mysql 搜寻附近N公里内数据的简单实例

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。

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 |
+----+-----------+----------------+---------------+

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