Question

这是我第一次使用Angularjs $资源，我正在尝试对我的应用程序进行一些测试。我正在尝试访问其中一个属性而不是显示整个对象。如何访问data.image或data.url？

来自天气地下的json数据

   {
   "response": {
   "version":"0.1",
    "termsofService":"http://www.wunderground.com/weather/api/d/terms.html",
    "features": {
    "conditions": 1
   }
   }
   ,    "current_observation": {
    "image": {
    "url":"http://icons.wxug.com/graphics/wu2/logo_130x80.png",
    "title":"Weather Underground",
    "link":"http://www.wunderground.com"
    },
    "display_location": {
    "full":"San Francisco, CA"
    }

天气工厂

app.factory('weatherService',['$resource', function($resource){
var factory={};
 factory.getWeather = function(){
     return $resource("http://api.wunderground.com/api/9eb7777065b59c55/conditions/q/CA/San_Francisco.json").get();
 }
return factory;
}]);

天气控制器

 app.controller('weather', ['$scope', '$http','weatherService',   function($scope, $http, weatherService){
 $scope.weather = weatherService.getWeather().get();
 }]);

Answer 1

您对$资源的实施是错误的，改变您的工厂 -

app.factory('weatherService',['$resource', function($resource){
var factory={};
factory.getWeather = function(){
     return $resource("http://api.wunderground.com/api/9eb7777065b59c55/conditions/q/CA/:city.json", {city: '@id'});
 }
return factory;
}]);

你应该改变你的控制器 -

 app.controller('weather', ['$scope', '$http','weatherService',   function($scope, $http, weatherService){
     $scope.weather = weatherService
     .getWeather()
     .get({city: 'San_Fransisco'}, function(){
         //other logic
     });
 }]);

$ scope.weather具有从服务器返回的json对象

Answer 2

也许您可以使用JSON.parse将您的JSON字符串转换为对象并从那里开始......

 $scope.weather = JSON.parse(weatherService.getWeather().get()); // JSON 
 $scope.weatherUrl = $scope.weather.url;

Angularjs $ resource：如何从json对象中获取特定数据？

2 个答案: