mysql 搜寻附近N公里内数据的实例

根据圆周率和地球半径系数以及搜寻点的经纬度，搜寻数据表中与搜寻点之间的距离为N公里内的数据。

转自 https://blog.csdn.net/fdipzone/article/details/52050471

1.创建测试表

CREATE TABLE `location` (
 `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
 `name` varchar(50) NOT NULL,
 `longitude` decimal(13,10) NOT NULL,
 `latitude` decimal(13,10) NOT NULL,
 PRIMARY KEY (`id`),
 KEY `long_lat_index` (`longitude`,`latitude`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

2.插入测试数据

insert into location(name,longitude,latitude) values
('广州东站',113.332264,23.156206),
('林和西',113.330611,23.147234),
('天平架',113.328095,23.165376);

mysql> select * from `location`;
+----+--------------+----------------+---------------+
| id | name         | longitude      | latitude      |
+----+--------------+----------------+---------------+
|  1 | 广州东站      | 113.3322640000 | 23.1562060000 |
|  2 | 林和西        | 113.3306110000 | 23.1472340000 |
|  3 | 天平架        | 113.3280950000 | 23.1653760000 |
+----+--------------+----------------+---------------+

3.搜寻1公里内的数据

搜寻点坐标：时代广场 113.323568, 23.146436

6370.996公里为地球的半径

计算球面两点坐标距离公式

C = sin(MLatA)sin(MLatB)cos(MLonA-MLonB) + cos(MLatA)cos(MLatB) 
Distance = RArccos(C)*Pi180

根据计算公式得到查询语句如下：

select * from `location` where (
acos(
sin(([#latitude#]*3.1415)/180) * sin((latitude*3.1415)/180) + 
cos(([#latitude#]*3.1415)/180) * cos((latitude*3.1415)/180) * cos(([#longitude#]*3.1415)/180 - (longitude*3.1415)/180)
)*6370.996
)<=1;

执行查询：

mysql> select * from `location` where (
    -> acos(
    -> sin((23.146436*3.1415)/180) * sin((latitude*3.1415)/180) + 
    -> cos((23.146436*3.1415)/180) * cos((latitude*3.1415)/180) * cos((113.323568*3.1415)/180 - (longitude*3.1415)/180)
    -> )*6370.996
    -> )<=1;
+----+-----------+----------------+---------------+
| id | name      | longitude      | latitude      |
+----+-----------+----------------+---------------+
|  2 | 林和西     | 113.3306110000 | 23.1472340000 |
+----+-----------+----------------+---------------+