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
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
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
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)
你要找的是哈弗辛公式。看这里。
还有其他的,但这是最常被引用的。
如果您正在寻找更健壮的东西,则可能需要考虑数据库的GIS功能。它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(区域、国家、大陆)中。
MS SQL版本在这里:
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
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.
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();
}