我有经纬度，我想从数据库中提取记录，它有最近的经纬度，如果这个距离比指定的长，就不检索它。
表结构:
id
latitude
longitude
place name
city
country
state
zip
sealevel
当前回答
这个问题最初的答案是好的，但是mysql的新版本(mysql 5.7.6上)支持地理查询，所以你现在可以使用内置的功能，而不是进行复杂的查询。
你现在可以这样做:
select *, ST_Distance_Sphere( point ('input_longitude', 'input_latitude'),
point(longitude, latitude)) * .000621371192
as `distance_in_miles`
from `TableName`
having `distance_in_miles` <= 'input_max_distance'
order by `distance_in_miles` asc
结果以米为单位返回。因此，如果你想计算KM，只需使用。001而不是。000621371192(这是英里)。
MySql文档在这里
其他回答
在极端情况下，这种方法会失败，但为了性能考虑，我跳过了三角函数，只是简单地计算对角线的平方。
根据文章Geo-Distance-Search-with-MySQL检查以下代码:
例如:找到半径10英里内离我目前位置最近的10家酒店。
#Please notice that (lat,lng) values mustn't be negatives to perform all calculations
set @my_lat=34.6087674878572;
set @my_lng=58.3783670308302;
set @dist=10; #10 miles radius
SELECT dest.id, dest.lat, dest.lng, 3956 * 2 * ASIN(SQRT(POWER(SIN((@my_lat -abs(dest.lat)) * pi()/180 / 2),2) + COS(@my_lat * pi()/180 ) * COS(abs(dest.lat) * pi()/180) * POWER(SIN((@my_lng - abs(dest.lng)) * pi()/180 / 2), 2))
) as distance
FROM hotel as dest
having distance < @dist
ORDER BY distance limit 10;
#Also notice that distance are expressed in terms of radius.
下面是我用PHP实现的完整解决方案。
该解决方案使用http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL中给出的Haversine公式。
值得注意的是，哈弗辛公式在极点处有弱点。这个答案展示了如何实现vincenty大圆距离公式来解决这个问题，但是我选择只使用Haversine，因为它足够适合我的目的。
我将纬度存储为DECIMAL(10,8)，经度存储为DECIMAL(11,8)。希望这能有所帮助!
showClosest.php
<?PHP
/**
* Use the Haversine Formula to display the 100 closest matches to $origLat, $origLon
* Only search the MySQL table $tableName for matches within a 10 mile ($dist) radius.
*/
include("./assets/db/db.php"); // Include database connection function
$db = new database(); // Initiate a new MySQL connection
$tableName = "db.table";
$origLat = 42.1365;
$origLon = -71.7559;
$dist = 10; // This is the maximum distance (in miles) away from $origLat, $origLon in which to search
$query = "SELECT name, latitude, longitude, 3956 * 2 *
ASIN(SQRT( POWER(SIN(($origLat - latitude)*pi()/180/2),2)
+COS($origLat*pi()/180 )*COS(latitude*pi()/180)
*POWER(SIN(($origLon-longitude)*pi()/180/2),2)))
as distance FROM $tableName WHERE
longitude between ($origLon-$dist/cos(radians($origLat))*69)
and ($origLon+$dist/cos(radians($origLat))*69)
and latitude between ($origLat-($dist/69))
and ($origLat+($dist/69))
having distance < $dist ORDER BY distance limit 100";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_assoc($result)) {
echo $row['name']." > ".$row['distance']."<BR>";
}
mysql_close($db);
?>
/资产/ db / db。php
<?PHP
/**
* Class to initiate a new MySQL connection based on $dbInfo settings found in dbSettings.php
*
* @example $db = new database(); // Initiate a new database connection
* @example mysql_close($db); // close the connection
*/
class database{
protected $databaseLink;
function __construct(){
include "dbSettings.php";
$this->database = $dbInfo['host'];
$this->mysql_user = $dbInfo['user'];
$this->mysql_pass = $dbInfo['pass'];
$this->openConnection();
return $this->get_link();
}
function openConnection(){
$this->databaseLink = mysql_connect($this->database, $this->mysql_user, $this->mysql_pass);
}
function get_link(){
return $this->databaseLink;
}
}
?>
资产/ db - dbSettings。php
<?php
$dbInfo = array(
'host' => "localhost",
'user' => "root",
'pass' => "password"
);
?>
根据上面“使用MySQL进行地理距离搜索”文章的建议，可以通过使用MySQL存储过程来提高性能。
我有一个约17,000个位置的数据库，查询执行时间为0.054秒。
试试这个，它显示最近的点提供的坐标(50公里内)。它工作得很完美:
SELECT m.name,
m.lat, m.lon,
p.distance_unit
* DEGREES(ACOS(COS(RADIANS(p.latpoint))
* COS(RADIANS(m.lat))
* COS(RADIANS(p.longpoint) - RADIANS(m.lon))
+ SIN(RADIANS(p.latpoint))
* SIN(RADIANS(m.lat)))) AS distance_in_km
FROM <table_name> AS m
JOIN (
SELECT <userLat> AS latpoint, <userLon> AS longpoint,
50.0 AS radius, 111.045 AS distance_unit
) AS p ON 1=1
WHERE m.lat
BETWEEN p.latpoint - (p.radius / p.distance_unit)
AND p.latpoint + (p.radius / p.distance_unit)
AND m.lon BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
ORDER BY distance_in_km
只需更改<table_name>。<userLat>和<userLon>
你可以在这里阅读更多关于这个解决方案:http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/
简单的一个;)
SELECT * FROM `WAYPOINTS` W ORDER BY
ABS(ABS(W.`LATITUDE`-53.63) +
ABS(W.`LONGITUDE`-9.9)) ASC LIMIT 30;
把坐标换成你需要的坐标。这些值必须存储为double类型。这是一个工作中的MySQL 5。x的例子。
干杯
