php设计模式为nealy相似的json响应

Question

我想在php中编写一个小类，它可以从不同的GEO-API（如Google Maps或osm nominatim）获取经度和纬度值。两个提供程序返回与json类似但具有不同键的结果。

在getCordinates（）中，您会看到解析Google的api。是否有一个很好的设计模式，我没有必要使用几个if语句来破坏每个新地理提供程序的方法并炸毁方法但保持分离？

public $googleEndPoint = "http://maps.google.com/maps/api/geocode/json?address=";
public $OSMendPoint =  "http://nominatim.openstreetmap.org/search?format=json&q=";
public $endpoint;
public $address;


function __construct($endpoint = 'google'){
    switch ($endpoint){
        case 'OSM':
            $this->endpoint = $this->OSMendPoint;
            break;
        default:
            $this->endpoint = $this->googleEndPoint;
        break; 
    }
}


function request(){
    $address = urlencode($this->address);
    $result = file_get_contents($this->endpoint.$address); 
    $response = json_decode($result, true);
    return $response;
}


function getCoordinates($address) {
    $this->address = $address;
    $response = $this->request();
    // this is google specific and does not fit for other provider
    if( $response['status']=='OK' ){
        $data_arr['lat'] = $response['results'][0]['geometry']['location']['lat']; 
        $data_arr['lon'] = $response['results'][0]['geometry']['location']['lng'];  
        $data_arr['formatted_address'] = $response['results'][0]['formatted_address'];   
            return $data_arr;            
        } else {
            return false;
        }
    }