13

Google json から $street、$city、$country の文字列を取得しようとしています。私の自宅の住所で動作します: http://maps.googleapis.com/maps/api/geocode/json?latlng=52.108662,6.307370&sensor=true

$url = "http://maps.googleapis.com/maps/api/geocode/json?latlng=".$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'];
    }

ただし、次の例のように、配列内により多くのデータがある別のアドレスの場合: http://maps.googleapis.com/maps/api/geocode/json?latlng=52.154184,6.199592&sensor=true があるため、機能しません。 json 配列にさらにデータを追加すると、州が国になります。

必要なタイプ (long_name) を選択するにはどうすればよいですか?

  • for street : long_name where "types" : [ "route" ]
  • for city : long_name where "types" : [ "locality", "political" ]
  • for country : long_name where "types" : [ "country", "political" ]

ジオコード JSON からの出力例:

{
   "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" ]
            }

私はそれを自分で修正したと思います。これが私のコードです：

// 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"];
        }
    }
}
4

4 に答える 4

3

あなたのコードはまったく問題ありませんが、foreach ループを繰り返すよりも、1 つの foreach 内でスイッチを使用した方がよいのではないでしょうか? まったく同じ配列を解析する方法は次のとおりです。

  $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;
    }

  }
于 2014-02-13T09:59:41.807 に答える
2

郵便番号を使用して住所を検索する場合、Google MAP API を使用して最近通り、市、国を生成したため、コードは次のとおりです。

$search_code = urlencode($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;
于 2015-03-18T06:16:47.200 に答える