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
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
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]和[startlng]是开始测量距离的位置。
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 ;
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');
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
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!
<?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);
?>
<?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;
}
}
?>
<?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。
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;
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)。
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
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>
和<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/
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。
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.
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)
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();
}
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
+----+-----------------------+---------+--------------+---------------+
| 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");
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。
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。
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;
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。