200

我有纬度和经度,我想从数据库中提取记录,该记录具有距离最近的纬度和经度,如果该距离大于指定的距离,则不要检索它。

表结构:

id
latitude
longitude
place name
city
country
state
zip
sealevel
4

18 回答 18

247
SELECT latitude, longitude, SQRT(
    POW(69.1 * (latitude - [startlat]), 2) +
    POW(69.1 * ([startlng] - longitude) * COS(latitude / 57.3), 2)) AS distance
FROM TableName HAVING distance < 25 ORDER BY distance;

其中[starlat][startlng]是开始测量距离的位置。

于 2011-04-05T07:58:22.633 回答
93

谷歌的解决方案:

创建表

创建 MySQL 表时,需要特别注意 lat 和 lng 属性。使用 Google 地图当前的缩放功能,您应该只需要小数点后 6 位的精度。为了将表所需的存储空间保持在最低限度,您可以指定 lat 和 lng 属性是大小为 (10,6) 的浮点数。这将使字段存储小数点后的 6 位数字,再加上小数点前的 4 位数字,例如 -123.456789 度。您的表还应该有一个 id 属性作为主键。

CREATE TABLE `markers` (
  `id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
  `name` VARCHAR( 60 ) NOT NULL ,
  `address` VARCHAR( 80 ) NOT NULL ,
  `lat` FLOAT( 10, 6 ) NOT NULL ,
  `lng` FLOAT( 10, 6 ) NOT NULL
) ENGINE = MYISAM ;

填充表格

创建表后,是时候用数据填充它了。下面提供的样本数据是针对分散在美国各地的大约 180 家披萨店。在 phpMyAdmin 中,您可以使用 IMPORT 选项卡来导入各种文件格式,包括 CSV(逗号分隔值)。Microsoft Excel 和 Google 电子表格都导出为 CSV 格式,因此您可以通过导出/导入 CSV 文件轻松地将数据从电子表格传输到 MySQL 表。

INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Frankie Johnnie & Luigo Too','939 W El Camino Real, Mountain View, CA','37.386339','-122.085823');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Amici\'s East Coast Pizzeria','790 Castro St, Mountain View, CA','37.38714','-122.083235');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Kapp\'s Pizza Bar & Grill','191 Castro St, Mountain View, CA','37.393885','-122.078916');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Round Table Pizza: Mountain View','570 N Shoreline Blvd, Mountain View, CA','37.402653','-122.079354');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Tony & Alba\'s Pizza & Pasta','619 Escuela Ave, Mountain View, CA','37.394011','-122.095528');
INSERT INTO `markers` (`name`, `address`, `lat`, `lng`) VALUES ('Oregano\'s Wood-Fired Pizza','4546 El Camino Real, Los Altos, CA','37.401724','-122.114646');

使用 MySQL 查找位置

要在标记表中查找位于给定纬度/经度的某个半径距离内的位置,您可以使用基于 Haversine 公式的 SELECT 语句。Haversine 公式通常用于计算球体上两对坐标之间的大圆距离。Wikipedia 给出了深入的数学解释,并且在 Movable Type 的网站上对与编程相关的公式进行了很好的讨论。

下面的 SQL 语句将找到距离 37, -122 坐标 25 英里半径范围内最近的 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,然后仅询问距离值小于 25 的行,按距离对整个查询进行排序,并将其限制为 20 个结果。要按公里而不是英里搜索,请将 3959 替换为 6371。

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 < 28 
ORDER BY distance LIMIT 0, 20;

这个是找28英里以内的经纬度。

另一种方法是在 28 到 29 英里的距离内找到它们:

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 < 29 and distance > 28 
ORDER BY distance LIMIT 0, 20;

https://developers.google.com/maps/articles/phpsqlsearch_v3#creating-the-map

于 2016-03-26T10:30:26.543 回答
31

这是我用 PHP 实现的完整解决方案。

此解决方案使用http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL中提供的 Haversine 公式。

应该注意的是,Haversine 公式在极点周围存在弱点。这个答案显示了如何实现vincenty Great Circle Distance 公式来解决这个问题,但是我选择只使用 Haversine,因为它对我的目的来说已经足够好了。

我将纬度存储为 DECIMAL(10,8),将经度存储为 DECIMAL(11,8)。希望这会有所帮助!

显示最近的.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);
?>

./assets/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;
    }
}
?>

./assets/db/dbSettings.php

<?php
$dbInfo = array(
    'host'      => "localhost",
    'user'      => "root",
    'pass'      => "password"
);
?>

正如上面发布的“Geo-Distance-Search-with-MySQL”文章所建议的那样,可以通过使用 MySQL 存储过程来提高性能。

我有一个约 17,000 个地点的数据库,查询执行时间为 0.054 秒。

于 2013-12-07T03:00:51.497 回答
26

以防万一你像我一样懒惰,这是一个从这个和其他关于 SO 的答案合并的解决方案。

set @orig_lat=37.46; 
set @orig_long=-122.25; 
set @bounding_distance=1;

SELECT
*
,((ACOS(SIN(@orig_lat * PI() / 180) * SIN(`lat` * PI() / 180) + COS(@orig_lat * PI() / 180) * COS(`lat` * PI() / 180) * COS((@orig_long - `long`) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS `distance` 
FROM `cities` 
WHERE
(
  `lat` BETWEEN (@orig_lat - @bounding_distance) AND (@orig_lat + @bounding_distance)
  AND `long` BETWEEN (@orig_long - @bounding_distance) AND (@orig_long + @bounding_distance)
)
ORDER BY `distance` ASC
limit 25;
于 2011-05-09T06:25:33.753 回答
24

该问题的原始答案很好,但较新版本的 mysql(MySQL 5.7.6 上)支持地理查询,因此您现在可以使用内置功能而不是执行复杂查询。

您现在可以执行以下操作:

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meters. 因此,如果您想KM简单地使用.001而不是.000621371192(以英里为单位)。

MySql 文档在这里

于 2018-10-02T19:27:51.670 回答
13

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;

只需将坐标替换为您需要的坐标即可。这些值必须存储为双精度值。这是一个有效的 MySQL 5.x 示例。

干杯

于 2012-10-14T12:52:36.513 回答
6

试试这个,它会显示离所提供坐标最近的点(50 公里内)。它完美地工作:

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

只是改变<table_name><userLat><userLon>

您可以在此处阅读有关此解决方案的更多信息:http ://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

于 2015-05-07T16:17:11.047 回答
5

您正在寻找诸如hasrsine formula之类的东西。也见这里

还有其他的,但这是最常被引用的。

如果您正在寻找更强大的功能,您可能需要查看您的数据库 GIS 功能。他们能够做一些很酷的事情,比如告诉你一个点(城市)是否出现在给定的多边形(地区、国家、大陆)内。

于 2010-02-10T03:28:21.477 回答
4

根据文章Geo-Distance-Search-with-MySQL检查此代码:

示例:查找距离我当前位置 10 英里半径范围内最近的 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.
于 2014-10-16T14:49:59.043 回答
4

查找离我最近的用户:

距离(米)

基于文森蒂公式

我有用户表:

+----+-----------------------+---------+--------------+---------------+
| id | email                 | name    | location_lat | location_long |
+----+-----------------------+---------+--------------+---------------+
| 13 | xxxxxx@xxxxxxxxxx.com | Isaac   | 17.2675625   | -97.6802361   |
| 14 | xxxx@xxxxxxx.com.mx   | Monse   | 19.392702    | -99.172596    |
+----+-----------------------+---------+--------------+---------------+

sql:

-- my location:  lat   19.391124   -99.165660
SELECT 
(ATAN(
    SQRT(
        POW(COS(RADIANS(users.location_lat)) * SIN(RADIANS(users.location_long) - RADIANS(-99.165660)), 2) +
        POW(COS(RADIANS(19.391124)) * SIN(RADIANS(users.location_lat)) - 
       SIN(RADIANS(19.391124)) * cos(RADIANS(users.location_lat)) * cos(RADIANS(users.location_long) - RADIANS(-99.165660)), 2)
    )
    ,
    SIN(RADIANS(19.391124)) * 
    SIN(RADIANS(users.location_lat)) + 
    COS(RADIANS(19.391124)) * 
    COS(RADIANS(users.location_lat)) * 
    COS(RADIANS(users.location_long) - RADIANS(-99.165660))
 ) * 6371000) as distance,
users.id
FROM users
ORDER BY distance ASC

地球半径:6371000(米)

于 2016-08-23T23:22:15.833 回答
3
simpledb.execSQL("CREATE TABLE IF NOT EXISTS " + tablename + "(id INTEGER PRIMARY KEY   AUTOINCREMENT,lat double,lng double,address varchar)");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2891001','70.780154','craftbox');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2901396','70.7782428','kotecha');");//22.2904718 //70.7783906
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2863155','70.772108','kkv Hall');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.275993','70.778076','nana mava');");
            simpledb.execSQL("insert into '" + tablename + "'(lat,lng,address)values('22.2667148','70.7609386','Govani boys hostal');");


    double curentlat=22.2667258;  //22.2677258
    double curentlong=70.76096826;//70.76096826

    double curentlat1=curentlat+0.0010000;
    double curentlat2=curentlat-0.0010000;

    double curentlong1=curentlong+0.0010000;
    double curentlong2=curentlong-0.0010000;

    try{

        Cursor c=simpledb.rawQuery("select * from '"+tablename+"' where (lat BETWEEN '"+curentlat2+"' and '"+curentlat1+"') or (lng BETWEEN         '"+curentlong2+"' and '"+curentlong1+"')",null);

        Log.d("SQL ", c.toString());
        if(c.getCount()>0)
        {
            while (c.moveToNext())
            {
                double d=c.getDouble(1);
                double d1=c.getDouble(2);

            }
        }
    }
    catch (Exception e)
    {
        e.printStackTrace();
    }
于 2016-03-03T10:23:58.863 回答
2

听起来您想在距离上进行一些最近的邻居搜索。据我所知,SQL 不支持这样的东西,您需要使用替代数据结构,例如R-treekd-tree

于 2010-02-10T03:29:22.800 回答
2

MS SQL 版在这里:

        DECLARE @SLAT AS FLOAT
        DECLARE @SLON AS FLOAT

        SET @SLAT = 38.150785
        SET @SLON = 27.360249

        SELECT TOP 10 [LATITUDE], [LONGITUDE], SQRT(
            POWER(69.1 * ([LATITUDE] - @SLAT), 2) +
            POWER(69.1 * (@SLON - [LONGITUDE]) * COS([LATITUDE] / 57.3), 2)) AS distance
        FROM [TABLE] ORDER BY 3
于 2018-06-13T05:54:06.643 回答
1
 +----+-----------------------+---------+--------------+---------------+
| id | email                 | name    | location_lat | location_long |
+----+-----------------------+---------+--------------+---------------+
| 7  | test@gmail.com        | rembo   | 23.0249256   |  72.5269697   |
| 25 | test1@gmail.com.      | Rajnis  | 23.0233221    | 72.5342112   |
+----+-----------------------+---------+--------------+---------------+

$lat = 23.02350629;

$long = 72.53230239;

DB:: SELECT (" SELECT * FROM ( SELECT , ( ( ( acos( sin(( ". $ lat ." * pi() / 180)) * sin(( lat* pi() / 180))) + cos(( ".$ lat ." pi() / 180 )) * cos(( lat* pi() / 180)) * cos((( ".$ long ." - LONG) * pi() / 180))) ) * 180 / pi() ) * 60 * 1.1515 * 1.609344 ) 作为距离 FROM users ) 用户 WHERE 距离 <= 2");

于 2020-10-09T12:49:12.757 回答
0

听起来您应该只使用 PostGIS、SpatialLite、SQLServer2008 或 Oracle Spatial。他们都可以使用空间 SQL 为您回答这个问题。

于 2010-02-10T04:24:45.673 回答
0

在极端情况下,这种方法会失败,但为了提高性能,我跳过了三角函数并简单地计算了对角线平方。

于 2017-07-03T13:29:56.333 回答
-1

Mysql查询具有距离限制和条件的搜索坐标

 SELECT id, ( 3959 * acos( cos( radians('28.5850154') ) * cos( radians(latitude) ) * cos( radians( longitude ) - radians('77.07207489999999') ) + sin( radians('28.5850154') ) * sin( radians( latitude ) ) ) ) AS distance FROM `vendors` HAVING distance < 5;
于 2021-01-01T06:13:18.967 回答
-20

这个问题一点也不难,但是如果你需要优化它会变得更加复杂。

我的意思是,您的数据库中有 100 个位置还是 1 亿个位置?它有很大的不同。

如果位置的数量很少,只需执行以下操作即可将它们从 SQL 中取出并放入代码中 ->

Select * from Location

将它们输入代码后,使用 Haversine 公式计算每个纬度/经度与原始数据之间的距离并对其进行排序。

于 2010-02-10T04:11:40.940 回答