Question

您好我需要在我的sencha touch应用程序中实现keep login

请参阅下面的代码：

Login.js - 用户点击登录后，会将“sessionToken”存储在本地存储中。然后它将转到主页面

   onBtnLoginClick: function(){

            var loginviewGetValue =  Ext.getCmp('loginview').getValues();
            var bbid =  Ext.getCmp('bbID').getValue();
            var bbpassword =  Ext.getCmp('bbPassword').getValue();
                var LoginLS =   Ext.getStore('LoginLS');





                                    LoginLS.add({
                                        sessionId: 'sadsadsadasd'
                                       ,deviceId:'1'
                                       ,bb_id :bbid
                                       });

                                       LoginLS.sync();      

                                   var mainForm= Ext.create('bluebutton.view.Main');
                                    Ext.Viewport.setActiveItem(mainForm);

App.js~每次启动功能都会检查localStorage中的sessionToken。如果Localstorage为空，则它将进入登录页面。否则它将转到主页面

        launch: function() {




        // Destroy the #appLoadingIndicator element
        Ext.fly('appLoadingIndicator').destroy();


        // Initialize the main view

             var LoginLS = Ext.getStore('LoginLS');
             LoginLS.load();

             var record =  LoginLS.getAt(0);


            if(record != undefined){
                var sessionId = record.get('sessionId');
               if (sessionId !=undefined){
                     Ext.Viewport.add(Ext.create('bluebutton.view.Main'));
               }
               else
                   Ext.Viewport.add(Ext.create('bluebutton.view.Login'));

            }
            else{
               Ext.Viewport.add(Ext.create('bluebutton.view.Login'));
               }

//        Ext.create('bluebutton.view.TopMenuList');

    },

Logout.js~Logout将清除sessionToken并再次转到登录页面

onLogoutClick: function scan() {
                var LoginLS = Ext.getStore('LoginLS');


                     Ext.Viewport.setMasked({
                        xtype: 'loadmask',
                        message: 'Loading...'
                    });



                 LoginLS.load();

                 var record =  LoginLS.getAt(0);
                   LoginLS.removeAll();
                    LoginLS.sync();
                   //Load a new view


//                   Ext.getCmp('tabpanel').destroy();






                var loginForm = Ext.create('bluebutton.view.Login');
                Ext.Viewport.setActiveItem(loginForm);   

                Ext.Viewport.setMasked(false); // hide the load screen

但我现在遇到了问题。我无法返回登录页面。它转到空白页面。请给我一些解决方案。感谢。

这是我得到的错误

    [WARN][Ext.data.Batch#runOperation] Your identifier generation strategy for the model does not ensure unique id's. Please use the UUID strategy, or implement your own identifier strategy with the flag isUnique. Console.js:35
[WARN][Ext.Component#constructor] Registering a component with a id (`loginview`) which has already been used. Please ensure the existing component has been destroyed (`Ext.Component#destroy()`. Console.js:35
[WARN][Ext.Component#constructor] Registering a component with a id (`bbID`) which has already been used. Please ensure the existing component has been destroyed (`Ext.Component#destroy()`. Console.js:35
[WARN][Ext.Component#constructor] Registering a component with a id (`bbPassword`) which has already been used. Please ensure the existing component has been destroyed (`Ext.Component#destroy()`. Console.js:35
[WARN][Ext.Component#constructor] Registering a component with a id (`btnLogin`) which has already been used. Please ensure the existing component has been destroyed (`Ext.Component#destroy()`. Console.js:35
[DEPRECATE][bluebutton.view.Login#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);

Answer 1

查看错误消息，很明显您正在尝试再次创建登录面板而不会破坏现有组件。出现错误是因为您不允许在应用程序中多次使用相同的id。

为避免这种情况，您不应多次创建相同的视图，您应该重复使用对性能有益的视图。还有一件事，当且仅当你不能没有它时，你应该给{n}元素{/ 1}}。

假设你无法避免id属性，你应该做以下两件事之一：

仅在新视图不存在时才创建
```
id
```

通过调用以阻止登录视图（隐藏/删除）视口：

var loginView = Ext.getCmp("loginview");
if(!loginView){
    loginView = Ext.create('bluebutton.view.Login');
}

Answer 2

使用itemId代替id而不是Controller，并在{{1}}中相应地引用它们。请查看此问题：Warning saying `Id` exist and should be destroyed

注销更改视图sencha touch

2 个答案: