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));

我想返回卢浮宫5 英里范围内的所有地方，这些地方应该返回 3 条记录(卢浮宫、埃菲尔铁塔、拉雪兹神父(

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((函数来实现上述目标，但目前我坚持使用5.6。

任何帮助表示赞赏。

Answer 1

您可以使用 Haversine 公式填充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 ;

被盗地址：https://www.bram.us/2018/03/01/mysql-st_distance_sphere-polyfill/