只想检索有采用了棱角分明JS

Question

我的JSON是从以下链接呈现特定字段的字段： http://maps.googleapis.com/maps/api/geocode/json?address=SFO

JSON渲染只能通过参数例如：ADRESS = SFO？ 它用SFO参数返回所有的值。

{ 
    "results" : [ 
     { 
     "address_components" : [ 
      { 
       "long_name" : "San Francisco International Airport", 
       "short_name" : "San Francisco International Airport", 
       "types" : [ "establishment", "point_of_interest" ] 
      }, 
      { 
       "long_name" : "San Francisco", 
       "short_name" : "SF", 
       "types" : [ "locality", "political" ] 
      }, 
      { 
       "long_name" : "San Mateo County", 
       "short_name" : "San Mateo County", 
       "types" : [ "administrative_area_level_2", "political" ] 
      }, 
      { 
       "long_name" : "California", 
       "short_name" : "CA", 
       "types" : [ "administrative_area_level_1", "political" ] 
      }, 
      { 
       "long_name" : "United States", 
       "short_name" : "US", 
       "types" : [ "country", "political" ] 
      }, 
      { 
       "long_name" : "94128", 
       "short_name" : "94128", 
       "types" : [ "postal_code" ] 
      } 
     ], 
     "formatted_address" : "San Francisco International Airport (SFO), San Francisco, CA 94128, USA", 
     "geometry" : { 
      "location" : { 
       "lat" : 37.6213129, 
       "lng" : -122.3789554 
      }, 
      "location_type" : "APPROXIMATE", 
      "viewport" : { 
       "northeast" : { 
        "lat" : 37.6226618802915, 
        "lng" : -122.3776064197085 
       }, 
       "southwest" : { 
        "lat" : 37.6199639197085, 
        "lng" : -122.3803043802915 
       } 
      } 
     }, 
     "place_id" : "ChIJVVVVVYx3j4ARP-3NGldc8qQ", 
     "types" : [ "airport", "establishment", "point_of_interest" ] 
     } 
    ], 
    "status" : "OK" 
} 

不过，我只想得到的formatted_address为类型===机场。 含义：我只想要机场的格式化地址。

<script> 

     var app = angular.module('myApp', ['ui.bootstrap']); 

     app.controller('myController', function($scope, $http){ 
      $http.get("https://raw.githubusercontent.com/vedvasa/airports/master/airports.json").then(function(response){$scope.airports = response.data.records;});  


      $scope.selected = undefined; 

      $scope.getLocation = function(val) { 
       return $http.get('http://maps.googleapis.com/maps/api/geocode/json', { 
        params: { 
        address: val, 
        sensor: false 
        } 
       }).then(function(res){ 
        var addresses = []; 
        angular.forEach(res.data.results, function(item){ 

         addresses.push(item.formatted_address); 

        }); 
        return addresses; 
       }); 
       }; 

       $scope.on_item_selected=function($item, $model, $label) 
       { 
        $scope.selected_item = $item; 
       } 
     }); 

</script> 

HTML：

<input type="text" class="form-control" id="source" ng-model="asyncSelected" placeholder="Enter Airport Code or City Name" typeahead="address for address in getLocation($viewValue)" typeahead-loading="loadingLocations" typeahead-on-select="on_item_selected($item, $model, $label)"> 

Answer 1

你为什么不尝试：

address for address in getLocation($viewValue) | filter: airportFilter 

然后在你的控制器：

$scope.airportFilter = function(obj) { 
    var isAirport = false; 
    if(typeof obj === 'array') { 
    obj.forEach(function(item) { 
     (item.toLowerCase() === 'airport') ? isAirport = true; 
    }); 
    } 
} 

Answer 2

您可以使用select和展示为下拉，

<select ng-model="selectedItem" ng-options="port as port for port in arrString"> 
</select> 

控制器

$scope.formated = response.data.results[0].formatted_address; 
$scope.arrString = new Array(); 
$scope.arrString = $scope.formated.split(','); 

DEMO

要在输入显示