6

给定以下数据，是否有可能，如果可能，哪种方法是确定第一个表中的位置“Shurdington”是否包含在第二个表中任何位置的给定半径内的最有效方法。

GeoData 列属于“地理”类型，因此可以选择使用 SQL Server 空间功能以及使用纬度和经度。

Location      GeoData       Latitude    Longitude
===========================================================
Shurdington   XXXXXXXXXX    51.8677979  -2.113189

ID  Location            GeoData     Latitude    Longitude   Radius
==============================================================================
1000    Gloucester      XXXXXXXXXX  51.8907127  -2.274598   10
1001    Leafield        XXXXXXXXXX  51.8360519  -1.537438   10
1002    Wotherton       XXXXXXXXXX  52.5975151  -3.061798   5
1004    Nether Langwith XXXXXXXXXX  53.2275276  -1.212108   20
1005    Bromley         XXXXXXXXXX  51.4152069  0.0292294   10

非常感谢任何帮助。

4

3 回答 3

9

创建数据

CREATE TABLE #Data (
    Id int,
    Location nvarchar(50),
    Latitude decimal(10,5),
    Longitude decimal(10,5),
    Radius int
)

INSERT #Data (Id,Location,Latitude,Longitude,Radius) VALUES 
(1000,'Gloucester', 51.8907127 ,-2.274598  , 20), -- Increased to 20
(1001,'Leafield', 51.8360519 , -1.537438  , 10),
(1002,'Wotherton', 52.5975151,  -3.061798  , 5),
(1004,'Nether Langwith', 53.2275276 , -1.212108  , 20),
(1005,'Bromley', 51.4152069 , 0.0292294  , 10)

测试

将您的兴趣点声明为POINT

DECLARE @p GEOGRAPHY = GEOGRAPHY::STGeomFromText('POINT(-2.113189 51.8677979)', 4326);

要确定它是否在另一个点的半径内： 

-- First create a Point.
DECLARE @point GEOGRAPHY = GEOGRAPHY::STGeomFromText('POINT(-2.27460 51.89071)', 4326);
-- Buffer the point (meters) and check if the 1st point intersects
SELECT @point.STBuffer(50000).STIntersects(@p)

将它们组合成一个查询：

select  *,
        GEOGRAPHY::STGeomFromText('POINT('+ 
            convert(nvarchar(20), Longitude)+' '+
            convert( nvarchar(20), Latitude)+')', 4326)
        .STBuffer(Radius * 1000).STIntersects(@p) as [Intersects]
from    #Data

给出：

Id      Location        Latitude    Longitude   Radius  Intersects
1000    Gloucester      51.89071    -2.27460    20      1
1001    Leafield        51.83605    -1.53744    10      0
1002    Wotherton       52.59752    -3.06180    5       0
1004    Nether Langwith 53.22753    -1.21211    20      0
1005    Bromley         51.41521    0.02923     10      0

回复：效率。通过一些正确的索引，似乎 SQL 的空间索引可以非常快

于 2013-02-04T10:37:29.117 回答
1

You calculate the distance between the two points and compare this distance to the given radius.

For calculating short distances, you can use the formula at Wikipedia - Geographical distance - Spherical Earth projected to a plane, which claims to be "very fast and produces fairly accurate result for small distances".

According to the formula, you need the difference in latitudes and longitudes and the mean latitude 

with geo as (select g1.id, g1.latitude as lat1, g1.longitude as long1, g1.radius,
                    g2.latitude as lat2, g2.longitude as long2
             from geography g1
             join geography g2 on g2.location = 'shurdington'
                               and g1.location <> 'shurdington')
     base as (select id,
                     (radians(lat1) - radians(lat2)) as dlat,
                     (radians(long1) - radians(long2)) as dlong,
                     (radians(lat1) + radians(lat2)) / 2 as mlat, radius
              from geo)
     dist as (select id,
                     6371.009 * sqrt(square(dlat) + square(cos(mlat) * dlong)) as distance,
                     radius
              from base)
select id, distance
from dist
where distance <= radius

I used the with selects as intermediate steps to keep the calculations "readable".

于 2013-02-04T10:43:27.653 回答
1

If you want to do the maths yourself, you could use Equirectangular approximation based upon Pythagoras. The formula is:

var x = (lon2-lon1) * Math.cos((lat1+lat2)/2); var y = (lat2-lat1); var d = Math.sqrt(x*x + y*y) * R;

In terms of SQL, this should give those locations in your 2nd table that contain your entry in the 1st within their radius:

SELECT *
FROM Table2 t2
WHERE EXISTS (
 SELECT 1 FROM Table1 t1
 WHERE 
  ABS (
  SQRT (
    (SQUARE((RADIANS(t2.longitude) - RADIANS(t1.longitude)) * COS((RADIANS(t2.Latitude) + RADIANS(t1.Latitude))/2))) +
    (SQUARE(RADIANS(t1.Latitude) - RADIANS(t2.Latitude)))
    ) * 6371 --Earth radius in km, use 3959 for miles
    )
    <= t2.Radius
)

Note that this is not the most accurate method available but is likely good enough. If you are looking at distances that stretch across the globe you may wish to Google 'haversine' formula.

It may be worth comparing this with Paddy's solution to see how well they agree and which performs best.

于 2013-02-04T10:43:32.093 回答