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查找包含给定坐标的圆的mysql查询

eboye • 4 年前 • 823 次点击  

我有一个正在工作的php脚本,它获取经度和纬度值,然后将它们输入到mysql查询中。我想把它单独做成MySQL。下面是我当前的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 ";
    }

有人知道怎么把它完全变成mysql吗?我浏览了一下因特网,但大多数有关它的文献都很混乱。

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823 次点击  
文章 [ 9 ]  |  最新文章 4 年前
IMRA
Reply   •   1 楼
IMRA    5 年前

在mysql中计算距离

 SELECT (6371 * acos(cos(radians(lat2)) * cos(radians(lat1) ) * cos(radians(long1) -radians(long2)) + sin(radians(lat2)) * sin(radians(lat1)))) AS distance

因此,距离值将被计算,任何人都可以根据需要申请。

David Harkness Sam Vloeberghs
Reply   •   2 楼
David Harkness Sam Vloeberghs    9 年前

我认为我的javascript实现是一个很好的参考:

/*
 * Check to see if the second coord is within the precision ( meters )
 * of the first coord and return accordingly
 */
function checkWithinBound(coord_one, coord_two, precision) {
    var distance = 3959000 * Math.acos( 
        Math.cos( degree_to_radian( coord_two.lat ) ) * 
        Math.cos( degree_to_radian( coord_one.lat ) ) * 
        Math.cos( 
            degree_to_radian( coord_one.lng ) - degree_to_radian( coord_two.lng ) 
        ) +
        Math.sin( degree_to_radian( coord_two.lat ) ) * 
        Math.sin( degree_to_radian( coord_one.lat ) ) 
    );
    return distance <= precision;
}

/**
 * Get radian from given degree
 */
function degree_to_radian(degree) {
    return degree * (Math.PI / 180);
}
Harish Lalwani
Reply   •   3 楼
Harish Lalwani    6 年前
 SELECT *, (  
    6371 * acos(cos(radians(search_lat)) * cos(radians(lat) ) *   
cos(radians(lng) - radians(search_lng)) + sin(radians(search_lat)) *         sin(radians(lat)))  
) AS distance  
FROM table  
WHERE lat != search_lat AND lng != search_lng AND distance < 25  
 ORDER BY distance  
FETCH 10 ONLY 

距离25公里

John Crenshaw
Reply   •   4 楼
John Crenshaw    11 年前

我无法评论上面的答案,但要小心@ Pavel Chuchuva的回答。如果两个坐标相同,则该公式不会返回结果。在这种情况下,distance为空,因此该行不会按原样随该公式返回。

我不是mysql专家,但这似乎对我有用:

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 OR distance IS NULL ORDER BY distance LIMIT 0 , 20;
Bo Persson
Reply   •   5 楼
Bo Persson    12 年前

我编写了一个可以计算相同的过程, 但必须在相应的表中输入纬度和经度。

drop procedure if exists select_lattitude_longitude;

delimiter //

create procedure select_lattitude_longitude(In CityName1 varchar(20) , In CityName2 varchar(20))

begin

    declare origin_lat float(10,2);
    declare origin_long float(10,2);

    declare dest_lat float(10,2);
    declare dest_long float(10,2);

    if CityName1  Not In (select Name from City_lat_lon) OR CityName2  Not In (select Name from City_lat_lon) then 

        select 'The Name Not Exist or Not Valid Please Check the Names given by you' as Message;

    else

        select lattitude into  origin_lat from City_lat_lon where Name=CityName1;

        select longitude into  origin_long  from City_lat_lon where Name=CityName1;

        select lattitude into  dest_lat from City_lat_lon where Name=CityName2;

        select longitude into  dest_long  from City_lat_lon where Name=CityName2;

        select origin_lat as CityName1_lattitude,
               origin_long as CityName1_longitude,
               dest_lat as CityName2_lattitude,
               dest_long as CityName2_longitude;

        SELECT 3956 * 2 * ASIN(SQRT( POWER(SIN((origin_lat - dest_lat) * pi()/180 / 2), 2) + COS(origin_lat * pi()/180) * COS(dest_lat * pi()/180) * POWER(SIN((origin_long-dest_long) * pi()/180 / 2), 2) )) * 1.609344 as Distance_In_Kms ;

    end if;

end ;

//

delimiter ;
O. Jones
Reply   •   6 楼
O. Jones    10 年前

我必须把这件事做得很详细,所以我要和大家分享我的结果。这使用了 zip 表与 latitude longitude 桌子。它不依赖于谷歌地图,而是可以将其适配到包含LAT/LUN的任何表中。

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

看看查询中间的这一行:

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

这将搜索 拉链 距纬度/经度点42.81/-70.81 50.0英里内的表。当你将其构建到一个应用程序中时,你可以在其中放置自己的点和搜索半径。

如果你想以公里而不是英里为单位工作,那就换 69 111.045 变化 3963.17 6378.10 在查询中。

这是一份详细的书面报告。我希望它能帮助别人。 http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

silvio
Reply   •   7 楼
silvio    10 年前

如果将辅助字段添加到坐标表中,则可以提高查询的响应时间。

这样地:

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`)
)    

如果您使用tokudb,如果添加集群,您将获得更好的性能 任何一个谓词上的索引,例如:

alter table Coordinates add clustering index c_lat(lat);
alter table Coordinates add clustering index c_lon(lon);

您将需要基本的LAT和Lon的程度,以及正弦(LAT)在弧度,COS(LAT)* COS(Lon)在弧度和COS(LAT)*SIN(Lon)在弧度为每个点。 然后创建一个mysql函数,smth如下:

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

这给了你距离。

不要忘记在LAT/Lon上添加索引,因此绑定装箱可以帮助搜索,而不是减慢它(在上面的CREATE表查询中已经添加了索引)。

INDEX `lat_lon_idx` (`lat`, `lon`)

给定一个只有LAT/Lon坐标的旧表,您可以设置一个脚本来更新它:(PHP使用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,以便仅在真正需要时进行距离计算,例如从内部和外部包围圆(well,oval)。 为此,您需要预先计算查询本身的几个指标:

// 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));

考虑到这些准备工作,查询将如下所示(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);

解释上面的查询可能会说它没有使用索引,除非有足够的结果触发这样的查询。当坐标表中有足够的数据时,将使用索引。 您可以添加 力指数(横向) 要选择,使其使用与表大小无关的索引,这样您就可以通过解释它是否正确工作来验证。

通过上面的代码示例,您应该能够以最小的错误实现按距离进行的对象搜索。

Jacco
Reply   •   8 楼
Jacco    6 年前

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

以弧度表示经纬度。

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;

是您的SQL查询

要得到以公里或英里为单位的结果,请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一个边界框。 如果你把你的坐标数据放在 spatial enabled MySQL column ,你可以使用 MySQL's build in functionality 查询数据。

SELECT 
  id
FROM spatialEnabledTable
WHERE 
  MBRWithin(ogc_point, GeomFromText('Polygon((0 0,0 3,3 3,3 0,0 0))'))
Yahel
Reply   •   9 楼
Yahel    10 年前

Google Code FAQ - Creating a Store Locator with PHP, MySQL & Google Maps :

下面的sql语句将找到距离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,然后只要求距离值小于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。要按公里而不是英里搜索,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;