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

这里是一个非常详细的描述地理距离搜索MySQL的一个解决方案的基础上实现的Haversine公式到MySQL。完整的解决方案描述,包括理论、实现和进一步的性能优化。尽管空间优化部分在我的案例中没有正确工作。 http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL

   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;
SELECT * FROM (SELECT *,(((acos(sin((43.6980168*pi()/180)) * 
sin((latitude*pi()/180))+cos((43.6980168*pi()/180)) * 
cos((latitude*pi()/180)) * cos(((7.266903899999988- longitude)* 
pi()/180))))*180/pi())*60*1.1515 ) as distance 
FROM wp_users WHERE 1 GROUP BY ID limit 0,10) as X 
ORDER BY ID DESC

这是MySQL中点与点之间的距离计算查询,我已经在一个长数据库中使用过它,它工作完美!注意:根据您的需求进行更改(数据库名称,表名称,列等)。

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);

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