Sql server 从GPS坐标组获取最大距离

所以我有一个包含多行GPS坐标的数据库。我知道如何计算从给定lat/lng到数据库中任意一行的距离，但我想做的基本上是查看一组行的坐标，并得到相距最远的两行。如果我能在SQL中做到这一点，我会很高兴，但如果我必须在我的应用程序代码中做到这一点，这将非常有用。下面是我计算两点之间距离的步骤：

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

我们试图做的是，查看特定用户的GPS数据，确保它们不会在全国范围内疯狂移动。用户的所有坐标最多应在几英里以内。如果坐标遍布全国，则表明我们的系统中存在恶意活动。因此，我希望能够快速浏览SPCIC用户的数据，并了解他们的最大距离

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我考虑在lat和lng上分别运行Max/Min，并设置一个可接受的内部阈值。也许这更容易，但如果我在第一部分中所问的是可能的，那将是最好的。

如果您有SQL Server 2008或更高版本，那么您可以使用地理位置来计算距离，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));

这使得这个问题变得非常简单

对于用户的一组lat/long，您需要计算每组之间的距离，然后返回最大距离。综上所述，您可能可以这样做：

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;

结果如下：

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

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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对其进行调整？

因此它看起来只比较了两行。如果我有50行，每行都有不同的lat/lng值呢。是否有一种方法可以将它们相互比较，并返回两行之间的最大距离。不，这将为用户比较所有行，它使用单尾方法，因此如果有三行，它将比较第1行、第2行和第3行，然后从第2排到第3排，然后返回距离最高的比赛。在我的示例中，第一个用户有四组坐标，它计算的最大距离为值3和值4之间。我希望我能理解它是如何做到的。我在我的答案中添加了一些注释，试图解释它是如何工作的。感谢您的帮助。现在就开始测试。