MySQL大圆距离（Haversine公式）

我有一个工作的PHP脚本获取经度和纬度值，然后将它们input到MySQL查询。我想完全使用MySQL。这是我目前的PHP代码：

if ($distance != "Any" && $customer_zip != "") { //get the great circle distance //get the origin zip code info $zip_sql = "SELECT * FROM zip_code WHERE zip_code = '$customer_zip'"; $result = mysql_query($zip_sql); $row = mysql_fetch_array($result); $origin_lat = $row['lat']; $origin_lon = $row['lon']; //get the range $lat_range = $distance/69.172; $lon_range = abs($distance/(cos($details[0]) * 69.172)); $min_lat = number_format($origin_lat - $lat_range, "4", ".", ""); $max_lat = number_format($origin_lat + $lat_range, "4", ".", ""); $min_lon = number_format($origin_lon - $lon_range, "4", ".", ""); $max_lon = number_format($origin_lon + $lon_range, "4", ".", ""); $sql .= "lat BETWEEN '$min_lat' AND '$max_lat' AND lon BETWEEN '$min_lon' AND '$max_lon' AND "; }

有谁知道如何使这完全MySQL？我已经浏览了一下互联网，但大部分的文献都相当混乱。

来自Google Code FAQ – 使用PHP，MySQL和Google地图创build商店定位器：

下面是SQL语句，它将查找距离37，-122坐标25英里范围内最近的20个位置。它根据该行的纬度/经度和目标纬度/经度计算距离，然后仅要求距离值小于25的行，按距离sorting整个查询，并将其限制为20个结果。以公里而不是英里search，用6371代替3959。

 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;

$greatCircleDistance = acos( cos($latitude0) * cos($latitude1) * cos($longitude0 - $longitude1) + sin($latitude0) * sin($latitude1));

以经度和纬度为单位。

所以

 SELECT acos( cos(radians( $latitude0 )) * cos(radians( $latitude1 )) * cos(radians( $longitude0 ) - radians( $longitude1 )) + sin(radians( $latitude0 )) * sin(radians( $latitude1 )) ) AS greatCircleDistance FROM yourTable;

是你的SQL查询

以公里或英里得出结果，将结果乘以地球的平均半径（ 3959英里， 6371公里或3440海里）

你在你的例子中计算的东西是一个边界框。如果您将坐标数据放入启用空间的MySQL列中，则可以使用MySQL的内置function来查询数据。

 SELECT id FROM spatialEnabledTable WHERE MBRWithin(ogc_point, GeomFromText('Polygon((0 0,0 3,3 3,3 0,0 0))'))

如果将辅助字段添加到坐标表中，则可以提高查询的响应时间。

喜欢这个：

 CREATE TABLE `Coordinates` ( `id` INT(10) UNSIGNED NOT NULL COMMENT 'id for the object', `type` TINYINT(4) UNSIGNED NOT NULL DEFAULT '0' COMMENT 'type', `sin_lat` FLOAT NOT NULL COMMENT 'sin(lat) in radians', `cos_cos` FLOAT NOT NULL COMMENT 'cos(lat)*cos(lon) in radians', `cos_sin` FLOAT NOT NULL COMMENT 'cos(lat)*sin(lon) in radians', `lat` FLOAT NOT NULL COMMENT 'latitude in degrees', `lon` FLOAT NOT NULL COMMENT 'longitude in degrees', INDEX `lat_lon_idx` (`lat`, `lon`) )

如果您使用的是TokuDB，如果您在任一谓词上添加聚簇索引，则会获得更好的性能，例如：

 alter table Coordinates add clustering index c_lat(lat); alter table Coordinates add clustering index c_lon(lon);

您需要以度为单位的基本纬度和经度，以弧度为单位的弧度（cos）（lat）* cos（lon）和弧度的cos（lat）* sin（lon）。然后你创build一个mysql函数，如下所示：

 CREATE FUNCTION `geodistance`(`sin_lat1` FLOAT, `cos_cos1` FLOAT, `cos_sin1` FLOAT, `sin_lat2` FLOAT, `cos_cos2` FLOAT, `cos_sin2` FLOAT) RETURNS float LANGUAGE SQL DETERMINISTIC CONTAINS SQL SQL SECURITY INVOKER BEGIN RETURN acos(sin_lat1*sin_lat2 + cos_cos1*cos_cos2 + cos_sin1*cos_sin2); END

这给你的距离。

不要忘记在纬度/经度上添加一个索引，这样边界框可以帮助search，而不是放缓速度（索引已经添加到上面的CREATE TABLE查询中）。

 INDEX `lat_lon_idx` (`lat`, `lon`)

给定一个只有纬度/经度坐标的旧桌子，你可以设置一个脚本来更新它：（php使用meekrodb）

 $users = DB::query('SELECT id,lat,lon FROM Old_Coordinates'); foreach ($users as $user) { $lat_rad = deg2rad($user['lat']); $lon_rad = deg2rad($user['lon']); DB::replace('Coordinates', array( 'object_id' => $user['id'], 'object_type' => 0, 'sin_lat' => sin($lat_rad), 'cos_cos' => cos($lat_rad)*cos($lon_rad), 'cos_sin' => cos($lat_rad)*sin($lon_rad), 'lat' => $user['lat'], 'lon' => $user['lon'] )); }

然后你优化实际的查询，只做真正需要的距离计算，例如通过从内部和外部界定圆（好，椭圆）。为此，您需要预先计算查询本身的几个指标：

 // assuming the search center coordinates are $lat and $lon in degrees // and radius in km is given in $distance $lat_rad = deg2rad($lat); $lon_rad = deg2rad($lon); $R = 6371; // earth's radius, km $distance_rad = $distance/$R; $distance_rad_plus = $distance_rad * 1.06; // ovality error for outer bounding box $dist_deg_lat = rad2deg($distance_rad_plus); //outer bounding box $dist_deg_lon = rad2deg($distance_rad_plus/cos(deg2rad($lat))); $dist_deg_lat_small = rad2deg($distance_rad/sqrt(2)); //inner bounding box $dist_deg_lon_small = rad2deg($distance_rad/cos(deg2rad($lat))/sqrt(2));

鉴于这些准备，查询就像这样（PHP）：

 $neighbors = DB::query("SELECT id, type, lat, lon, geodistance(sin_lat,cos_cos,cos_sin,%d,%d,%d) as distance FROM Coordinates WHERE lat BETWEEN %d AND %d AND lon BETWEEN %d AND %d HAVING (lat BETWEEN %d AND %d AND lon BETWEEN %d AND %d) OR distance <= %d", // center radian values: sin_lat, cos_cos, cos_sin sin($lat_rad),cos($lat_rad)*cos($lon_rad),cos($lat_rad)*sin($lon_rad), // min_lat, max_lat, min_lon, max_lon for the outside box $lat-$dist_deg_lat,$lat+$dist_deg_lat, $lon-$dist_deg_lon,$lon+$dist_deg_lon, // min_lat, max_lat, min_lon, max_lon for the inside box $lat-$dist_deg_lat_small,$lat+$dist_deg_lat_small, $lon-$dist_deg_lon_small,$lon+$dist_deg_lon_small, // distance in radians $distance_rad);

上面的查询的EXPLAIN可能会说，它没有使用索引，除非有足够的结果来触发这样的。索引将在坐标表中有足够的数据时使用。您可以将FORCE INDEX（lat_lon_idx）添加到SELECT以使其与索引无关，因此您可以使用EXPLAINvalidation它是否正常工作。

使用上面的代码示例，您应该有一个工作和可扩展的对象search的实现距离与最小的错误。

我必须详细解决这个问题，所以我会分享一下我的结果。这使用一个带有latitude和longitude表的zip表。它不依赖Google地图; 相反，你可以适应任何包含lat / long的表格。

 SELECT zip, primary_city, latitude, longitude, distance_in_mi FROM ( SELECT zip, primary_city, latitude, longitude,r, (3963.17 * ACOS(COS(RADIANS(latpoint)) * COS(RADIANS(latitude)) * COS(RADIANS(longpoint) - RADIANS(longitude)) + SIN(RADIANS(latpoint)) * SIN(RADIANS(latitude)))) AS distance_in_mi FROM zip JOIN ( SELECT 42.81 AS latpoint, -70.81 AS longpoint, 50.0 AS r ) AS p WHERE latitude BETWEEN latpoint - (r / 69) AND latpoint + (r / 69) AND longitude BETWEEN longpoint - (r / (69 * COS(RADIANS(latpoint)))) AND longpoint + (r / (69 * COS(RADIANS(latpoint)))) ) d WHERE distance_in_mi <= r ORDER BY distance_in_mi LIMIT 30

查看该查询中间的这一行：

  SELECT 42.81 AS latpoint, -70.81 AS longpoint, 50.0 AS r

这将在纬度/长点42.81 / -70.81的50.0英里范围内searchzip表中30个最近的条目。当你把它构build到一个应用程序中，这就是你放置自己的点和search半径的地方。

如果要以千米而不是英里工作， 111.045 69更改为111.045 ，并将3963.17更改为6378.10 。

这是一个详细的写作。我希望它能帮助别人。 http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

我已经写了一个可以计算相同的程序，但是你必须在相应的表中input纬度和经度。

 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 ;

我不能评论上述答案，但要小心@Pavel Chuchuva的答案。如果两个坐标相同，那么该公式不会返回结果。在这种情况下，距离是空的，所以这行不会像原来一样返回。

我不是一个MySQL专家，但这似乎是为我工作：

 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 OR distance IS NULL ORDER BY distance LIMIT 0 , 20;

  SELECT *, ( 6371 * acos(cos(radians(search_lat)) * cos(radians(lat) ) * cos(radians(lng) - radians(search_lng)) + sin(radians(search_lat)) * (radians(lat))) ) AS distance FROM table WHERE lat != search_lat AND lng != search_lng AND distance < 25 ORDER BY distance FETCH 10 ONLY

距离25公里

我以为我的JavaScript实现将是一个很好的参考：

 /* * Check to see if the second coord is within the precision ( meters ) * of the first coord and return accordingly */ function checkWithinBound(coord_one, coord_two, precision) { var distance = 3959000 * Math.acos( Math.cos( degree_to_radian( coord_two.lat ) ) * Math.cos( degree_to_radian( coord_one.lat ) ) * Math.cos( degree_to_radian( coord_one.lng ) - degree_to_radian( coord_two.lng ) ) + Math.sin( degree_to_radian( coord_two.lat ) ) * Math.sin( degree_to_radian( coord_one.lat ) ) ); return distance <= precision; } /** * Get radian from given degree */ function degree_to_radian(degree) { return degree * (Math.PI / 180); }