我目前在mysql数据库中有不到一百万个位置,都有经度和纬度信息。

我试图通过查询找到一个点和许多其他点之间的距离。它没有我想要的那么快,尤其是每秒100+次。

有没有比mysql更快的查询或更快的系统?我使用这个查询:

SELECT 
  name, 
   ( 3959 * acos( cos( radians(42.290763) ) * cos( radians( locations.lat ) ) 
   * cos( radians(locations.lng) - radians(-71.35368)) + sin(radians(42.290763)) 
   * sin( radians(locations.lat)))) AS distance 
FROM locations 
WHERE active = 1 
HAVING distance < 10 
ORDER BY distance;

注:提供的距离单位为英里。如果需要公里,请使用6371而不是3959。


当前回答

如果你使用的是MySQL 5.7。*,那么你可以使用st_distance_sphere(POINT, POINT)。

Select st_distance_sphere(POINT(-2.997065, 53.404146 ), POINT(58.615349, 23.56676 ))/1000  as distcance

其他回答

下面的MySQL函数发布在这篇博文上。我还没有对它进行太多测试,但从我从帖子中收集到的内容来看,如果你的纬度和经度字段被索引了,这可能对你很有用:

DELIMITER $$

DROP FUNCTION IF EXISTS `get_distance_in_miles_between_geo_locations` $$
CREATE FUNCTION get_distance_in_miles_between_geo_locations(
  geo1_latitude decimal(10,6), geo1_longitude decimal(10,6), 
  geo2_latitude decimal(10,6), geo2_longitude decimal(10,6)) 
returns decimal(10,3) DETERMINISTIC
BEGIN
  return ((ACOS(SIN(geo1_latitude * PI() / 180) * SIN(geo2_latitude * PI() / 180) 
    + COS(geo1_latitude * PI() / 180) * COS(geo2_latitude * PI() / 180) 
    * COS((geo1_longitude - geo2_longitude) * PI() / 180)) * 180 / PI()) 
    * 60 * 1.1515);
END $$

DELIMITER ;

示例用法:

假设有一个名为places的表,其中包含纬度和经度字段:

SELECT get_distance_in_miles_between_geo_locations(-34.017330, 22.809500, AS distance_from_input FROM places;

   select
   (((acos(sin(('$latitude'*pi()/180)) * sin((`lat`*pi()/180))+cos(('$latitude'*pi()/180)) 
    * cos((`lat`*pi()/180)) * cos((('$longitude'- `lng`)*pi()/180))))*180/pi())*60*1.1515) 
    AS distance
    from table having distance<22;

使用mysql

SET @orig_lon = 1.027125;
SET @dest_lon = 1.027125;

SET @orig_lat = 2.398441;
SET @dest_lat = 2.398441;

SET @kmormiles = 6371;-- for distance in miles set to : 3956

SELECT @kmormiles * ACOS(LEAST(COS(RADIANS(@orig_lat)) * 
 COS(RADIANS(@dest_lat)) * COS(RADIANS(@orig_lon - @dest_lon)) + 
 SIN(RADIANS(@orig_lat)) * SIN(RADIANS(@dest_lat)),1.0)) as distance;

参见:https://andrew.hedges.name/experiments/haversine/

参见:https://stackoverflow.com/a/24372831/5155484

参见:http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

注意:LEAST用于避免null值,如https://stackoverflow.com/a/24372831/5155484上建议的注释

我真的很喜欢@Māris kiseovs的解决方案,但我喜欢许多其他人可能会从他的例子中得到Lat和lng的POINTS。在概括它时,我想我会分享它。在我的情况下,我需要找到所有的开始点,在end_point的一定半径内。

我希望这能帮助到一些人。

SELECT @LAT := ST_X(end_point), @LNG := ST_Y(end_point) FROM routes  WHERE route_ID = 280;
SELECT 
  *,
  (6371e3 * ACOS(COS(RADIANS(@LAT)) * COS(RADIANS(ST_X(start_point))) 
  * COS(RADIANS(ST_Y(start_point)) - RADIANS(@LNG)) + SIN(RADIANS(@LAT))
  * SIN(RADIANS(ST_X(start_point))))) AS distance 
FROM routes
WHERE MBRContains
 (
  LineString
    (
    Point (
            @LNG + 15 / (111.320 * COS(RADIANS(@LAT))),
            @LAT + 15 / 111.133
    ),
    Point (
    @LNG - 15 / (111.320 * COS(RADIANS(@LAT))),
        @LAT - 15 / 111.133
    )
 ),
 POINT(ST_Y(end_point),ST_X(end_point))
)
HAVING distance < 100
ORDER By distance;
set @latitude=53.754842;
set @longitude=-2.708077;
set @radius=20;

set @lng_min = @longitude - @radius/abs(cos(radians(@latitude))*69);
set @lng_max = @longitude + @radius/abs(cos(radians(@latitude))*69);
set @lat_min = @latitude - (@radius/69);
set @lat_max = @latitude + (@radius/69);

SELECT * FROM postcode
WHERE (longitude BETWEEN @lng_min AND @lng_max)
AND (latitude BETWEEN @lat_min and @lat_max);