谷歌地图 - 使用PHP无法运行的Haversine公式

Question

我对hasrsine公式的查询是：

$query = "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;";

使用此公式时，它返回一个空结果集。未找到结果。让我知道正确的解决方案。

另外，使用

select * FROM (
        SELECT *, ( 3959 * acos( cos( radians(37) ) * cos( radians( -33.898113 ) ) * cos( radians( 151.174469 ) - radians(-122) ) + sin( radians(37) ) * sin( radians( -33.898113 ) ) ) ) AS Distance FROM markers 
        ) as T 
    WHERE T.Distance < 25 
    ORDER BY T.Distance 
    LIMIT 0 , 20 

这一个，但返回相同的结果。

Answer 1

我有几个Stored Procedures找到最近的标记等似乎运作良好〜我尝试编辑一个为您的目的 - 猜测两个相关的列是lat和lng - 也许它会有所帮助

CREATE DEFINER=`root`@`localhost` PROCEDURE `spNearestMarkers`(
    IN `p_lat` DOUBLE,
    IN `p_lng` DOUBLE,
    IN `p_rad` INT
)
LANGUAGE SQL
NOT DETERMINISTIC
CONTAINS SQL
SQL SECURITY DEFINER
COMMENT ''
proc:begin
    declare latitude double default 0;
    declare longitude double default 0;
    declare radius float default 0;
    declare earth_radius integer default 0;
    declare longitude_1 float;
    declare latitude_1 float;
    declare longitude_2 float;
    declare latitude_2 float;


    set @latitude=p_lat;
    set @longitude=p_lng;
    set @radius=p_rad;


    set @earth_radius=3956;
    set @longitude_1 = @longitude - @radius/abs( cos( radians(@latitude) ) * 69 );
    set @longitude_2 = @longitude + @radius/abs( cos( radians(@latitude) ) * 69 );
    set @latitude_1 = @latitude - (@radius/69);
    set @latitude_2 = @latitude + (@radius/69);



    select distinct
        *,
        round( @earth_radius * 2 * asin( sqrt( power( sin( ( @latitude - m.`lat` ) * pi()/180 / 2 ), 2 ) + cos( @latitude * pi()/180 ) * cos( m.`lat` * pi()/180) *power( sin( ( @longitude - m.`lng`) * pi()/180 / 2 ), 2) ) ) , 2 ) as 'distance'
    from `markers` m
    where
        ( m.`lat` between @latitude_1 and @latitude_2 ) 
            and 
        ( m.`lng` between @longitude_1 and @longitude_2 )

    group by m.`name`
    having `distance` <= @radius
    order by `distance`;

    set @latitude=null;
    set @longitude=null;
    set @radius=null;
    set @longitude_1 = null;
    set @longitude_2 = null;
    set @latitude1 = null;
    set @latitude2 = null;

end

在php中调用该过程

$lat=-33.74788;
$lng=151.23526;
$radius=10;

$sql="call `spNearestMarkers`('$lat','$lng','$radius');`";

或者，在mysql（gui）中调用该过程

set @_lat=-33.74788;
set @_lng=151.23526;
set @_radius=10;

call `spNearestMarkers`(@_lat,@_lng,@_radius);

以上回报：

"id","name","address","lat","lng","distance",
"1","Heir Apparel","Crowea Pl, Frenchs Forest NSW 2086","-33.737885","151.235260","0.69",
"4","The Legacy","Charlotte Ln, Chatswood NSW 2067","-33.796669","151.183609","4.49",
"6","Trish & Tash","Lincoln St, Lane Cove West NSW 2066","-33.812222","151.143707","6.88",
"10","Moda Couture","Northcote Rd, Glebe NSW 2037","-33.873882","151.177460","9.31",

假设你的数据库有latitude和longitude列，你试图从PHP调用sql查询然后

/* the lat/lng of the location you are trying to find places nearby */
$lat=56.25;
$lng=-2.65;

$sql="SELECT id, ( 3959 * acos( cos( radians( $lat ) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians($lng) ) + sin( radians($lat) ) * sin( radians( latitude ) ) ) ) AS distance 
FROM markers 
HAVING distance < 25 
ORDER BY distance 
LIMIT 0,20;";