MySQL 5.6查找距离内的点并按距离ASC排序

Question

我有下表：

CREATE TABLE `places` (
 `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
 `name` varchar(256) DEFAULT NULL,
 `coordinates` point DEFAULT NULL,
 PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `places` ( `name`, `coordinates`)
VALUES ("Louvre ", POINT(48.861105, 2.335337));

INSERT INTO `places` ( `name`, `coordinates`)
VALUES ("Eiffel Tower", POINT(48.858271, 2.293795));

INSERT INTO `places` ( `name`, `coordinates`)
VALUES ("Pere Lachaise", POINT(48.861131, 2.394683));

INSERT INTO `places` ( `name`, `coordinates`)
VALUES ("Brooklyn", POINT(40.711089, -73.948391));

我想归还卢浮宫5英里范围内的所有地点，这应该返回3条记录（卢浮宫，艾菲尔铁塔，佩雷拉查兹）

我运行以下查询，但最终得到所有4条记录。我做错了什么？

 SET @radius = 5 * 1609.344; -- convert miles to meters

 SELECT name, ST_Distance(coordinates,POINT(48.861105, 2.335337)) AS distance 
   FROM places
  WHERE ST_Within(coordinates, ST_Buffer(POINT(48.861105, 2.335337),@radius))
ORDER BY distance ASC

注意：我知道MySQL 5.7具有ST_Distance_Sphere（）函数来实现上述功能，但目前我仍然使用5.6。

感谢任何帮助。

Answer 1

您可以使用Haversine公式来填充ST_Distance_Sphere：

DELIMITER $$

CREATE FUNCTION `ST_Distance_Sphere` (point1 POINT, point2 POINT)

    RETURNS FLOAT
    no sql deterministic
    BEGIN
        declare R INTEGER DEFAULT 6371000;
        declare `φ1` float;
        declare `φ2` float;
        declare `Δφ` float;
        declare `Δλ` float;
        declare a float;
        declare c float;
        set `φ1` = radians(y(point1));
        set `φ2` = radians(y(point2));
        set `Δφ` = radians(y(point2) - y(point1));
        set `Δλ` = radians(x(point2) - x(point1));

        set a = sin(`Δφ` / 2) * sin(`Δφ` / 2) + cos(`φ1`) * cos(`φ2`) * sin(`Δλ` / 2) * sin(`Δλ` / 2);
        set c = 2 * atan2(sqrt(a), sqrt(1-a));

        return R * c;
    END$$

DELIMITER ;

从https://www.bram.us/2018/03/01/mysql-st_distance_sphere-polyfill/处偷来的