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How can I do it? If I add an item to a viewport with : 

Ext.define('Me.view.Viewport', {
extend: 'Ext.container.Viewport',
layout: 'fit',

requires: [
    'Me.view.Params',
    'Me.view.DrawChart',
'Me.view.ParamsBtn'
    'Me.view.ChartWin'
],

initComponent: function() {
    this.items = {
       items : [{ xtype : params, align : center},{xtype : paramsbtn, align...}]};

Well, if I add my views in "requires" I need to instantiate them somewhere in my viewport, or else it gives me this error : 

Uncaught TypeError: Cannot read property 'isInstance' of undefined

Per example I have a window that I need to show after a button click, but my app cannot recognize it unless it's in the viewport, but I don't want it to be in the viewport, I just want to show it... I've asked this question : adding item to window Extjs 4 where you can see the code of my ChartWin where i added an item which is the view DrawChart. Now i just need to show that ChartWin after a button click.

Please any help would be much much much appreciated.

Community
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salamey
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1 Answers1

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If it is a window you want to display there is no need to add it to the viewport at all. Windows are special case of a Panel component that are automatically floated. So all you have to do is define your window by extending from Window component. When you create an instance of that window you can call show() to display it. There are plenty of examples on their website: http://docs.sencha.com/ext-js/4-1/#!/example/window/layout.html

dbrin
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  • Thank you for your answer, but I'm using the MVC model, my window is in my "view" folder and i'd like to call it with my button click in my controller, how would I do that ? win.show() won't work since it's not a global variable... – salamey Oct 24 '12 at 21:20
  • use Ext.create('MyApp.view.whateverICalledIt',{any:options}).show(); – dbrin Oct 24 '12 at 21:55