查询以查找给定lat/lng内的位置

Question

所以我试图显示给定lat/lng范围内的位置列表.我对此没有任何问题:

一英里内的地方(地点列表......)

使用类似的东西

SELECT * FROM places WHERE lat < $latmax AND lat > $latmin AND lng < $lngmax AND lng > $lngmin

但后来我想列出两英里内的地方,但不在一英里内 - 也就是说,我不想重复第一次查询的结果.

这是我尝试过的一个版本:

$milesperdegree = 0.868976242 / 60.0 * 1.2;

// 1 mile -- this works
$degrees = $milesperdegree * 1;
$latmin = $lat - $degrees;
$latmax = $lat + $degrees;
$lngmin = $lng - $degrees;
$lngmax = $lng + $degrees;

$query = "SELECT * FROM places WHERE lat < $latmax AND lat > $latmin AND lng < $lngmax AND lng > $lngmin";

// 2 miles -- this doesn't work
$degrees_2 = $milesperdegree * 2;
$latmin_2 = $lat - $degrees_2;
$latmax_2 = $lat + $degrees_2;
$lngmin_2 = $lat - $degrees_2;
$lngmax_2 = $lat + $degrees_2;

$query = "SELECT * FROM places WHERE ";
$query .= "lat BETWEEN $latmax AND $latmax_2 AND lng BETWEEN $lngmax AND $lngmax_2 OR ";
$query .= "lat BETWEEN $latmin AND $latmin_2 AND lng BETWEEN $lngmin AND $lngmin_2 OR ";
$query .= "lat BETWEEN $latmax AND $latmax_2 AND lng BETWEEN $lngmin AND $lngmin_2 OR ";
$query .= "lat BETWEEN $latmin AND $latmin_2 AND lng BETWEEN $lngmax AND $lngmax_2";

那不是这样的.我猜这只是一些逻辑我周日下午无法解决,但我也可能做错了.任何输入都非常感谢.

Answer 1

我们或多或少地像下面的代码一样实现它（免责声明：我从文件中剪下了它，并删除了与当前问题无关的代码。我没有运行这个，但你应该能够理解这个想法。

$maxLat = $city->latitude + ($max_distance / 69); // 69 Miles/Degree
$minLat = $city->latitude - ($max_distance / 69);

$maxLon = $city->longitude + ($max_distance / (69.172 * cos($city->latitude * 0.0174533)));
$minLon = $city->longitude - ($max_distance / (69.172 * cos($city->latitude * 0.0174533)));

// Simplify terms to speed query
$originLatRadCos = cos($city->latitude * 0.0174533);
$originLatRadSin = sin($city->latitude * 0.0174533);
$originLonRad = $city->longitude * 0.0174533;

$city_distance_query = "
SELECT city_id, 
  3963 * acos(($originLatRadSin * sin( latitude * 0.0174533)) + 
  ($originLatRadCos * cos(latitude * 0.0174533) * cos((longitude * 0.0174533) -
  $originLonRad))) AS distanceFromOrigin
FROM cities
WHERE
 latitude < $maxLat AND latitude > $minLat AND longitude < $maxLon AND longitude > $minLon";

其余查询

SELECT cities.city_name, CityDistance.distanceFromOrigin,
FROM cities 
INNER JOIN ($city_distance_query) AS CityDistance ON CityDistance.city_id=cities.city_id
WHERE (distanceFromOrigin < $distance OR distanceFromOrigin IS NULL)