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    Default Unanswered: How can I know which window is on front?

    Unanswered: How can I know which window is on front?


    Hi,

    I'm looking for something like "window.isOnFront() == true". Which doesn't exist.. =/

    Is there something similar that can tell me if the window is on front or not?

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    If nothing else, you can compare the Window's zindex to some other component to determine if it's 'on top':

    Code:
    if (window1.getEl().zindex > someOtherComponent.getEl().zindex) {
      // do something.
    }

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    I'm doing a web desktop, so I can open multiple windows.
    With the ZIndex, can I know easily which of the windows is on front (if there is one), or must I test each against each ?

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    Take a look at Ext.WindowManager. Ext.WindowManager.getActive() should give you what you are looking for.

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    Hi!
    i guess you have to first register your windows in WindowManager before using getActive() method of it.

    Sample:-

    Code:
    var win1 = Ext.create ('Ext.window.Window', {
      title: 'First window' ,
      id:'first',
      ..
    });
    
    var win2 = Ext.create ('Ext.window.Window', {
      title: 'Second window' ,
      id:'second',
      ..
    });
    Code:
    // Register your windows to the WindowManager
    Ext.WindowManager.register (win1);
    Ext.WindowManager.register (win2);
    Code:
    alert (Ext.WindowManager.getActive().getId ());
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    I don't think you need to manually register the windows. See the API for http://docs.sencha.com/ext-js/4-1/#!...ethod-register

    "Registers a floating Ext.Component with this ZIndexManager. This should not need to be called under normal circumstances. Floating Components (such as Windows, BoundLists and Menus) are automatically registered with a zIndexManager at render time."

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