[游戏开发]php 使用地图计算两点之间的距离

    public function get_distance_list(){

       $param = $this->request->param();

        $EARTH = 6378.137; //地球半径
        $PI = 3.1415926535898; //PI值
        $lng = $param['lng'];  //经度
        $lat = $param['lat'];   //纬度
        $res = Db::name('maps')->alias('mp')
            ->field("sp.*,mp.*,(2 * $EARTH* ASIN(SQRT(POW(SIN($PI*(" . $lat . "-lat)/360),2)+COS($PI*" . $lat . "/180)* COS(lat * $PI/180)*POW(SIN($PI*(" . $lng . "-lng)/360),2)))) as distance")
            ->join('shop sp','sp.id = mp.shop_id')
            ->order('distance asc')
            ->select();
        if ($res) {
            foreach ($res as $k => $v) {
                $res[$k]['distance'] = $this->get_distance($lng, $lat, $v['lng'], $v['lat']);
//                $res[$k]['image'] = $this->request->domain() . $v['image'];
            }
        }
        $this->success('数据成功', $res);
    }

##获取单个
 public  function get_distance($lng1, $lat1, $lng2,$lat2) {
        // 将角度转为狐度
        $radLat1 = deg2rad($lat1); //deg2rad()函数将角度转换为弧度
        $radLat2 = deg2rad($lat2);
        $radLng1 = deg2rad($lng1);
        $radLng2 = deg2rad($lng2);
        $a = $radLat1 - $radLat2;
        $b = $radLng1 - $radLng2;
        $s = 2 * asin(sqrt(pow(sin($a / 2), 2) + cos($radLat1) * cos($radLat2) * pow(sin($b / 2), 2))) * 6378.137 * 1000;
        return number_format($s/1000,1);
    }