When I create a new modal window in qooxdoo it gets a z-index of 100000.
A second window will get a z-index of 100002. etc.
How can I set an initial value to 1000 ?
The automatic zIndex assignment is done by a qx.ui.window.Manager instance. See http://www.qooxdoo.org/5.0.1/pages/desktop/window_management.html
qx.ui.window.Manager has a private member _minZIndex which is initated with a value of 10000. In order to change that, you have to derive your own window manager class from qx.ui.window.Manager and change that value in the overridden constructor, like this:
qx.Class.define("qx.ui.window.MyManager",
{
extend : qx.ui.window.Manager,
construct : function() {
this.base(arguments);
this._minZIndex = 1000;
}
});
(Please change the class path qx.ui.window.MyManager to your needs).
You could then replace the window manager of your application by using the following code, before creating the first window. A good place would be at the beginning of the overridden main method of qx.application.Standalone in your application
(see: http://www.qooxdoo.org/devel/api/#qx.application.AbstractGui~main):
var oldWindowManager = this.getRoot().getWindowManager();
var desktop = oldWindowManager.getDesktop();
var windowManager = new qx.ui.window.MyManager();
windowManager.setDesktop(desktop);
this.getRoot().setWindowManager(windowManager);
qx.ui.core.queue.Widget.remove(oldWindowManager);
oldWindowManager.dispose();
The window is a qx.ui.core.Widget. It inherits the zIndex property. So use setZIndex http://www.qooxdoo.org/current/apiviewer/#qx.ui.core.Widget~setZIndex
Related
I use the MVVM Light Toolkit to define the association between the view-model and the view.
The container is instructed to register a view-model as a singleton instance. Thus, the same instance will always be returned when the GagaViewModel is required:
public GagaViewModel GagaViewModel
{
get
{
var vm = ServiceLocator.Current.GetInstance<GagaViewModel>();
vm.Setup(); //Clear the ObservableCollection
return vm;
}
}
You can click on a thumbnail item on PriorGaga.xml. The self-chosen item is then selected in the GridView "MyGridView" in Gaga.xaml. Code-behind file of Gaga.xaml:
protected override async void LoadState(Object navigationParameter, Dictionary<String, Object> pageState)
{
var itemId = navigationParameter as String;
if (String.IsNullOrEmpty(itemId))
{
throw new ArgumentException("navigationParameter was either null or empty");
}
await ((GagaViewModel)DataContext).Init(itemId); //Busy(-Indicator) while loading data from server, filling the ObservableCollection and writing the selected item down
BringItemIntoView();
}
private void BringItemIntoView()
{
var vm = (GagaViewModel)DataContext;
Dispatcher.RunAsync(CoreDispatcherPriority.Normal,
() => MyGridView.ScrollIntoView(vm.SelectedItem));
}
That works fine. As a sample: Item #45 appears within the viewport immediately (correct viewport position from the beginning).
But when you click the back button and return to Gaga.xaml by selecting an arbitrarily thumbnail item (let's just say #29), you will see item #1 and then the switch to #29 (the viewport is moving over the container). Do someone know what's going on under there? Are there any virtualized items in the container from the preceding Gaga.xaml visit?
My understanding is that the lifespan of the instance of your Gaga page is determined by its NavigationCacheMode property. By default, it is set to Disabled. Assuming that you haven't changed this property, you should be seeing a new instance of your Gaga page every time you navigate to it. You can verify this behavior by setting a breakpoint in its constructor. Consequently, I would think that each time you navigate to Gaga, the behavior of the UI should be identical, because everything is fresh.
(I wanted to add this as a comment, since I haven't actually answered your question, but sadly I do not have enough rep. I apologize in advance; please do not smite me down!）
In ShellViewModel I have below command binding to open a new window "Remote View"
public ICommand RemoteViewCommand
{
get { return new RelayCommand(RemoteViewExecute, CanRemoteViewExecute); }
}
private void RemoteViewExecute()
{
if (!CanRemoteViewExecute())
{
return;
}
var shellRemoteView = Application._Container.Resolve<ShellRemoteView>();
if (_ShellRemoteView.DataContext==null)
_ShellRemoteView.DataContext = Application._Container.Resolve<ShellRemoteViewModel>();
shellRemoteView.Show();
}
On startup I have already registered both "ShellRemoteView" and "ShellRemoteViewModel" using lifetime managers to have singleton instance.
_Container.RegisterType<ShellRemoteView>(new ContainerControlledLifetimeManager());
_Container.RegisterType<ShellRemoteViewModel>(new ContainerControlledLifetimeManager());
When shellRemoteView.Show() executed and I close the form, then again on calling shellRemoteView.Show() I'm getting Invalid Operation excepton:Cannot set Visibility or call Show, ShowDialog, or WindowInteropHelper.EnsureHandle after a Window has closed.
Is there is any work-around in Unity to get window instance again if its closed.
This line is your problem:
return new RelayCommand(RemoteViewExecute, CanRemoteViewExecute);
Basically you are creating a new view each time you call the Get command. The way to fix this is to put a variable outside your GET statement that is scoped at the ViewModel level. Have it store a reference to the view and return that reference instead of creating a new reference each time. Look at the Singleton pattern for how best to do this.
You should register your View with LifetimeManager to create only one instance. Look at Using Lifetime Managers.
I create some RibbonButtons dynamically and add them to a group according to an xml file. The follwoing function is carried out as often as entries found in the xml file.
private void ExtAppsWalk(ExternalAppsXml p, AppsWalkEventArgs args)
{
RibbonButton rBtn = new RibbonButton();
rBtn.Name = args.Name;
Binding cmdBinding = new Binding("ExtAppCommand");
rBtn.SetBinding(RibbonButton.CommandProperty, cmdBinding);
Binding tagBinding = new Binding("UrlTag");
tagBinding.Mode = BindingMode.OneWayToSource;
rBtn.SetBinding(RibbonButton.TagProperty, tagBinding);
rBtn.Label = args.Haed;
rBtn.Tag = args.Url;
rBtn.Margin = new Thickness(15, 0, 0, 0);
MyHost.ribGrpExtern.Items.Add(rBtn);
}
I tried to use the Tag property to store the Url's to be started when the respective button is clicked. Unfortunately the binding to the Tag property gives me the last inserted Url only.
What would be the best way to figure out which button is hit or to update the Tag property.
The datacontext is by default the context of the Viewmodel. The RibbonGroup to which the Buttons are added is created in the xaml file at designtime. I use that construct:
MyHost.ribGrpExtern.Items.Add(rBtn);
to add the buttons. It maight not really be conform with the mvvm pattern. May be someone else has a better idea to carry that out.
I foud a solution for my problem here and use the RelayCommand class. So I can pass objects (my Url) to the CommandHandler.
RibbonButton rBtn = new RibbonButton();
rBtn.Name = args.Name;
Binding cmdBinding = new Binding("ExtAppCommand");
rBtn.SetBinding(RibbonButton.CommandProperty, cmdBinding);
rBtn.CommandParameter = (object)args.Url;
private void ExtAppFuncExecute(object parameter)
{
if (parameter.ToString().....//myUrl
have the following in my CodeBehind (class name MainHostWindow)
private object _hostContent = null;
public object HostContent
{
get { return _hostContent; }
set { _hostContent = value;}
}
this binds into a ContentControl of my View.
in a different class I do the following:
MainHostWindow host = new MainHostWindow();
{
host.HostContent = MyView; // this is an instance of a UserControl
host.Owner = this._mainWindow;
host.DataContext = viewModel;
}
host.ShowDialog();
first time it shows the MainHostWindow with the correct Content, 2nd time I get the following exception:
Specified element is already the logical child of another element. Disconnect it first.
It looks as if you are trying to add the same UserControl (not a new instance of it) to another instance of your MainHostWindow. The error is correct because the same element cannot be the child of two different containers (what would UserControl.Parent return?). You will need to create a new instance of your UserControl.
host.HostContent = new MyView();
are you able to set MyView declaratively in the XAML for MainHostWindow as this would always create a new instance when the Control is created.
I'm having difficulty adding the inline of specific type InlineUIContainer into the InlineCollection (Content property) of a TextBlock. It appears the .Add() method of InlineCollection doesn't accept this type, however you can clearly set it through XAML without explicitly marking the content as a InlineContainer, as demonstrated in many examples:
http://msdn.microsoft.com/en-us/library/system.windows.documents.inlineuicontainer.aspx
Is it possible to programatically add one of these as in the following?
Target.Inlines.Add(new Run() { Text = "Test" });
Target.Inlines.Add(new InlineUIContainer() {
Child = new Image() { Source = new BitmapImage(new Uri("http://example.com/someimage.jpg")) } });
Target.Inlines.Add(new Run() { Text = "TestEnd" });
I have a feeling what's going on is that Silverlight is using a value converter to create the runs when specified in XAML as in the example which doesn't use InlineContainer, but I'm not sure where to look to find out.
The specific error I'm getting is as follows:
Cannot add value of type 'System.Windows.Documents.InlineUIContainer' to a 'InlineCollection' in a 'System.Windows.Controls.TextBlock'.
As pointed out by Jedidja, we need to use RichTextBox to do this in Silverlight.
You can't Add() Runs directly, but you can add Spans containing Runs.
Interestingly, you can also do this:
textBlock.Inlines.Clear();
textBlock.Inlines.Add(new Span());
textBlock.Inlines[0] = new Run();
Not that it's a good idea to hack around what the framework is actively trying to prevent you from doing.
P.S. If you can't figure out what XAML is doing, inspect the visual tree.