从纬度和经度中获取PHP时区名称？

Question

有没有办法通过纬度和经度获得用户的时区？而不只是偏移量,而是它们所处的实际时区.

基本上,我正在搜索DateTimeZone :: getLocation的极性相反,它返回某个时区的纬度和经度.

Answer 1

对于那些想要从国家代码,纬度和经度获得时区的人.(如果您的服务器上安装了geoip模块,则很容易获得)

试试这个,我添加了一个距离计算 - 仅适用于那些有多个时区的国家.啊,国家代码是两个字母的ISO代码.

// ben@jp

function get_nearest_timezone($cur_lat, $cur_long, $country_code = '') {
    $timezone_ids = ($country_code) ? DateTimeZone::listIdentifiers(DateTimeZone::PER_COUNTRY, $country_code)
                                    : DateTimeZone::listIdentifiers();

    if($timezone_ids && is_array($timezone_ids) && isset($timezone_ids[0])) {

        $time_zone = '';
        $tz_distance = 0;

        //only one identifier?
        if (count($timezone_ids) == 1) {
            $time_zone = $timezone_ids[0];
        } else {

            foreach($timezone_ids as $timezone_id) {
                $timezone = new DateTimeZone($timezone_id);
                $location = $timezone->getLocation();
                $tz_lat   = $location['latitude'];
                $tz_long  = $location['longitude'];

                $theta    = $cur_long - $tz_long;
                $distance = (sin(deg2rad($cur_lat)) * sin(deg2rad($tz_lat))) 
                + (cos(deg2rad($cur_lat)) * cos(deg2rad($tz_lat)) * cos(deg2rad($theta)));
                $distance = acos($distance);
                $distance = abs(rad2deg($distance));
                // echo '<br />'.$timezone_id.' '.$distance; 

                if (!$time_zone || $tz_distance > $distance) {
                    $time_zone   = $timezone_id;
                    $tz_distance = $distance;
                } 

            }
        }
        return  $time_zone;
    }
    return 'unknown';
}
//timezone for one NY co-ordinate
echo get_nearest_timezone(40.772222,-74.164581) ;
// more faster and accurate if you can pass the country code 
echo get_nearest_timezone(40.772222, -74.164581, 'US') ;

Answer 2

Geonames应该很好地完成这项工作:

http://www.geonames.org/

他们还有一个php库.