sencha-architect

can anybody tell How to show label with line and display label outside in extjs4.1.I want to display label outside of pie and show line to map the label eg / \ \ test test1 test2 You've got the display option of the label config , that you can set to 'outside'. As for the lines, you'll have to draw them yourself... Which would be a quite involved hack IMO. If you're interested in trying, that takes place in Ext.chart.Label#renderLabels() , and uses Ext.chart.series.Pie 's onCreateLabel() and onPlaceLabel() methods. Don't forget to take your trigonometry with you before diving in there! 来源：

I'm writing a sencha touch app using sencha architect. Because my app do lot of ajax request, most of it need to send 'token' in request header for authentication. So I think of create child class base on Ext.Ajax which always has 'token' in request header. Then I can use this child class without care of the header. MyApp.override.Ajax.request({ ... }) I try define this in app/override/Ajax.js Ext.define('Myapp.override.Ajax', { override: 'Ext.Ajax', headers: { 'token': 'test' } }); I also set this as 'requires' in Application. But get error when try to call Myapp.override.Ajax.request({ ... }

I'm having a dumb problem and I would like you to give me a hand.Thanks in advance. The situatios is as follows: I have 2 wiews (both created with sencha architect 2.0), one for login, and another for general purposes. And I would like to load the second view on successful response when trying to log in, this is, after any successful login. The main problem is that I've tried with Ex.create, Ext.Viewport.add, Ext.Viewport.setActiveItem, but I can't manage to make the second view to appear on screen, the login screen just keeps there and the app does not load the other view. Another thing, I

I have a grid with lots of records in it. so I want to add filters in each column header to filter the required data easily. I have seen the example given in the sencha docs. But How to implement it in the Extjs 4.2 version. How to use UX classes in it? Is there any plugin available? I am greatly thankful to you. Please help me. Anand This is an example: Ext.define('Webdesktop.view.admin.List', { extend: 'Ext.grid.Panel', alias: 'widget.admin_casoslist', initComponent: function() { var me = this, filters = { ftype: 'filters', encode: false, local: true }; Ext.apply(me, { title: 'gridTitle',