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我有這個代碼,我無法弄清楚什麼是錯的。我在回送中使用來自MySql的數據。使用UI路由器的路由問題<a>標籤
<ul id="main-menu" class="main-menu" ng-class="{'mobile-is-visible': layoutOptions.sidebar.isMenuOpenMobile}" >
<li ng-class="sideClass('{{ menu.link }}')" ng-repeat="menu in myMenus">
<a ui-sref="{{ menu.link }}">
<i class="fa fa-{{ menu.icon }}" aria-hidden="true"></i>
<span class="title ng-binding">{{ menu.label }}</span>
</a>
</li>
</ul>
我的角碼是:
app.config(['$stateProvider','$urlRouterProvider', function($stateProvider,$urlRouterProvider){
$stateProvider
.state('home',{
url: '/home',
authenticate: true,
views: {
'[email protected]': {
templateUrl: 'views/home.html'
}
}
})
.state('users', {
url: '/users',
parent: 'home',
views: {
'[email protected]' : {
templateUrl: 'views/users.html',
controller: 'usersCtrl'
}
}
})
});
Menus.find({
}, function(res){
$scope.myMenus = res;
});
用它來很好地工作。我不知道爲什麼會突然它給我安的錯誤
Error: Invalid state ref '' ...
的$ scope.myMenus:
[
{
"label": "HOME",
"icon": "dashboard",
"link": "home"
},{
"label": "Users",
"icon": "users",
"link": "users"
}
]
錯誤消息
Error: Invalid state ref ''
at parseStateRef (http://localhost:3000/vendor/angular-ui-router.js:3358:45)
at link (http://localhost:3000/vendor/angular-ui-router.js:3440:17)
at ea (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:73:293)
at D (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:62:190)
at g (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:55:105)
at D (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:62:134)
at g (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:55:105)
at https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:54:249
at https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:56:79
at k (https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js:60:377) <a ui-sref="{{ menu.link }}">
什麼是'$ scope.myMenus'的內容? – Mistalis
它擁有菜單屬性,我已經將它添加到我的問題 – oded
你的一個'menu.link'可能是'void'或'''',並導致這個錯誤。 – Mistalis