为什么这个简单的代码不起作用:
angular.module('menuApp', [])
.factory('menuService', function($http){
var Menus = {};
Menus.get = function() {
return $http.get('/api/foods');
};
return Menus;
})
.controller('menuCtrl', function($http, Menus){
var vm = this;
vm.headline = "Menu Card";
Menus.get()
.success(function(data) {
vm.menus = data;
});
});
HTML:
<div ng-controller="menuCtrl as menu">
<h1>{{menu.headline}}</h1>
<p>Search by catergory:</p>
<input type="text" ng-model="search.category">
<div class="ui divider"></div>
<div class="ui grid">
<div class="four wide column" ng-repeat="menu in menu.menus | filter: search">
<div class="ui segment">
<h3>{{menu.category}}</h3>
<h2>{{menu.name}}</h2>
<p>{{menu.desc}}</p>
</div>
</div>
</div>
</div>
这是控制台错误:
错误:[$ injector:unpr] http://errors.angularjs.org/1.3.14/ $ injector / unpr?p0 = MenusProvider%20%3C-%20Menus%20%3C-%20menuCtrl 在错误(本机) 在https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:6:417 在https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:38:7 在Object.d [as get](https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:36:13) 在https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:38:81 在d(https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:36:13) at Object.e [as invoke](https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:36:283) at $ get.w.instance(https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:75:451) 在https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:58:476 在s(https://ajax.googleapis.com/ajax/libs/angularjs/1.3.14/angular.min.js:7:408)
答案 0 :(得分:0)
Angular将通过匹配工厂注册的名称来查找您的服务,因此您的控制器参数必须具有名称menuService
。
此外,如果您计划最小化代码,我建议您use array notation。
您的控制器行应如下所示:
.controller('menuCtrl', ['$http', 'menuService', function($http, menuService){
...
}]);