Javascript 访问Angular.js中的另一个控制器
到目前为止,我的基本代码是:Javascript 访问Angular.js中的另一个控制器,javascript,json,angularjs,Javascript,Json,Angularjs,到目前为止,我的基本代码是: function UsersCtrl($scope) { $scope.users = [ { "id": 1, "usersName": "Matt Dance", "favorites": { "id":1, "lodge_id":1 } }, {
function UsersCtrl($scope) {
$scope.users = [
{
"id": 1,
"usersName": "Matt Dance",
"favorites": {
"id":1,
"lodge_id":1
}
},
{
"id":2,
"usersName": "Jennifer Dance",
"favorites": {
"id":1,
"lodge_id":2
}
}
];
};
function LodgesCtrl($scope) {
$scope.lodges = [
{
id: 1,
lodgesName: 'Matt\'s Awesome Lodge',
lodgeOwnersName: 'Matthew D',
lodgeStreetAddress: '5555 J lane',
lodgeCity: 'Salt Lake city',
lodgeZip: '11111',
lodgeState: 'Utah'
},
{
id: 2,
lodgesName: 'Some Historic Lodge',
lodgeOwnersName: 'Santa Claus',
lodgeStreetAddress: '1 Clearwater Road',
lodgeCity: 'Grand Marais',
lodgeZip: '55555',
lodgeState: 'Minnesota'
},
{
id: 3,
lodgesName: 'Yellowstone Lodge',
lodgeOwnersName: 'The Feds',
lodgeStreetAddress: '1 Yellowstone Way',
lodgeCity: 'Yellowstone',
lodgeZip: '55555',
lodgeState: 'Wyoming'
},
{
id: 4,
lodgesName: 'Cool Lodge',
lodgeOwnersName: 'Awesome Guy',
lodgeStreetAddress: '1 awesome Road',
lodgeCity: 'Eagle Mountain',
lodgeZip: '55555',
lodgeState: 'Utah'
},
{
id: 5,
lodgesName: 'The last lodge',
lodgeOwnersName: 'Adam V',
lodgeStreetAddress: '444 Last Road',
lodgeCity: 'San Francisco',
lodgeZip: '55555',
lodgeState: 'California'
}
];
};
我希望用户能够看到他们喜欢的小屋的名字。例如,用户喜欢id为1的小屋。那么,我如何循环遍历每个用户,并显示该用户喜欢的小屋的名称呢?基本上,我需要在每个用户的收藏夹中使用lodge\u id
来显示小屋名称
到目前为止,我的html是:
<div ng-controller="UsersCtrl">
<ul class="unstyled">
<li ng-repeat="user in users">
{{user.usersName}}<br>
</li>
</ul>
</div>
-
{{user.usersName}}
我对Angular(一般来说是JS框架)是新手,不知道该怎么做。我研究过其他框架,如Ember,但还不能以“简单”的方式解决它。最简单的解决方案是构造一个映射对象并将其绑定到$rootScope。例如:
$rootScope.mapping = {
userId: [lodge1, lodge2...]
}
我对您的代码做了一些更改,它现在正在运行,但我认为您需要以更好的方式构造数据,以便以更简单的方式进行迭代 您还可以在此处看到一个正在工作的plunker()
文件
var myApp=angular.module('myApp',[]);
myApp.controller('UsersCtrl',['$scope',函数($scope){
$scope.users=[
{
“id”:1,
“用户名称”:“马特舞蹈”,
“收藏夹”:{
“id”:1,
“洛奇id”:1
}
},
{
“id”:2,
“usersName”:“Jennifer Dance”,
“收藏夹”:{
“id”:1,
“洛奇id”:2
}
}
];
$scope.lodges=[
{
id:1,
小屋名称:“马特的很棒的小屋”,
房主姓名:“Matthew D”,
酒店地址:“5555J巷”,
住宿城市:“盐湖城”,
lodgeZip:'11111',
“犹他州”
},
{
id:2,
lodgesName:“一些历史悠久的小屋”,
房主姓名:“圣诞老人”,
酒店地址:“清水路1号”,
lodgeCity:“大马来斯”,
lodgeZip:'55555',
“明尼苏达州”
},
{
id:3,
旅馆名称:“黄石旅馆”,
房主姓名:“联邦调查局”,
lodgeStreetAddress:“黄石大道1号”,
住宿城市:“黄石公园”,
lodgeZip:'55555',
洛德盖斯特:“怀俄明州”
},
{
id:4,
旅馆名称:'酷旅馆',
房主姓名:“帅哥”,
lodgeStreetAddress:“一条很棒的路”,
住宿城市:“鹰山”,
lodgeZip:'55555',
“犹他州”
},
{
id:5,
旅馆名称:'最后的旅馆',
房主姓名:“亚当五世”,
旅馆地址:“最后一条路444号”,
洛奇城:“旧金山”,
lodgeZip:'55555',
“加州”
}
];
}]);
-
{{user.usersName}}
{{user.favorites.lodge_id}}
{{lodge.lodgesName}
您应该使用服务。由于服务是单例的,因此它们是在控制器之间共享内容的最简单方式。事实上,这是一种有角度的方式:
不要将控制器用于:
跨控制器共享无状态或有状态代码-使用
而是服务
我相信mdance是在问两个控制器是如何相互交谈的。您的解决方案只有一个控制器。请注意,您引用的文本涉及共享“代码”,而不是共享数据。事实上,AngularJS FAQ指出:“相反,不要创建一个在生活中唯一目的就是存储和返回数据位的服务”。FAQ实际上告诉我们,$rootScope可能也应该用于此目的。不过,我不确定我是否完全同意他们的建议,其他人也对此表示怀疑。那么,什么是最佳实践呢?最终,我是在共享数据,而不是代码。到目前为止还不清楚最好的方法是什么。这基本上是将一个用户id映射到几个小屋id。。。对的
<!doctype html>
<html lang="en" ng-app="myApp">
<head>
<meta charset="UTF-8">
<title>Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.0.8/angular.min.js"></script>
<script>
var myApp = angular.module('myApp', []);
myApp.controller('UsersCtrl', ['$scope', function ($scope) {
$scope.users = [
{
"id": 1,
"usersName": "Matt Dance",
"favorites": {
"id":1,
"lodge_id":1
}
},
{
"id":2,
"usersName": "Jennifer Dance",
"favorites": {
"id":1,
"lodge_id":2
}
}
];
$scope.lodges = [
{
id: 1,
lodgesName: 'Matt\'s Awesome Lodge',
lodgeOwnersName: 'Matthew D',
lodgeStreetAddress: '5555 J lane',
lodgeCity: 'Salt Lake city',
lodgeZip: '11111',
lodgeState: 'Utah'
},
{
id: 2,
lodgesName: 'Some Historic Lodge',
lodgeOwnersName: 'Santa Claus',
lodgeStreetAddress: '1 Clearwater Road',
lodgeCity: 'Grand Marais',
lodgeZip: '55555',
lodgeState: 'Minnesota'
},
{
id: 3,
lodgesName: 'Yellowstone Lodge',
lodgeOwnersName: 'The Feds',
lodgeStreetAddress: '1 Yellowstone Way',
lodgeCity: 'Yellowstone',
lodgeZip: '55555',
lodgeState: 'Wyoming'
},
{
id: 4,
lodgesName: 'Cool Lodge',
lodgeOwnersName: 'Awesome Guy',
lodgeStreetAddress: '1 awesome Road',
lodgeCity: 'Eagle Mountain',
lodgeZip: '55555',
lodgeState: 'Utah'
},
{
id: 5,
lodgesName: 'The last lodge',
lodgeOwnersName: 'Adam V',
lodgeStreetAddress: '444 Last Road',
lodgeCity: 'San Francisco',
lodgeZip: '55555',
lodgeState: 'California'
}
];
}]);
</script>
</head>
<body>
<div ng-controller="UsersCtrl">
<ul>
<li ng-repeat="user in users">
{{user.usersName}}<br>
{{user.favorites.lodge_id}}<br>
<p ng-repeat="lodge in lodges | filter: { id: user.favorites.lodge_id }">
{{lodge.lodgesName}}
</p>
</li>
</ul>
</div>
</body>
</html>