MySQL大圆距离(Haversine公式)

Question

我有一个工作的PHP脚本,它获取经度和纬度值,然后将它们输入到MySQL查询中.我想把它做成MySQL.这是我目前的PHP代码:

if ($distance != "Any" && $customer_zip != "") { //get the great circle distance

    //get the origin zip code info
    $zip_sql = "SELECT * FROM zip_code WHERE zip_code = '$customer_zip'";
    $result = mysql_query($zip_sql);
    $row = mysql_fetch_array($result);
    $origin_lat = $row['lat'];
    $origin_lon = $row['lon'];

    //get the range
    $lat_range = $distance/69.172;
    $lon_range = abs($distance/(cos($details[0]) * 69.172));
    $min_lat = number_format($origin_lat - $lat_range, "4", ".", "");
    $max_lat = number_format($origin_lat + $lat_range, "4", ".", "");
    $min_lon = number_format($origin_lon - $lon_range, "4", ".", "");
    $max_lon = number_format($origin_lon + $lon_range, "4", ".", "");
    $sql .= "lat BETWEEN '$min_lat' AND '$max_lat' AND lon BETWEEN '$min_lon' AND '$max_lon' AND ";
    }

有谁知道如何完全成为MySQL？我已经浏览了一下互联网,但大部分关于它的文献都令人困惑.

Answer 1

来自Google Code FAQ - 使用PHP,MySQL和Google Maps创建商店定位器:

这是SQL语句,它将找到距离37,-122坐标25英里半径范围内最近的20个位置.它根据该行的纬度/经度和目标纬度/经度计算距离,然后仅询问距离值小于25的行,按距离对整个查询进行排序,并将其限制为20个结果.要以公里而不是里程搜索,请将3959替换为6371.

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) 
* cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin(radians(lat)) ) ) AS distance 
FROM markers 
HAVING distance < 25 
ORDER BY distance 
LIMIT 0 , 20;

Answer 2

$greatCircleDistance = acos( cos($latitude0) * cos($latitude1) * cos($longitude0 - $longitude1) + sin($latitude0) * sin($latitude1));

纬度和纬度经度.

所以

SELECT 
  acos( 
      cos(radians( $latitude0 ))
    * cos(radians( $latitude1 ))
    * cos(radians( $longitude0 ) - radians( $longitude1 ))
    + sin(radians( $latitude0 )) 
    * sin(radians( $latitude1 ))
  ) AS greatCircleDistance 
 FROM yourTable;

是你的SQL查询

以Km或英里为单位获得结果,将结果乘以地球的平均半径(3959英里,6371公里或3440海里)

您在示例中计算的是边界框.如果将坐标数据放在启用空间的MySQL列中,则可以使用MySQL的内置功能来查询数据.

SELECT 
  id
FROM spatialEnabledTable
WHERE 
  MBRWithin(ogc_point, GeomFromText('Polygon((0 0,0 3,3 3,3 0,0 0))'))

Answer 3

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.

Answer 4

我必须详细解决这个问题,所以我会分享我的结果.这使用带zip表latitude和longitude表的表.它不依赖于Google地图; 相反,你可以将它适应任何包含lat/long的表.

SELECT zip, primary_city, 
       latitude, longitude, distance_in_mi
  FROM (
SELECT zip, primary_city, latitude, longitude,r,
       (3963.17 * ACOS(COS(RADIANS(latpoint)) 
                 * COS(RADIANS(latitude)) 
                 * COS(RADIANS(longpoint) - RADIANS(longitude)) 
                 + SIN(RADIANS(latpoint)) 
                 * SIN(RADIANS(latitude)))) AS distance_in_mi
 FROM zip
 JOIN (
        SELECT  42.81  AS latpoint,  -70.81 AS longpoint, 50.0 AS r
   ) AS p 
 WHERE latitude  
  BETWEEN latpoint  - (r / 69) 
      AND latpoint  + (r / 69)
   AND longitude 
  BETWEEN longpoint - (r / (69 * COS(RADIANS(latpoint))))
      AND longpoint + (r / (69 * COS(RADIANS(latpoint))))
  ) d
 WHERE distance_in_mi <= r
 ORDER BY distance_in_mi
 LIMIT 30

查看该查询中间的这一行:

    SELECT  42.81  AS latpoint,  -70.81 AS longpoint, 50.0 AS r

这zip将在纬度/长点42.81/-70.81的50.0英里内搜索表中最近的30个条目.当您将其构建到应用程序中时,您可以在此处放置自己的点和搜索半径.

如果你想在公里的工作,而不是英里,改69到111.045和改变3963.17,以6378.10在查询中.

这是一篇详细的文章.我希望它对某人有所帮助. http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/