所以我有一个包含多行GPS坐标的数据库。我知道如何计算数据库中任意一个给定lat / lng的距离,但我想要做的基本上是查看一组行的坐标并获得最远的两行。我喜欢它,如果我可以在SQL中这样做,但如果我必须在我的应用程序代码中这样做。以下是我计算两点之间距离的方法:
ROUND(( 3960 * acos( cos( radians( :lat ) ) *
cos( radians( p.latitude ) ) * cos( radians( p.longitude ) - radians( :lng ) ) +
sin( radians( :lat ) ) * sin( radians( p.latitude ) ) ) ),1) AS distance
我们要做的是,查看特定用户的GPS数据,并确保他们不会在全国范围内疯狂地移动。用户的所有坐标应彼此相距几英里。如果坐标遍布全国,我们系统中存在恶意活动的标志。因此,我希望能够为spcicic用户快速浏览数据并知道他们的最大距离是什么。
我考虑过分别在lat和lng上运行Max / Min,并为可接受的内容设置内部阈值。也许这更容易,但如果我在第一部分中提出的问题是可能的,那将是最好的。
答案 0 :(得分:1)
如果您有SQL Server 2008或更高版本,则可以使用GEOGRAPHY计算距离,例如:
DECLARE @lat1 DECIMAL(19,6) = 44.968046;
DECLARE @lon1 DECIMAL(19,6) = -94.420307;
DECLARE @lat2 DECIMAL(19,6) = 44.33328;
DECLARE @lon2 DECIMAL(19,6) = -89.132008;
SELECT GEOGRAPHY::Point(@lat1, @lon1, 4326).STDistance(GEOGRAPHY::Point(@lat2, @lon2, 4326));
这使问题变得非常简单?
对于用户的一组拉特/长,您需要计算每组之间的距离,然后返回最高距离。把这一切放在一起,你可能会做这样的事情:
DECLARE @UserGPS TABLE (
UserId INT, --the user
GPSId INT, --the incrementing unique id associated with this GPS reading (could link to a table with more details, e.g. time, date)
Lat DECIMAL(19,6), --lattitude
Lon DECIMAL(19,6)); --longitude
INSERT INTO @UserGPS SELECT 1, 1, 44.968046, -94.420307; --User #1 goes on a very long journey
INSERT INTO @UserGPS SELECT 1, 2, 44.33328, -89.132008;
INSERT INTO @UserGPS SELECT 1, 3, 34.12345, -92.21369;
INSERT INTO @UserGPS SELECT 1, 4, 44.978046, -94.430307;
INSERT INTO @UserGPS SELECT 2, 1, 44.968046, -94.420307; --User #2 doesn't get far
INSERT INTO @UserGPS SELECT 2, 2, 44.978046, -94.430307;
--Make a working table to store the distances between each set of co-ordinates
--This isn't strictly necessary; we could change this into a common-table expression
DECLARE @WorkTable TABLE (
UserId INT, --the user
GPSIdFrom INT, --the id of the first set of co-ordinates
GPSIdTo INT, --the id of the second set of co-ordinates being compared
Distance NUMERIC(19,6)); --the distance
--Get the distance between each and every combination of co-ordinates for each user
INSERT INTO
@WorkTable
SELECT
c1.UserId,
c1.GPSId,
c2.GPSId,
GEOGRAPHY::Point(c1.Lat, c1.Lon, 4326).STDistance(GEOGRAPHY::Point(c2.Lat, c2.Lon, 4326))
FROM
@UserGPS c1
INNER JOIN @UserGPS c2 ON c2.UserId = c1.UserId AND c2.GPSId > c1.GPSId;
--Note this is a self-join, but single-tailed. So we compare each set of co-ordinates to each other set of co-ordinates for a user
--This is handled by the "c2.GPSID > c1.GPSId" in the JOIN clause
--As an example, say we have three sets of co-ordinates for a user
--We would compare set #1 to set #2
--We would compare set #1 to set #3
--We would compare set #2 to set #3
--We wouldn't compare set #3 to anything (as we already did this)
--Determine the maximum distance between all the GPS co-ordinates per user
WITH MaxDistance AS (
SELECT
UserId,
MAX(Distance) AS Distance
FROM
@WorkTable
GROUP BY
UserId)
--Report the results
SELECT
w.UserId,
g1.GPSId,
g1.Lat,
g1.Lon,
g2.GPSId,
g2.Lat,
g2.Lon,
md.Distance AS MaxDistance
FROM
MaxDistance md
INNER JOIN @WorkTable w ON w.UserId = md.UserId AND w.Distance = md.Distance
INNER JOIN @UserGPS g1 ON g1.UserId = md.UserId AND g1.GPSId = w.GPSIdFrom
INNER JOIN @UserGPS g2 ON g2.UserId = md.UserId AND g2.GPSId = w.GPSIdTo;
结果是:
UserId GPSId Lat Lon GPSId Lat Lon MaxDistance
1 3 34.123450 -92.213690 4 44.978046 -94.430307 1219979.460185
2 1 44.968046 -94.420307 2 44.978046 -94.430307 1362.820895
现在我对你持有的数据做了很多假设,因为在你的问题中没有关于这个细节的信息。您可能需要在某种程度上适应这种情况吗?