如何使用 MySQL 中的半正矢公式测量距离？

Question

我从 Google Maps Reverse-Geocoding API 获取纬度和经度，然后我需要这样的东西：

mysql_query("SELECT users.*, ".mysql_distance_column($lat,$lng)." FROM users ORDER BY DISTANCE";

function mysql_distance_column($lat=40 , $lng=-73) {

   $defaultLatitudeColumn = 'user_lat'; 
   $defaultLongitudeColumn='user_lng';
   $defaultColumnName='user_distance';
    return  "(( 
(3956 * 2 * ASIN(SQRT( POWER(SIN(({$lat} - abs({$defaultLatitudeColumn})) 
* pi()/180 / 2), 2) + COS({$lat} * pi()/180 ) 
* COS(abs({$defaultLatitudeColumn}) * pi()/180) 
* POWER(SIN(({$lng} - {$defaultLongitudeColumn}) * pi()/180 / 2), 2) ))
 )) ) as {$defaultColumnName} ";

}

更新 我无法让这个工作

delimiter //
CREATE FUNCTION `GeoDistMiles`( lat1 FLOAT (10,6), lon1 FLOAT (10,6), lat2 FLOAT (10,6), lon2 FLOAT (10,6) ) 
RETURNS FLOAT 
DETERMINISTIC 
NO SQL
BEGIN
  DECLARE pi, q1, q2, q3 FLOAT (10,6);
  DECLARE rads FLOAT (10,6) DEFAULT 0;
  SET pi = PI();
  SET lat1 = lat1 * pi / 180;
  SET lon1 = lon1 * pi / 180;
  SET lat2 = lat2 * pi / 180;
  SET lon2 = lon2 * pi / 180;
  SET q1 = COS(lon1-lon2);
  SET q2 = COS(lat1-lat2);
  SET q3 = COS(lat1+lat2);
  SET rads = ACOS( 0.5*((1.0+q1)*q2 - (1.0-q1)*q3) );
  RETURN 3963.346 * rads;
END

Answer 1

这是我使用的公式。请记住，地球不是一个完美的球体，因此结果永远不会是完美的。

CREATE DEFINER=`root`@`localhost` FUNCTION `GeoDistMiles`( lat1 FLOAT, lon1 FLOAT, lat2 FLOAT, lon2 FLOAT ) RETURNS float
BEGIN
  DECLARE pi, q1, q2, q3 FLOAT;
  DECLARE rads FLOAT DEFAULT 0;
  SET pi = PI();
  SET lat1 = lat1 * pi / 180;
  SET lon1 = lon1 * pi / 180;
  SET lat2 = lat2 * pi / 180;
  SET lon2 = lon2 * pi / 180;
  SET q1 = COS(lon1-lon2);
  SET q2 = COS(lat1-lat2);
  SET q3 = COS(lat1+lat2);
  SET rads = ACOS( 0.5*((1.0+q1)*q2 - (1.0-q1)*q3) );
  RETURN 3963.346 * rads;
END

Answer 2

我假设您正在尝试使用 http://en.wikipedia.org/wiki/Haversine_formula 。

这是经过轻微测试的，但我认为你的公式应该是：（

(ROUND((3956 * 2 * ASIN(SQRT(POWER(SIN(({$lat} - {$defaultLatitudeColumn}) * pi() / 180 / 2), 2) + COS({$lat} * pi()/180 ) * COS({$defaultLatitudeColumn} * pi()/180) *POWER(SIN(({$lng} - {$defaultLongitudeColumn}) * pi()/180 / 2), 2) )) )*{$magicNumber}) )/{$magicNumber}

我删除了 abs 调用。）

Answer 3

您好，我有一个简单的程序，您可以将其用于您的工作。

程序非常简单，计算两个城市之间的距离。您可以按照自己的方式修改。

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 ;

Answer 4

SELECT ACOS(COS(RADIANS(lat)) *
COS(RADIANS(lon)) * COS(RADIANS(34.7405350)) * COS(RADIANS(-92.3245120)) +
COS(RADIANS(lat)) * SIN(RADIANS(lon)) * COS(RADIANS(34.7405350)) * 
SIN(RADIANS(-92.3245120)) + SIN(RADIANS(lat)) * SIN(RADIANS(34.7405350))) * 
3963.1 AS Distance
FROM Stores 
WHERE 1
HAVING Distance <= 50

以下是我在 PHP 中使用它的方法：

// Find rows in Stores within 50 miles of $lat,$lon
$lat = '34.7405350';
$lon = '-92.3245120';

$sql = "SELECT Stores.*, ACOS(COS(RADIANS(lat)) *
COS(RADIANS(lon)) * COS(RADIANS($lat)) * COS(RADIANS($lon)) +
COS(RADIANS(lat)) * SIN(RADIANS(lon)) * COS(RADIANS($lat)) * 
SIN(RADIANS($lon)) + SIN(RADIANS(lat)) * SIN(RADIANS($lat))) * 
3963.1 AS Distance
FROM Stores 
WHERE 1
HAVING Distance <= 50";

Answer 5

3956 * 2 * ASIN(SQRT( POWER(SIN(($latitude -( cp.latitude)) * pi()/180 / 2),2) + COS($latitude * pi()/180 ) * COS( abs( cp.latitude) *  pi()/180) * POWER(SIN(($longitude - cp.longitude) *  pi()/180 / 2), 2) )) as distance

返回里程还是公里？