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    Question Unanswered: Console.js:35[DEPRECATE][view.Layout#show] Call show() on a component that doesn't ..

    Unanswered: Console.js:35[DEPRECATE][view.Layout#show] Call show() on a component that doesn't ..


    Code:
        launch: function() {
    
    		Ext.widget('mainLayout').show();
    I have tried Ext.Viewport.add('mainLayout'); before the line above but I get an error so I am after the correct usage please anyone...

    Console.js:35[DEPRECATE][view.Layout#show] Call show() on a component that doesn't currently belong to any container. Please add it to the the Viewport first, i.e: Ext.Viewport.add(component);

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    Try
    PHP Code:
    var mainView Ext.widget('mainLayout');
    Ext.Viewport.add(mainView); 

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    If I create using items {} fashion will I still need to add and show() the view in the actual class ?
    Thanks - at least my main view is showing.

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    Aaah, Ext.Viewport.add expects an array of components. The following doesn't generate the warning:

    var mainView = Ext.widget('mainLayout');
    Ext.Viewport.add([mainView]);

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    you are try this code

    first define refs of this view class

    then automatically generate getter and setter method
    and try

    var sheet = this.getTagSheet(); //this is refs var ex:tagSheet:'#someId'
    if(typeof sheet== 'undefined'){
    var sheet= Ext.widget('addtag');
    Ext.Viewport.add(sheet);
    }
    sheet.show();


    it's work for me

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