Mysql: LBS实现查找附近的人 (两经纬度之间的距离)

来源:转载

1. 利用GeoHash封装成内置数据库函数的简易方案;


A:Mysql 内置函数方案,适合于已有业务,新增加LBS功能,增加经纬度字段方可,避免数据迁移


B:Mongodb 内置函数方案,适合中小型应用,快速实现LBS功能,性能优于A(推荐)


方案A: (MySQL Spatial)


1、先简历一张表:(MySQL 5.0 以上 仅支持 MyISAM 引擎)


CREATE TABLE address (address CHAR(80) NOT NULL,address_loc POINT NOT NULL,PRIMARY KEY(address));


空间索引:

ALTER TABLE address ADD SPATIAL INDEX(address_loc);

插入数据:(注:此处Point(纬度,经度) 标准写法)


INSERT INTO address VALUES('Foobar street 12', GeomFromText('POINT(30.620076 104.067221)'));INSERT INTO address VALUES('Foobar street 13', GeomFromText('POINT(31.720076 105.167221)'));


查询: 查找(30.620076,104.067221)附近 10 公里

SELECT*
FROMaddress
WHERE MBRContains
(
LineString
(
Point
(
30.620076 + 10 / ( 111.1 / COS(RADIANS(104.067221))),
104.067221 + 10 / 111.1
),
Point
(
30.620076 - 10 / ( 111.1 / COS(RADIANS(104.067221))),
104.067221 - 10 / 111.1
)
),
address_loc
)



方案B:


1、先建立一张简单的表user,两条数据


{
"_id": ObjectId("518b1f1a83ba88ca60000001"),
"account": "[email protected]",
"gps": [
104.067221,
30.620076
]
}{
"_id": ObjectId("518b1dae83ba88d660000000"),
"account": "[email protected]",
"gps": [
104.07958,
30.653936
]
}

其中,gps为二维数组,分别为经度,纬度


(注:此处必须按照(经度,纬度)顺序存储。我们平时表示经纬度,都是(纬度,精度),此处这种方式有木有很亲民)


2、使用之前,先建立二维索引


//建立索引 最大范围在经度-180~180


db.user.ensureIndex({"gps":"2d"},{"min":-180,"max":180})

//删除索引


db.user.dropIndex({"gps":"2d"})

3、Mongodb有两中方式可以查找附近的XXX;其中方案2)会返回距离(推荐)


1)标准查询,为地球经纬度查询内置;参数一为查询条件利用$near查找附近,参数二$maxDistance为经纬弧度(1° latitude = 111.12 kilometers)即 1/111.12,表示查找附近一公里。


db.user.find({ gps :{ $near : [104.065847, 30.657554] , $maxDistance : 1/111.12} })

2)执行命名方式,模拟成一个圆球;参数一指定geoNear方式和表名;参数二坐标,参数三是否为球形,参数四弧度(弧度=弧长/半径 一千米的弧度1000/6378000),参数五指定球形半径(地球半径)


db.runCommand({geoNear:'user', near:[104.065847, 30.657554],
spherical:true, maxDistance:1000/6378000, distanceMultiplier:6378000});

转自:http://stackoverflow.com/a/1006668/4484798


2 利用谷歌方案


The SQL statement that will find the closest 20 locations that are within a radius of 30 miles to the 78.3232, 65.3234 coordinate. It calculates the distance based on the latitude/longitude of that row and the target latitude/longitude, and then asks for only rows where the distance value is less than 30 miles, orders the whole query by distance, and limits it to 20 results. To search by kilometers instead of miles, replace 3959 with 6371.


3959是地球半径的英里,6371是地球半径的千米:http://baike.baidu.com/view/758812.htm



SELECT
id, (
3959 * acos (
cos ( radians(78.3232) )
* cos( radians( lat ) )
* cos( radians( lng ) - radians(65.3234) )
+ sin ( radians(78.3232) )
* sin( radians( lat ) )
)
) AS distance
FROM markers
HAVING distance < 30
ORDER BY distance
LIMIT 0 , 20;


This is using the Google Maps API v3 with a MySQL backend which your already have.

https://developers.google.com/maps/articles/phpsqlsearch_v3#findnearsql


转自:http://gis.stackexchange.com/a/31629

3 其他


Anyways, here’s the PHP formula for calculating the distance between two points (along with Mile vs. Kilometer conversion) rounded to two decimal places:


function getDistanceBetweenPointsNew($latitude1, $longitude1, $latitude2, $longitude2,
$unit = 'Mi') {
$theta = $longitude1 - $longitude2;
$distance = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) +
(cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * cos(deg2rad($theta)));
$distance = acos($distance);
$distance = rad2deg($distance);
$distance = $distance * 60 * 1.1515; switch($unit) {
case 'Mi': break; case 'Km' : $distance = $distance * 1.609344;
}
return (round($distance,2));
}

It’s also possible to use MySQL to do a calculation to find all records within a specific distance. In this example, I’m going to query MyTable to find all the records that are less than or equal to variable $distance (in Miles) to my location at $latitude and $longitude:


$qry = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) * sin((`Latitude`*pi()/180))+cos((".$latitude."*pi()/180)) *
cos((`Latitude`*pi()/180)) *
cos(((".$longitude."- `Longitude`)*pi()/180))))*
180/pi())*60*1.1515) as distance
FROM `MyTable`
WHERE distance >= ".$distance."

For Kilometers:


$qry = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) * sin((`Latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((".$longitude."- `Longitude`)*pi()/180))))*180/pi())*60*1.1515*1.609344) as distance
FROM `MyTable`
WHERE distance >= ".$distance."

转自:


http://stackoverflow.com/a/8599305/4484798 &https://www.marketingtechblog.com/calculate-distance/


更对参考:


Google Maps V3: 导航到指定地址 Draw (Plot) route between User’s current location and Specified location


Google Maps V3: 通过经纬度获取地址信息 Get address from Latitude and Longitude


jQuery地图插件-jqvmap


谷歌地图查询两地开车路线 Google Maps API V3: DirectionsService (Driving Directions) example


谷歌地图添加点击事件 Google Maps API V3: Add click event listener to all (multiple) markers


Google Maps API v3: Remove Markers 移除标记


点击谷歌地图后获取经纬度 Get Latitude and Longitude (Location Coordinates) using Google Maps OnClick event


GMaps.js:让你快速集成 Google Maps 服务的 jQuery 插件


Google Maps API V3: 通过邮编获取经纬度 Get Location (Latitude and Longitude) from Zip Code (Pin Code) using JavaScript


Google Maps API 以某一经纬度为中心,以某一长度位半径画圆 Draw the radius of a circle

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