通过使用谷歌反向地理编码获取街道，城市和国家

Question

我试图从谷歌json获得$street，$City和$Country字符串。它适用于我的家庭地址：http://maps.googleapis.com/maps/api/geocode/json?latlng=52.108662，6.307370

$url = "http://maps.googleapis.com/maps/api/geocode/json?latlng=".$lat.",".$lng."&sensor=true";
    $data = @file_get_contents($url);
    $jsondata = json_decode($data,true);
    if(is_array($jsondata) && $jsondata['status'] == "OK")
    {
          $city = $jsondata['results']['0']['address_components']['2']['long_name'];
          $country = $jsondata['results']['0']['address_components']['5']['long_name'];
          $street = $jsondata['results']['0']['address_components']['1']['long_name'];
    }

但对于数组中包含更多数据的不同地址，如本例所示：http://maps.googleapis.com/maps/api/geocode/json?latlng=52.154184,6.199592

如何选择所需的类型（长名称）？

• 对于街道：long_名称，其中“类型”：[“路线”]
• 对于城市：long_名称，其中“类型”：[“地区”、“政治”]
• 对于国家：long_名称，其中“类型”：[“国家”，“政治”]

地理代码JSON的输出示例：

{
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "89",
               "short_name" : "89",
               "types" : [ "street_number" ]
            },
            {
               "long_name" : "Wieck De",
               "short_name" : "Wieck De",
               "types" : [ "establishment" ]
            },
            {
               "long_name" : "Industrieweg",
               "short_name" : "Industrieweg",
               "types" : [ "route" ]
            },
            {
               "long_name" : "Zutphen",
               "short_name" : "Zutphen",
               "types" : [ "locality", "political" ]
            },
            {
               "long_name" : "Zutphen",
               "short_name" : "Zutphen",
               "types" : [ "administrative_area_level_2", "political" ]
            },
            {
               "long_name" : "Gelderland",
               "short_name" : "GE",
               "types" : [ "administrative_area_level_1", "political" ]
            },
            {
               "long_name" : "Nederland",
               "short_name" : "NL",
               "types" : [ "country", "political" ]
            },
            {
               "long_name" : "7202 CA",
               "short_name" : "7202 CA",
               "types" : [ "postal_code" ]
            }

我想是我自己修好的，我的代码是：

// street
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("route", $address["types"])) {
            $street = $address["long_name"];
        }
    }
}
// city
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("locality", $address["types"])) {
            $city = $address["long_name"];
        }
    }
}
// country
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("country", $address["types"])) {
            $country = $address["long_name"];
        }
    }
}

Answer 1

如果您使用邮政编码来查找地址，因为我最近使用Google MAP API生成了街道，城市，国家代码是：

$search_code = urlencode($postcode);
        $url = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $search_code . '&sensor=false';
        $json = json_decode(file_get_contents($url));
        if($json->results == []){
            return '';
        }
        $lat = $json->results[0]->geometry->location->lat;
        $lng = $json->results[0]->geometry->location->lng;

        //Now build the actual lookup
        $address_url = 'http://maps.googleapis.com/maps/api/geocode/json?latlng=' . $lat . ',' . $lng . '&sensor=false';
        $address_json = json_decode(file_get_contents($address_url));

        $address_data = $address_json->results[0]->address_components;
        //return $address_data = $address_json->results[0]->formatted_address;

        $street = str_replace('Dr', 'Drive', $address_data[1]->long_name);
        $town = $address_data[2]->long_name;
        $county = $address_data[3]->long_name;

        return $street.', '. $town. ', '.$county;

Answer 2

您的代码非常好，但是在1 foreach中使用开关而不是重复的foreach循环不是更好吗？下面是我如何解析完全相同的数组：

  $location = array();

  foreach ($result['address_components'] as $component) {

    switch ($component['types']) {
      case in_array('street_number', $component['types']):
        $location['street_number'] = $component['long_name'];
        break;
      case in_array('route', $component['types']):
        $location['street'] = $component['long_name'];
        break;
      case in_array('sublocality', $component['types']):
        $location['sublocality'] = $component['long_name'];
        break;
      case in_array('locality', $component['types']):
        $location['locality'] = $component['long_name'];
        break;
      case in_array('administrative_area_level_2', $component['types']):
        $location['admin_2'] = $component['long_name'];
        break;
      case in_array('administrative_area_level_1', $component['types']):
        $location['admin_1'] = $component['long_name'];
        break;
      case in_array('postal_code', $component['types']):
        $location['postal_code'] = $component['long_name'];
        break;
      case in_array('country', $component['types']):
        $location['country'] = $component['long_name'];
        break;
    }

  }

Answer 3

您可以将数据转换为关联数组，并像

 $data = array();
 foreach($jsondata['results']['0']['address_components'] as $element){
     $data[ implode(' ',$element['types']) ] = $element['long_name'];
 }
 print_r($data);

 echo 'route: ' . $data['route'] . "\n";
 echo 'country: ' . $data['country political'];