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
查找离我最近的用户:
距离(米)
根据文森特提的公式
i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(单位:米)
其他回答
查找离我最近的用户:
距离(米)
根据文森特提的公式
i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(单位:米)
这个问题一点也不难，但是如果你需要优化它，它就会变得更加复杂。
我的意思是，你的数据库中有100个地点还是1亿个?这有很大的不同。
如果位置的数量很小，只需执行->，就可以将它们从SQL中取出并放入代码中
Select * from Location
一旦你把它们转换成代码，用哈弗辛公式计算出每一个纬度/长度与原始值之间的距离，然后排序。
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。
+----+-----------------------+---------+--------------+---------------+
| 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:: 选择 (“ 选择 * 从 （ 选择 ， （ (acos(sin)。决。”* pi(美元)/ 180)* sin (lat * pi(+) - 180)因为(”。拉丁语“圆周率(美元)/ 180)* cos (lat * pi () / 180) * cos(("。龙。”- long) * pi(美元)/ 180))* 180 / pi() * 60 * 1515 1。1 . 609344 ) 距离美国 从 \用户 ) \用户 在哪里 距离<= 2");
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.
