How to show existing window in Ext Js using Sencha designer

Question

I am new in Ext js. I started Ext js with ext designer 1.2 now I use sencha designer 2.0, I created panel and button on it, and window. Now I want to use click event to show my window on click but I do not know how, I searched in many sites and especially in sencha.com but I could not find, I can show new window by creating it onclik function but I want to show existing one, if anyone knows could you please tell me how. I am interested in controllers in secha designer.

How to do it without controllers on directly inserting code in onclickbutton function?

Answer 1

Here is what I have done for this use case:

   /**
     * Check to see if window already open with the specified ID.
     * If window exists, bring it to front and expand if collapsed.
     * @param id
     */
    windowAlreadyOpen: function(id){
        //Check for already opened window with the same  ID
        var windowOpen = Ext.ComponentQuery.query('window[id='+id+']'); //always returns array
        if (windowOpen && windowOpen[0]){
                try {
                    windowOpen[0].toFront();            //bring to front
                    if (windowOpen[0].collapsed){
                       windowOpen[0].toggleCollapse();  //expand window if collapsed
                    }
                } catch (e) {
                    console.error(e);
                }
                return windowOpen[0];
        }
        return false;
    }