Ionic1应用程序在单击按钮时会重新加载

Question

我有一个带有侧边菜单的离子应用程序。我的要求是在侧面菜单页面之前有一个登录页面。 通过单击登录按钮第一次应用程序重新加载，在第二次单击时导航到侧面菜单页面。 这是我的代码

app.ts

config(function ($stateProvider, $urlRouterProvider) {
$stateProvider

  .state('login', {
    url: '/login',
    templateUrl: 'templates/login.html',
    controller: 'loginCtrl'
  })

  .state('app', {
    url: '/app',
    abstract: true,
    templateUrl: 'templates/menu.html',
    controller: 'AppCtrl'
  })

  .state('app.home', {
    url: '/home',
    cache: true,
    views: {
      'menuContent': {
        templateUrl: 'templates/home.html',
        controller: "homeController"
      }
    }
  })

<ion-view>
<ion-content>
<div class="login-main-heading">

  <h3 class="all-elements-center login-heading">Welcome</h3>

<div class="form-elements responsive-sm">
  <h5 class="all-elements-center login-heading">Login</h5>
  <form action="">
    <div class="list list-inset">
      <label class="item item-input">
             <i class="icon ion-person placeholder-icon"></i>
            <input type="text" placeholder="User Name">
          </label>
      <label class="item item-input">
             <i class="icon ion-lock-combination placeholder-icon"></i>
            <input type="password" placeholder="Password">
          </label>
      <div class="responsive-sm">
        <button class="button button-large button-assertive button-login">
           Login
        </button>
        <button class="button button-large button-assertive button-signup">
            Sign Up
        </button>
      </div>
      <div>
        <button class="button button-block button-positive button-facebook-    login" ng-click="login()">
           Login 
        </button>
      </div>
    </div>
  </form>
</div>
</ion-content>
</ion-view>

登录-controller.js

angular.module('starter')
.controller('loginCtrl', function ($scope, $state,$ionicSideMenuDelegate) {
$scope.login = function () {
$state.go('app.home');
}
});

应用-controller.js

angular.module('starter.controllers', [])

.controller('AppCtrl', function ($scope, $state) { 
});