192

我有一个有效的 PHP 脚本,它获取经度和纬度值,然后将它们输入到 MySQL 查询中。我想只做MySQL。这是我当前的 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 ";
    }

有谁知道如何使这个完全 MySQL ?我浏览了一些互联网,但大多数关于它的文献都很混乱。

4

9 回答 9

374

来自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;
于 2009-02-22T11:04:31.847 回答
34

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

以弧度为单位的纬度和经度。

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

要获得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。如果您将坐标数据放在启用了空间的 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))'))
于 2009-02-22T11:19:48.393 回答
14

如果将帮助字段添加到坐标表中,可以提高查询的响应时间。

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`)
)    

如果您使用的是 TokuDB,如果您在任一谓词上添加集群索引,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);

对于每个点,您需要以度数为单位的基本纬度和经度以及以弧度表示的 sin(lat)、以弧度表示的 cos(lat)*cos(lon) 和以弧度表示的 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

这给了你距离。

不要忘记在纬度/经度上添加索引,这样边界框可以帮助搜索而不是减慢搜索速度(索引已经添加到上面的 CREATE TABLE 查询中)。

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

给定一个只有纬度/经度坐标的旧表,您可以设置一个脚本来更新它,如下所示:(使用 meekrodb 的 php)

$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,您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形(嗯,椭圆形)。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));

鉴于这些准备工作,查询如下所示(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);

对上述查询的解释可能会说它不使用索引,除非有足够的结果来触发它。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,您应该有一个按距离进行对象搜索的工作且可扩展的实现,并且误差最小。

于 2011-09-02T13:54:04.557 回答
10

我不得不在一些细节上解决这个问题,所以我会分享我的结果。这使用zip带有latitude和表的longitude表。它不依赖于谷歌地图;相反,您可以将其调整为任何包含纬度/经度的表格。

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

这将搜索zip表中距离纬度/经度点 42.81/-70.81 50.0 英里范围内最近的 30 个条目。当您将其构建到应用程序中时,您可以在其中放置自己的点和搜索半径。

如果您想以公里而不是英里为单位工作,请在查询中更改69111.045和更改3963.176378.10

这里有详细的写法。我希望它可以帮助某人。 http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

于 2013-07-25T12:19:02.870 回答
4
 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

于 2015-05-02T11:24:39.797 回答
3

我已经编写了一个可以计算相同的程序,但是您必须在相应的表格中输入纬度和经度。

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 ;
于 2011-08-29T04:08:21.610 回答
3

我无法对上述答案发表评论,但要小心@Pavel Chuchuva 的答案。如果两个坐标相同,该公式将不会返回结果。在这种情况下,距离为空,因此该行不会按原样返回该公式。

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;
于 2012-11-24T19:38:43.923 回答
2

我认为我的 javascript 实现将是一个很好的参考:

/*
 * 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);
}
于 2012-04-15T07:57:09.300 回答
2

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

因此将计算距离值,任何人都可以根据需要申请。

于 2019-02-06T08:58:52.153 回答