silvio..
13
如果将辅助字段添加到坐标表,则可以缩短查询的响应时间.
像这样:
CREATE TABLE `Coordinates` (
`id` INT(10) UNSIGNED NOT NULL COMMENT 'id for the object',
`type` TINYINT(4) UNSIGNED NOT NULL DEFAULT '0' COMMENT 'type',
`sin_lat` FLOAT NOT NULL COMMENT 'sin(lat) in radians',
`cos_cos` FLOAT NOT NULL COMMENT 'cos(lat)*cos(lon) in radians',
`cos_sin` FLOAT NOT NULL COMMENT 'cos(lat)*sin(lon) in radians',
`lat` FLOAT NOT NULL COMMENT 'latitude in degrees',
`lon` FLOAT NOT NULL COMMENT 'longitude in degrees',
INDEX `lat_lon_idx` (`lat`, `lon`)
)
如果您正在使用TokuDB,那么如果在任一谓词上添加聚类索引,您将获得更好的性能,例如,如下所示:
alter table Coordinates add clustering index c_lat(lat);
alter table Coordinates add clustering index c_lon(lon);
你需要以度为单位的基本lat和lon以及以弧度为单位的sin(lat),以弧度为单位的cos(lat)*cos(lon)和以弧度为单位的cos(lat)*sin(lon).然后你创建一个mysql函数,像这样:
CREATE FUNCTION `geodistance`(`sin_lat1` FLOAT,
`cos_cos1` FLOAT, `cos_sin1` FLOAT,
`sin_lat2` FLOAT,
`cos_cos2` FLOAT, `cos_sin2` FLOAT)
RETURNS float
LANGUAGE SQL
DETERMINISTIC
CONTAINS SQL
SQL SECURITY INVOKER
BEGIN
RETURN acos(sin_lat1*sin_lat2 + cos_cos1*cos_cos2 + cos_sin1*cos_sin2);
END
这给你的距离.
不要忘记在lat/lon上添加索引,这样边界装箱可以帮助搜索而不是减慢速度(索引已经添加到上面的CREATE TABLE查询中).
INDEX `lat_lon_idx` (`lat`, `lon`)
给定一个只有lat/lon坐标的旧表,你可以设置一个脚本来更新它:(php using meekrodb)
$users = DB::query('SELECT id,lat,lon FROM Old_Coordinates');
foreach ($users as $user)
{
$lat_rad = deg2rad($user['lat']);
$lon_rad = deg2rad($user['lon']);
DB::replace('Coordinates', array(
'object_id' => $user['id'],
'object_type' => 0,
'sin_lat' => sin($lat_rad),
'cos_cos' => cos($lat_rad)*cos($lon_rad),
'cos_sin' => cos($lat_rad)*sin($lon_rad),
'lat' => $user['lat'],
'lon' => $user['lon']
));
}
然后优化实际查询以仅在真正需要时进行距离计算,例如通过从内部和外部限制圆(井,椭圆).为此,您需要为查询本身预先计算几个指标:
// assuming the search center coordinates are $lat and $lon in degrees
// and radius in km is given in $distance
$lat_rad = deg2rad($lat);
$lon_rad = deg2rad($lon);
$R = 6371; // earth's radius, km
$distance_rad = $distance/$R;
$distance_rad_plus = $distance_rad * 1.06; // ovality error for outer bounding box
$dist_deg_lat = rad2deg($distance_rad_plus); //outer bounding box
$dist_deg_lon = rad2deg($distance_rad_plus/cos(deg2rad($lat)));
$dist_deg_lat_small = rad2deg($distance_rad/sqrt(2)); //inner bounding box
$dist_deg_lon_small = rad2deg($distance_rad/cos(deg2rad($lat))/sqrt(2));
鉴于这些准备,查询就像这样(PHP):
$neighbors = DB::query("SELECT id, type, lat, lon,
geodistance(sin_lat,cos_cos,cos_sin,%d,%d,%d) as distance
FROM Coordinates WHERE
lat BETWEEN %d AND %d AND lon BETWEEN %d AND %d
HAVING (lat BETWEEN %d AND %d AND lon BETWEEN %d AND %d) OR distance <= %d",
// center radian values: sin_lat, cos_cos, cos_sin
sin($lat_rad),cos($lat_rad)*cos($lon_rad),cos($lat_rad)*sin($lon_rad),
// min_lat, max_lat, min_lon, max_lon for the outside box
$lat-$dist_deg_lat,$lat+$dist_deg_lat,
$lon-$dist_deg_lon,$lon+$dist_deg_lon,
// min_lat, max_lat, min_lon, max_lon for the inside box
$lat-$dist_deg_lat_small,$lat+$dist_deg_lat_small,
$lon-$dist_deg_lon_small,$lon+$dist_deg_lon_small,
// distance in radians
$distance_rad);
解析上面的查询可能会说它没有使用索引,除非有足够的结果来触发这样的.当坐标表中有足够的数据时,将使用索引.您可以将FORCE INDEX(lat_lon_idx)添加到SELECT以使其使用索引而不考虑表大小,因此您可以使用EXPLAIN验证它是否正常工作.
使用上面的代码示例,您应该通过距离以最小的错误实现对象搜索的工作和可伸缩实现.