除了你可能会考虑使用笛卡尔坐标(x,y和z)其他答案而不是用于db存储的纬度/经度,因为生成的查询表达式在数据库服务器上的加载/时间比纬度/经度距离的可能查询更简单。
一个PHP实现的例子下可以找到:
http://headers-already-sent.com/geodistance/
的方法“getCartesian”将纬度/经度转换成直角坐标系和方法“getDistanceByCartesian”说明了如何计算实际距离。你需要做的是将这个距离计算从PHP转移到SQL查询中(这应该不那么复杂)。
编辑,因为我发现给一个更实际的例子
基于类,你可以在上面的链接我设置了2演示表为我公司的地点和所有MC多纳尔兹餐厅下找到时间在我们的周围,并转换纬度/经度从谷歌地图笛卡尔X,Y,Z:
CREATE TABLE `locations` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`title` varchar(255) NOT NULL DEFAULT '',
`lat` double NOT NULL,
`lng` double NOT NULL,
`x` double NOT NULL,
`y` double NOT NULL,
`z` double NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `locations` (`id`, `title`, `lat`, `lng`, `x`, `y`, `z`)
VALUES
(1,'Ida-Ehre-Platz 10, 20095 Hamburg',53.55053,9.99949,3727600.05477,657242.251356,5124712.81705),
(2,'Kieler Straße 191-193, 22525 Hamburg',53.57731,9.93686,3725956.4981,652753.812254,5126481.40905),
(3,'Reeperbahn 42, 20359 Hamburg',53.549951,9.964937,3728046.74189,655003.113578,5124674.56664),
(4,'Theodor-Heuss-Platz 3, 20354 Hamburg',53.56083,9.99038,3726797.15378,656489.722425,5125393.17725),
(5,'Mundsburger Damm 67, 22087 Hamburg',53.57028,10.02642,3725550.98379,658686.623655,5126017.24553),
(6,'Paul-Nevermann-Platz 1, 22765 Hamburg',53.552602,9.936678,3728135.78521,653123.397726,5124849.69505),
(7,'Friedrich-Ebert-Damm 101, 22047 Hamburg',53.58753,10.08958,3723303.02881,662522.688778,5127156.05819),
(8,'Amsinckstraße 73, 20097 Hamburg',53.54271,10.02654,3727978.07563,659123.791421,5124196.16112),
(9,'Eiffestraße 440, 20537 Hamburg',53.55214,10.04638,3726919.13256,660267.521487,5124819.17553);
CREATE TABLE `user` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL DEFAULT '',
`lat` double NOT NULL,
`lng` double NOT NULL,
`x` double NOT NULL,
`y` double NOT NULL,
`z` double NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `user` (`id`, `name`, `lat`, `lng`, `x`, `y`, `z`)
VALUES
(1,'Ministry.BBS, Cremon 36, 20457 Hamburg',53.545943,9.988761,3728127.10678,656615.385203,5124409.77226),
(2,'BBS, Dorotheenstraße 60, 22301 Hamburg',53.583231,10.008315,3724617.80169,657307.963226,5126872.28974);
基于这两个表的SQL查询找到一定的距离(2000内的所有位置(餐馆),在米在这个例子)到每个用户(我们公司的办公室)将是:
SELECT locations.*,
2 * 6371000.785 *
asin(
sqrt(
pow(locations.x - user.x, 2)
+ pow(locations.y - user.y, 2)
+ pow(locations.z - user.z, 2)
)/(2 * 6371000.785)
) AS distance
FROM locations, user
HAVING distance < 2000
ORDER BY distance ASC
如果你需要比“米”比你将不得不改变约为地球半径的东西。 6371000.785(以米为单位)以满足您的需要,并且还可以将所需的2000年的距离更改为您喜欢的或存储在用户表格中的每个用户的任何内容。
对于2来说,第二个缺点就是无论何时添加新的事件/客户,您都必须维护或重新生成该列表。 – xQbert
@xQbert你说得对,我没有说清楚。我会再补充一点。 – ean5533