angularjs – 保持角度控制器薄

此刻我正致力于巨大的角度SPA应用.我试着保持我的控制器很薄：

<div ng-controller='HomeController as home'>
   <div ng-repeat='var item in home.items' ng-bind='item' ></div>
   <button ng-click='home.remove(1)' >remove</button>
</div>

function HomeController (homeService){
    var vm = this; //$scope
    vm.items = [1,2,3,4,5];
    vm.remove = remove;

    function remove (id){
        homeService.remove({ items: vm.items,targetId: id });
    }

    //a lot of other logic here
}


angular.module('my').service('homeService',homeService);
function homeService (){
    this.remove = remove;

    function remove (param){
        for(var a = 0; a < param.items.length; a++){
            if(param.items[a] == param.targetId){
                param.items.splice(a,1);
                break;
            }
        }
    }
}

好处：

>逻辑不在控制器之外

缺点：

>服务变更范围状态

你保持控制器薄的方法是什么？

What is your approach to keep controllers thin?

我个人喜欢在工厂/服务中保留与应用程序模型相关的任何内容.因此,不会在控制器中定义代码中的remove和item.在控制器内部,我将设置对模型的引用,以满足指令的任何需要,即可通过$scope访问.

例如,考虑一个带有实体和方法数组的模型,以在数组中添加/删除/查找实体.我首先创建一个工厂,公开我的模型和方法来使用它：

angular.module('myApp').factory('model',function() {

    // private helpers
    var add = function(array,element) {...}
    var remove = function(array,element) {...}
    var find = function(array,id) {...}

    return {
        Entity: function(id) {
            this.id = id;
        },entities: {
            entities: [],find: function(id) {
                return find(this.entities,id);
            },add: function(entity) {
                add(this.entities,entity);
            },remove: function(entity) {
                remove(this.entities,entity);
            }       
        }
});

然后将模型传递给我的控制器：

angular.module('myApp').controller('ctrl',function($scope,model) {
    $scope["model"] = model; // set reference to the model if i have to
    var entity = new model.Entity('foo'); // create a new Entity
    model.entities.add(entity); // add entity to entities
    model.entities.find('foo'); // find entity with id 'foo'
});

等等