我无法理解错误的来源,因为html方面很好地选择list[3].main.temp
之类的东西,但是在generateList函数的第二个for循环中,我在$scope.list[i].main.temp
上得到错误说
TypeError:无法读取undefined = \
的属性“0”
该代码应该包含30个城市的列表,随机选择10并显示其当前温度。
var WeatherApp = angular.module("WeatherApp", ["ngRoute", "ngResource"]).
config(function ($routeProvider) {
$routeProvider.
when('/', { controller: ListCtrl, templateUrl: 'list.html' }).
otherwise({ redirectTo: '/' });
});
WeatherApp.factory('City', function ($resource) {
return $resource('/api/City/:id', { id: '@id' }, {update: { method: 'PUT'}});
});
var ListCtrl = function ($scope, $location, City, $http) {
$scope.city = City.query();
$scope.units = 'metric';
$scope.appId = '';
$scope.displayNum = 10;
$scope.display = [];
$scope.display.temp = [];
$scope.generateList = function () {
$scope.exp = City.query(function (exp) {
shuffle(exp);
$scope.cityIdAr = [];
for (var i = 0; i < $scope.displayNum; ++i) {
$scope.display.push($scope.exp[i]);
$scope.cityIdAr.push($scope.exp[i].CityId);
};
$scope.cityId = $scope.cityIdAr.join();
$scope.getWeather();
for (var i = 0; i < $scope.displayNum; ++i) {
$scope.display.temp.push($scope.list[i].main.temp);
};
});
};
function shuffle(ob) {
for (var j, x, i = ob.length; i; j = Math.floor(Math.random() * i), x = ob[--i], ob[i] = ob[j], ob[j] = x);
return ob;
};
$scope.getWeather = function () {
var url = 'http://api.openweathermap.org/data/2.5/group';
$http.jsonp(url, {
params: {
id: $scope.cityId,
APPID: $scope.appId,
units: $scope.units,
callback : 'JSON_CALLBACK'
}
}).success(function (data, status, headers, config) {
$scope.data = data;
$scope.list = data.list;
});
};
$scope.generateList();
};
答案 0 :(得分:0)
问题可能是在运行回调之前$scope.list
未定义。你可以从$scope.getWeather
返回一个promise并在$scope.generateList
中解析它,然后在检索数据时(在回调中)执行for循环,例如。
从$scope.getWeather
返回承诺:
$scope.getWeather = function () {
...
return $http.jsonp(...)
}
然后在$scope.generateList
:
...
$scope.getWeather().success(function(data, status, headers, config) {
$scope.data = data;
$scope.list = data.list;
for (var i = 0; i < $scope.displayNum; ++i) {
$scope.display.temp.push($scope.list[i].main.temp);
};
}
或沿着这些方向的东西。
答案 1 :(得分:0)
$ scope.display是一个List使用另一个变量
{{1}}