I have a user control sitting on my main window, It is a menu.
When a user tries editing objects on a page I want to disable this user control while the user is on edit mode so they do not navigate off the page they are editing.
I am having trouble accessing the user control(placed on the main window) from a page
User Control Name - MenuView
The code below is on the page load of the page
MainWindow mainWindow = new MainWindow();
mainWindow.MenuView.IsEnabled = false;
This does not seem to work though
You need to get a reference to the existing instance of the MainWindow instead of creating a new one.
If you don't care about MVVM, the easiest way to do this would probably be to use the static Application.Current.Windows or App.Current.MainWindow property:
MainWindow mainWindow = Application.Current.Windows.OfType<MainWindow>().FirstOrDefault();
if (mainWindow != null)
mainWindow.MenuView.IsEnabled = false;
Note that this approach does create a coupling between your classes, but this is inevitable if you access a field of the window from in the Page class.
You probably want to look into the MVVM design pattern but that's another story.
Related
So I've started looking at the Catel MVVM framework and it looks like it will solve a couple of problems I have encountered, but I have one really silly issue. I think I'm just missing something small.
I took one of my smaller WPF projects to switch over the Catel as a way for me to learn it. I have a simple 'Player Registration' form, with fields like name and surname. I recreated my original view model by using the vm codesnippet and all is good, all the properties and attributes I've set up as I've read in the documentation.
I then changed the UserControl I used for 'Player Registration' (PlayerRegistrationView) to a catel:UserControl. I placed PlayerRegistrationView on a standard WPF Window (nothing else, just a xmlns for the View and the view as the only content on the window, no attributes)
But here is my problem:
I have a MainWindow with a button on to open the Window for the player registration. The on-click event simply is this:
private void ButtonPlayerClick(object sender, RoutedEventArgs e)
{
var playerRegistration = new PlayerRegistrationDialog
{
Owner = this,
DataContext = new PlayerRegistrationViewModel(),
};
playerRegistration.Show();
}
Running my program and then clicking on the button results in an NotSupportedException on my PlayerRegistrationView: The view model of the view could not be resolved. Use either the GetViewModelType() method or IViewModelLocator
I tried making the ViewModel a static resource on the window and setting the datacontext there, but it produces the same error.
I am at a loss. What have I missed?
Thanks
The whole point of Catel is that it automatically wires up all the views and view models. The "complex" thing that you are trying to achieve is that you have a view which is placed on a window. You want the window to have the same data context as the view in order to do some stuff in the window as well.
In Catel, it is possible to place any view with datacontext management on a DataWindow (window in Catel). Then it will work like this:
DataWindow
|=> View
If the DataWindow and the View share the same view model type, then they share the same view model. For example:
PlayerRegistrationWindow => derives from catel:DataWindow
PlayerRegistrationView => derives from catel:UserControl
Since both start with PlayerRegistration, they will both be resolved to PlayerRegistrationViewModel automatically.
To show the window, the only thing you have to do is this:
var viewModel = new PlayerRegistrationViewModel();
var uiVisualizerService = ServiceLocator.Default.ResolveType<IUIVisualizerService>();
uiVisualizerService.Show(viewModel);
All will work automatically and you don't have to worry about setting any datacontext yourself.
I am using ShowDialog() with WindowStyle = WindowStyle.SingleBorderWindow; to open a modal window in my WPF (MVVM) application, but it lets me navigate to parent window using the Windows taskbar (Windows 7).
I've found an answer here: WPF and ShowDialog() but it isn't suitable for me because I don't need an "always on top" tool window.
Thanks in advance
Try setting the Owner property of the dialog. That should work.
Window dialog = new Window();
dialog.Owner = mainWindow;
dialog.ShowDialog();
Edit:
I had a similar problem using this with MVVM. You can solve this by using delegates.
public class MainWindowViewModel
{
public delegate void ShowDialogDelegate(string message);
public ShowDialogDelegate ShowDialogCallback;
public void Action()
{
// here you want to show the dialog
ShowDialogDelegate callback = ShowDialogCallback;
if(callback != null)
{
callback("Message");
}
}
}
public class MainWindow
{
public MainWindow()
{
// initialize the ViewModel
MainWindowViewModel viewModel = new MainWindowViewModel();
viewModel.ShowDialogCallback += ShowDialog;
DataContext = viewModel;
}
private void ShowDialog(string message)
{
// show the dialog
}
}
I had this problem but as the Window was being opened from a view model I didn't have a reference to the current window. To get round it I used this code:
var myWindow = new MyWindowType();
myWindow.Owner = Application.Current.Windows.OfType<Window>().SingleOrDefault(x => x.IsActive);
You can use: myWindow.Owner = Application.Current.MainWindow;
However, this method causes problems if you have three windows open like this:
MainWindow
|
-----> ChildWindow1
|
-----> ChildWindow2
Then setting ChildWindow2.Owner = Application.Current.MainWindow will set the owner of the window to be its grandparent window, not parent window.
When the parent window makes (and shows) the child window, that is where you need to set the owner.
public partial class MainWindow : Window
{
private void openChild()
{
ChildWindow child = new ChildWindow ();
child.Owner = this; // "this" is the parent
child.ShowDialog();
}
}
Aditionally, if you don't want an extra taskbar for all the children... then
<Window x:Class="ChildWindow"
ShowInTaskbar="False" >
</Window>
Much of the reason for the MVVM pattern is so that your interaction logic can be unit tested. For this reason, you should never directly open a window from the ViewModel, or you'll have dialogs popping up in the middle of your unit tests.
Instead, you should raise an event that the View will handle and open a dialog for you. For example, see this article on Interaction Requests: https://msdn.microsoft.com/en-us/library/gg405494(v=pandp.40).aspx#sec12
The problem seems to be related to Window.Owner, and indeed if you judge by previous knowledge that you might have of the Win32 API and WinForms, a missing owner would be the typical cause of such a problem, but as many have pointed out, in the case of WPF that's not it. Microsoft keeps changing things to keep things interesting.
In WPF you can have a dialog with a specific owner and you can still have the dialog appear in the taskbar. Because why not. And that's the default behavior. Because why not. Their rationale is that modal dialogs are not kosher anymore, so you should not be using them; you should be using modeless dialogs, which make sense to show as separate taskbar icons, and in any case the user can then decide whether they want to see different app windows as separate icons, or whether they want to see them grouped.
So, they are trying to enforce this policy with complete disregard to anyone who might want to go against their guidelines and create a modal dialog. So, they force you to explicitly state that you do not want a taskbar icon to appear for your dialog.
To fix this problem, do the following in the constructor of your view class:
ShowInTaskbar = false;
(This may happen right after InitializeComponent();
This is equivalent to Xcalibur37's answer, though the way I figure things, since WPF forces you to have both a .cs file and a .xaml file, you might as well put things that are unlikely to change in the .cs file.
Add "ShowInTaskbar" and set it to false.
Even if this post is a bit old, I hope it is OK that I post my solution.
All the above results are known to me and did not exactly yield the desired result.
I am doing it for the other googlers :)
Lets say f2 is your window that you want to display on top of f1 :
f2.Owner = Window.GetWindow(this);
f2.ShowDialog();
That's it , I promise it will not disappear !
HTH
Guy
how to make something like this:(Screen is not from my app)
When is change the button in dialogBox, it automatically change the Bitmap in MainWindow...
I hope you understand my prob ;d
So how can i make this dialog window ?
Or how can i get access to Mainwindow variables from dialogWindow?
If you have no architecture to support that (like MVVM), you can simply use the Application object in you dialog code-behind:
MainWindow mainWindow = Application.Current.MainWindow as MainWindow;
mainWindow.button1.Content = "Dialog rename me";
Application object is set on your application execution and is global.
Define a class where you put all information relevant to your item.
All properties uses NotifyPropertyChanged.
One of this property is the Title.
So now in your main window you have a TextBlock bound your object title (it might be within a control that draws the object and write the title above), and when you click on a button it opens another window. In the constructor( new()) of the second window, you give the drawn object as argument.
So when you change the title in the second window, it gets updated in the first.
In my application is required the user to select an item from a list before continues uses the application.
To do this, I have a window with the desired items and I display it when the MainWindow displays.
public MainWindow()
{
InitializeComponent();
var itemsWindow = new ItemsWindow();
itemsWindow.Show();
}
The problem is that the window opens in background. How can I open the window in foreground?
The preferable would be to open the itemsWindow on applications start up and onClose event of itemsWindow to display the mainWindow, but I think this approach is far away from my knowledge. Nevertheless, I would appreciate it if someone could post something for how to achieve this.
Thanks
Use ShowDialog() method instead. That way, you won't have to worry about activating that window and user won't be able to interact with MainWindow until ItemsWindow has been closed.
Example:
var itemsWindow = new ItemsWindow();
itemsWindow.ShowDialog();
i've done this before but i cannot find my old code.
how do you embed a window inside a window.
let say i created a custom form and saved it as Window1.xaml, and want to embed it in Window2.xaml, without copy and pasting the xaml code.. TIA
EDIT: i think my question is somewhat misleading, i'll rephrase it.
i have this Window1.xaml i added custom headers and background images/colors.
then in Window2.xaml, i want Window1 to be a custom control and embed it here.
not sure if its Content Presenters, still googling for the answer :)
You can't host a WPF Window inside another WPF Window, but you could move the content from one Window to another:
var window1 = new Window1();
var window2 = new Window2();
var content = window1.Content;
window1.Content = null;
window2.Content = content;
Note that you set window1.Content to null or else you get an exception, since the content will have a visual parent otherwise.
UPDATE
It appears all you need to do is to copy all the XAML between the <Window></Window> tags in Window1 into a new UserControl, then host that user control in Window2.
I believe you should make use of Pages or usercontrols in such cases. This way you can navigate to other parts/pages/controls defined in application. CodeKaizen is right , you can't host a window inside another window
I'm not sure you can do that - however, you shouldn't put the user interface directly into a window, use a normal control (either custom or user) instead and reuse that in your windows.
I know you can do it in code behind
//Window you want to show
Window1 child = new Window1();
object content = child.Content;
child.Content = null;
//Where to show
this.grid1.Children.Clear();
this.grid1.Children.Add((UIElement)content);
Hope helps!
It sounds like you really want a UserControl. Change Window1's type from Window to UserControl and then put that UserControl in Window2.