2013-02-12 51 views
0

我正在尝试使用半径搜索来获取地理点的公共数据库。 我已经找到了关于这个主题的几个很好的教程,但我最终失败了。haversine公式php/mysql

主要教程在这里:http://janmatuschek.de/LatitudeLongitudeBoundingCoordinates

的基本公式,在SQL查询的形式,是

SELECT * FROM Places 
WHERE (Lat => 1.2393 AND Lat <= 1.5532) AND (Lon >= -1.8184 AND Lon <= 0.4221) 
AND acos(sin(1.3963) * sin(Lat) + cos(1.3963) * cos(Lat) * cos(Lon - (-0.6981))) 
    <= 0.1570; 

我实现了这一个简单的PHP测试页面是这样的:

$R = 6371; // radius of Earth in KM 

$lat = '46.98025235521883'; // lat of center point 
$lon = '-110.390625'; // longitude of center point 
$distance = 1000; // radius in KM of the circle drawn 
$rad = $distance/$R; // angular radius for query 
$query = ''; 

// rough cut to exclude results that aren't close 
$max_lat = $lat + rad2deg($rad/$R); 
$min_lat = $lat - rad2deg($rad/$R); 
$max_lon = $lon + rad2deg($rad/$R/cos(deg2rad($lat))); 
$min_lon = $lon - rad2deg($rad/$R/cos(deg2rad($lat))); 
// this part works just fine! 
$query .= '(latitude > ' . $min_lat . ' AND latitude < ' . $max_lat . ')'; 
$query .= ' AND (longitude > ' . $min_lon . ' AND longitude < ' . $max_lon . ')'; 
// refining query -- this part returns no results 
$query .= ' AND acos(sin('.$lat.') * sin(latitude) + cos('.$lat.') * cos(latitude) * 
    cos(longitude - ('.$lon.'))) <= '.$rad; 

我失去了一些东西在这里?我认为我完全遵循方法,但是我无法获得“微调”查询来返回任何结果。

+0

你可以把一些预期的输入/输出鼓捣? – ehime 2013-02-12 18:52:26

回答

1

不知道,但:

$R = 6371; // radius of Earth in KM 

$lat = '46.98025235521883'; // lat of center point 
$lon = '-110.390625'; // longitude of center point 
$distance = 1000; // radius in KM of the circle drawn 
$rad = $distance/$R; // angular radius for query 
$query = ''; 

// rough cut to exclude results that aren't close 
$radR = rad2deg($rad/$R); 
$max_lat = $lat + radR; 
$min_lat = $lat - radR; 
$radR = rad2deg($rad/$R/cos(deg2rad($lat))); 
$max_lon = $lon + radR; 
$min_lon = $lon - radR; 
// this part works just fine! 
$query .= '(latitude > ' . $min_lat . ' AND latitude < ' . $max_lat . ')'; 
$query .= ' AND (longitude > ' . $min_lon . ' AND longitude < ' . $max_lon . ')'; 
// refining query -- this part returns no results 
$query .= ' AND acos(sin('.deg2rad($lat).') * sin(radians(latitude)) + cos('.deg2rad($lat).') * cos(radians(latitude)) * 
    cos(radians(longitude) - ('.deg2rad($lon).'))) <= '.$rad; 
+0

感谢您的帮助!事实上,在二次搜索中经纬度必须以弧度为单位。 – user101289 2013-02-12 21:34:53

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