MS SQL版本在这里:

        DECLARE @SLAT AS FLOAT
        DECLARE @SLON AS FLOAT

        SET @SLAT = 38.150785
        SET @SLON = 27.360249

        SELECT TOP 10 [LATITUDE], [LONGITUDE], SQRT(
            POWER(69.1 * ([LATITUDE] - @SLAT), 2) +
            POWER(69.1 * (@SLON - [LONGITUDE]) * COS([LATITUDE] / 57.3), 2)) AS distance
        FROM [TABLE] ORDER BY 3

查找离我最近的用户:

距离(米)

根据文森特提的公式

i有用户表:

+----+-----------------------+---------+--------------+---------------+
| id | email                 | name    | location_lat | location_long |
+----+-----------------------+---------+--------------+---------------+
| 13 | xxxxxx@xxxxxxxxxx.com | Isaac   | 17.2675625   | -97.6802361   |
| 14 | xxxx@xxxxxxx.com.mx   | Monse   | 19.392702    | -99.172596    |
+----+-----------------------+---------+--------------+---------------+

sql:

-- my location:  lat   19.391124   -99.165660
SELECT 
(ATAN(
    SQRT(
        POW(COS(RADIANS(users.location_lat)) * SIN(RADIANS(users.location_long) - RADIANS(-99.165660)), 2) +
        POW(COS(RADIANS(19.391124)) * SIN(RADIANS(users.location_lat)) - 
       SIN(RADIANS(19.391124)) * cos(RADIANS(users.location_lat)) * cos(RADIANS(users.location_long) - RADIANS(-99.165660)), 2)
    )
    ,
    SIN(RADIANS(19.391124)) * 
    SIN(RADIANS(users.location_lat)) + 
    COS(RADIANS(19.391124)) * 
    COS(RADIANS(users.location_lat)) * 
    COS(RADIANS(users.location_long) - RADIANS(-99.165660))
 ) * 6371000) as distance,
users.id
FROM users
ORDER BY distance ASC

地球半径:6371000(单位:米)

简单的一个;)

SELECT * FROM `WAYPOINTS` W ORDER BY
ABS(ABS(W.`LATITUDE`-53.63) +
ABS(W.`LONGITUDE`-9.9)) ASC LIMIT 30;

把坐标换成你需要的坐标。这些值必须存储为double类型。这是一个工作中的MySQL 5。x的例子。

干杯

以防你像我一样懒，这里有一个解决方案，由这个和其他关于SO的答案合并而成。

set @orig_lat=37.46; 
set @orig_long=-122.25; 
set @bounding_distance=1;

SELECT
*
,((ACOS(SIN(@orig_lat * PI() / 180) * SIN(`lat` * PI() / 180) + COS(@orig_lat * PI() / 180) * COS(`lat` * PI() / 180) * COS((@orig_long - `long`) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS `distance` 
FROM `cities` 
WHERE
(
  `lat` BETWEEN (@orig_lat - @bounding_distance) AND (@orig_lat + @bounding_distance)
  AND `long` BETWEEN (@orig_long - @bounding_distance) AND (@orig_long + @bounding_distance)
)
ORDER BY `distance` ASC
limit 25;

试试这个，它显示最近的点提供的坐标(50公里内)。它工作得很完美:

SELECT m.name,
    m.lat, m.lon,
    p.distance_unit
             * DEGREES(ACOS(COS(RADIANS(p.latpoint))
             * COS(RADIANS(m.lat))
             * COS(RADIANS(p.longpoint) - RADIANS(m.lon))
             + SIN(RADIANS(p.latpoint))
             * SIN(RADIANS(m.lat)))) AS distance_in_km
FROM <table_name> AS m
JOIN (
      SELECT <userLat> AS latpoint, <userLon> AS longpoint,
             50.0 AS radius, 111.045 AS distance_unit
     ) AS p ON 1=1
WHERE m.lat
BETWEEN p.latpoint  - (p.radius / p.distance_unit)
    AND p.latpoint  + (p.radius / p.distance_unit)
    AND m.lon BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
    AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
ORDER BY distance_in_km

只需更改<table_name>。<userLat>和<userLon>

你可以在这里阅读更多关于这个解决方案:http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

 +----+-----------------------+---------+--------------+---------------+
| id | email                 | name    | location_lat | location_long |
+----+-----------------------+---------+--------------+---------------+
| 7  | test@gmail.com        | rembo   | 23.0249256   |  72.5269697   |
| 25 | test1@gmail.com.      | Rajnis  | 23.0233221    | 72.5342112   |
+----+-----------------------+---------+--------------+---------------+

$lat = 23.02350629;

$long = 72.53230239;

DB:: 选择 (“ 选择 * 从 （ 选择 ， （ (acos(sin)。决。”* pi(美元)/ 180)* sin (lat * pi(+) - 180)因为(”。拉丁语“圆周率(美元)/ 180)* cos (lat * pi () / 180) * cos(("。龙。”- long) * pi(美元)/ 180))* 180 / pi() * 60 * 1515 1。1 . 609344 ) 距离美国 从 \用户 ) \用户 在哪里 距离<= 2");