Ensure one stacked Dock is "visible"

Byte-Welt - Projekte / Projects DockingFrames
1

Hello,

I have some docks stacked. Only one is “visible” (had the focus, is selected) at the time; the other ones are behind the selected one.

How can I select/choose at run time the “selected/focused” one (in a way to make it ‘visible’ in front of the other stacked ones) ?

If I use only the method “setVisible(true)”, if the dock is already in a stack, but behind others ones, the dock doesn’t become the “selected” one from the stack.

Thanks for your work on this project.

Christophe.

2

Make it focused, e.g. if using Common call “yourDockable.toFront()”.

3

Well … I have no “toFront()” method to call …
I explain a little bit more:

I have a couple of “SingleCDockable”. Some of them are “staked” (and shown in a kind of tabpane so).

During a process, I want to ensure that one of the dock is “visible by the user”. So I would like to :

  • Know if my dock is stacked in a tabbed pane.
  • If so … I should “select the tab” to make it “selected/visible to the user”.

I have a CControl but this object does not provide a kind of “toFront(CDockable)” and my “SingleCDockable” do not have a “toFront()” method.

4

The “toFront” method is defined in AbstractCDockable, so any DefaultSingleCDockable or DefaultCDockable does have it.

Another solution: use CDockable#intern to get a CommonDockable. Use CommonDockable#getDockParent to get its parent DockStation. If the parent station is an instance of StackDockStation: then the Dockable is stacked. In any case, calling “DockStation#setFrontDockable” will move the Dockable to the front.

5

you are right …

the cause was that my variables have a “too low type” : they are typed as “CSingleCDockable” but I instanciate “DefaultSingleCDockable” … which, this ones have the toFront() method !

And … it works perfectly !

Thank you very much.
You made a really good job on this project !

Christophe.