SharedForm\Default.cshtml" />
If you want to use a PageModel, you could do so in a BasePageModel class that inherits from PageModel and that all your pages inherit from. You can use a named handler (https://www.learnrazorpages.com/razor-pages/handler-methods#named-handler-methods) in the BasePageModel class to process the form submission. Add the form directly to the Layout, which will need an #model directive for your BasePageModel type.
public class BasePageModel : PageModel
{
[BindProperty]
public string SearchString { get; set; }
public void OnPostBaseSearch()
{
// process the search
}
}

dialogPostRun method in RunBaseReport

I have a RunBaseReport which contains overrided dialog method where I'm adding couple of my controls. One of those controls is a combobox.
Controls enabled() property should be changed when I'm modifying combobox.
So basically I need to know when the value of my dfReportType dialog field changes.
public Object dialog(Object dialog)
{
DialogRunbase dialog = dialog;
;
//adding my combobox
dfReportType = dialog.addFieldValue(typeid(ReportType), ReportType:DefaultType);
//adding some other controls here
return dialog;
}
According to many articles that I found I need to override dialogPostRun Method and do something like this:
public void dialogPostRun(DialogRunbase dialog)
{
super(dialog);
dialog.dialogForm().formRun().controlMethodOverload(true);
dialog.dialogForm().formRun().controlMethodOverloadObject(this);
}
But unfortunately I don't have this method in RunBaseReport class.
Which should be there according to msdn .
Are there any other workarounds?
I'm currently on AX 2012 but I still looked at it. I have the method available in the context menu, but not on the first column. I have to go over "Plus..." to find the method in the second column.
Well, there is no dialogPostRun method in Report object that inherits RunBaseReport, but we have this method in Class that inherits RunBaseReport.
So that was my mistake. I used report object instead of class.
If you want to make custom dialog for the report but you also want to use all default controls you should:
Create class
Inherit RunBaseReport
Override dialog, getFromDialog etc.
Override lastValueElementName method
public identifiername lastValueElementName()
{
//just put name of your report object
return reportStr(YourReportName);
}
Don't forget to add main() method if you going to make call from menuItem.

How to extend the context menu inside the rehosted workflow designer?

We are using a rehosted designer (currently WF 4.0) with a lot of custom activities, all of them have custom designers. For a bunch of them, I would like to add entries to the context menu of the designer when in design mode. I'm talking about this menu:
E.g. for a XAML coded activity I would like to have an "Open source..." entry which will load the XAML source of that specific activity into a new designer. To do that I must add the entry to the menu, and when clicked figure out on which activity it was clicked. Both parts are unclear to me. How can I achieve that?
In WF 3 there was the ActivityDesignerVerb class to do that. In WF 4 there seems to be workflowDesigner.Context.Services.Publish<ICommandService>(...), but I can't figure out how to use that to add a custom action to the context menu. How can I do that?
This SO entry shows something for internal debugger commands, but I want to add a completely new command.
Solving it in the host
If you want to solve this on the workflow designer host, and not in the individual activities, it's quite simple and straightforward to do this.
When you host the workflow designer and create the workflow designer, you can simply access its ContextMenu property and modify its Items collection.
var wfd = new WorkflowDesigner();
wfd.ContextMenu.Items.Add(new MenuItem() { Header = "Hello", Command = yourCommand, });
If you want different menu items for each activity, you can subscribe to the SelectionChanged event:
wfd.Context.Items.Subscribe<Selection>(SelectionChanged);
And then implement your own logic:
private void SelectionChanged(Selection selection)
{
// Remove old menu item
if (oldMenuItem != null)
{
wfd.ContextMenu.Items.Remove(oldMenuItem);
oldMenuItem = null;
}
var modelItem = selection.PrimarySelection;
if (selection.SelectionCount == 1 && modelItem != null)
{
// Get activity type
var activityType = modelItem.ItemType;
var menuItem = new MenuItem() { /* ... */ };
wfd.ContextMenu.Items.Add(menuItem);
oldMenuItem = menuItem;
}
}
Solving it in the activity designer
If you want to always show a specific context menu item regardless of where your workflow designer UI is hosted, you can create a custom item in your activity designer XAML:
<sap:ActivityDesigner.ContextMenu>
<ContextMenu>
<MenuItem Header="Show" Command="{Binding YourCommand}"/>
</ContextMenu>
</sap:ActivityDesigner.ContextMenu>
OK so all you need to do is implement the ICommand interface on your custom activity.
So, for example - expose a custom command property from the customer activity class, then in the constructor apply a delegate event to the command handler -
/// <summary>
/// Custom activity
/// </summary>
public partial class CustomActivityDesigner
{
/// <summary>
/// Command used to display a dialog at design time
/// </summary>
public ICommand ShowCustomDialog{ get; set; }
public CustomSchedulerDesigner()
{
InitializeComponent();
ShowCustomDialog= new DelegateCommand(x =>
//Do some stuff here that will display your dialog
//you may want to consider passing the `this.ModelItem`
//to your dialog so it can then interact with the ModelTrees etc
//for example
var dialog = new MyDialog(this.ModelItem);
dialog.ShowDialog();
);
}
}
Finally,
hook the new command up to the UI via the activity designer xaml.
<sap:ActivityDesigner.ContextMenu>
<ContextMenu>
<MenuItem Header="Show" Command="{Binding ShowCustomDialog}"/>
</ContextMenu>
</sap:ActivityDesigner.ContextMenu>

Partial Views with view models different to the main view

My home-screen has a view model of DashboardViewModel. It has PartialViews with their own ViewModels such as CustomerSearchViewModel and SelectProductViewModel.
All three ViewModels are separate.
When I run the application I get this error:
The model item passed into the dictionary is of type
'Invoice.Web.ViewModels.DashboardViewModel', but this dictionary
requires a model item of type
'Invoice.Web.ViewModels.SearchCustomerWindowVM'.
I wonder what should I do to resolve this issue.
As planned, the Home screen will eventually integrate a lot of PartialViews with their own view models. Do I declare the Partial-view-models inside the DashboardViewModel? or do I simply have a single big DashboardViewModel for all partialViews to come?
You can have your partial view viewmodels as properties of your main viewmodel and call Html.Partial and pass these properties.
public class DashBoardVM
{
public string Name { set;get;}
public CustomerSearchVM CustomerSearch { set; get;}
public DashBoardVM()
{
CustomerSearch =new CustomerSerachVM();
}
}
In your dashboard view,
#model DashBoardVM
<h2>#Model.Name</h2>
#Html.Partial("CustomerSearch",Model.CustomerSearch)
Assuming CustomerSearch partial view is strongly typed o CustomerSearchVM class.
Another option is to use Html.Action() or Html.RenderAction(). This allows you to call a completely separate controller from your parent view, and return a completely different, non associated model. Here is a pretty good explanation on both rendering Partial Views and Actions. http://www.midnight-coding.com/2013/01/partial-vs-action-vs-renderpartial-vs-renderaction.html

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