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我正在尝试使用半径搜索获得一个通用的地理点数据库。我找到了几个关于这个主题的好教程，但最后我失败了。

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

SQL查询形式的基本公式是

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;

我已经在一个简单的 PHP 测试页面中实现了这一点，如下所示：

$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;

我在这里错过了什么吗？我认为我完全遵循该方法，但我无法获得“微调”查询来返回任何结果。

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1 回答 1

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不确定，但是：

$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;
于 2013-02-12T19:00:57.373 回答