Sql server 查找点的最远距离
在这个场景中,ID 1和ID 2有一组四个点。我需要找到并选择两个距离设置点最远且ID相同的点 该表的脚本如下所示:Sql server 查找点的最远距离,sql-server,tsql,spatial,Sql Server,Tsql,Spatial,在这个场景中,ID 1和ID 2有一组四个点。我需要找到并选择两个距离设置点最远且ID相同的点 该表的脚本如下所示: ID name point id_point ==================================================================== 1 DN700 POINT (-550493.96 -1218974.69)
ID name point id_point
====================================================================
1 DN700 POINT (-550493.96 -1218974.69) 1
1 DN700 POINT (-550513.92733318976 -1218929.5493905835) 2
1 DN700 POINT (-550490.62291509821 -1218980.7209425652) 3
1 DN700 POINT (-550512.43436134933 -1218933.2777663434) 4
2 DN700 POINT (-550235.5039543492 -1219120.0737476321) 5
2 DN700 POINT (-550278.61165674869 -1219099.6880138929) 6
2 DN700 POINT (-550301.89265282557 -1219088.8117909778) 7
2 DN700 POINT (-550330.76399739366 -1219075.4882849427) 8
对于ID 1,最远的点是ID_点2和3。我将在另一个过程中使用它,在这里我必须定义起点和终点,在本例中正好是第2点和第3点
我知道解决方法是使用STDistance函数比较具有相同ID的点,然后从结果中选择一个最大值,但我被困在这里
非常感谢您的帮助,非常感谢
这是我的一个查询示例,我在评论中描述了它。林上的points_是我之前编写的一个表。唯一的区别是点存储为几何体:
DECLARE @max_id int, @current_id int = 1;
SET @max_id = (SELECT MAX(ID_point) FROM #points_on_lin)
WHILE @current_id <= @max_id
BEGIN
DECLARE @point1 geometry, @point2 geometry, @ID int, @IDP int;
SET @point1 = (SELECT points FROM #points_on_lin WHERE ID_point = @current_id)
SET @point2 = (SELECT points FROM #points_on_lin WHERE ID_point = @current_id + 1)
SET @ID = (SELECT ID FROM #points_on_linia WHERE ID_point = @current_id)
SET @name = (SELECT name FROM #points_on_lin WHERE ID_point = @current_id)
SET @IDP = (SELECT ID_point FROM #points_on_lin WHERE ID_point = @current_id)
INSERT INTO #points_dist
SELECT @ID, @name, @point1.STDistance(@point2), @IDP
SET @current_ID = @current_ID +1
END
我这样做:
create table #MainTable
(
id int,
name nvarchar(10),
point nvarchar(max),
id_point int
)
insert into #MainTable
values
(1 ,'DN700','POINT (-550493.96 -1218974.69)' ,1),
(1 ,'DN700','POINT (-550513.92733318976 -1218929.5493905835)' ,2),
(1 ,'DN700','POINT (-550490.62291509821 -1218980.7209425652)' ,3),
(1 ,'DN700','POINT (-550512.43436134933 -1218933.2777663434)' ,4),
(2 ,'DN700','POINT (-550235.5039543492 -1219120.0737476321)' ,5),
(2 ,'DN700','POINT (-550278.61165674869 -1219099.6880138929)' ,6),
(2 ,'DN700','POINT (-550301.89265282557 -1219088.8117909778)' ,7),
(2 ,'DN700','POINT (-550330.76399739366 -1219075.4882849427)' ,8)
create table #Result
(
id int,
name nvarchar(10),
distance float,
id_point int
)
declare @minId as int=(select min(id) from #MainTable)
declare @maxId as int=(select max(id) from #MainTable)
declare @p1 as int=0
declare @p2 as int=0
DECLARE @point1 geometry, @point2 geometry
declare @Distance as float=0
while @minId<=@maxId
begin
set @p1=(select min(id_point) from #MainTable where id=@minId)
set @p2=(select max(id_point) from #MainTable where id=@minId)
while @p1<@p2
begin
set @point1=(select geometry::STGeomFromText(point,0) from #MainTable where id_point=@p1)
set @point2=(select geometry::STGeomFromText(point,0) from #MainTable where id_point=@p1+1)
set @Distance=@point1.STDistance(@point2)
insert into #Result (id,name,distance,id_point)
select id,name,@Distance,@p1 from #MainTable where id_point=@p1
insert into #Result (id,name,distance,id_point)
select id,name,@Distance,@p1+1 from #MainTable where id_point=@p1+1
set @p1=@p1+1
end
set @minId=@minId+1
end
select r1.*
from #Result r1
inner join (select id,max(distance)maxDistance
from #Result
group by id)r2 on r1.id=r2.id
and r1.distance=r2.maxDistance
order by id
drop table #MainTable,#Result
结果如下:
希望对您有所帮助。试试这个:
;WITH CTE AS
(
SELECT ID FROM #points_on_lin GROUP BY ID
)
SELECT CTE.ID, Point1, Point2, Distance FROM CTE
CROSS APPLY
(
SELECT TOP 1
T1.id_point AS Point1, T2.id_point AS Point2, T1.Point.STDistance(T2.Point) AS Distance
FROM
#points_on_lin T1 join #points_on_lin T2 on T1.ID = T2.ID AND T1.id_point < T2.id_point
WHERE
T1.ID = CTE.ID
ORDER BY
Distance DESC
) A
那么,当您尝试实现STDistance和MAX时,您的尝试是什么?为什么不起作用?@Larnu使用这些点,我创建了一个ID为1的迭代,但只从一个ID_点到下一个点。所以我只知道从一个到另一个的距离。但是,我必须选择最远的点,在本例中是id_点2和3,但距离计算中的最大值在点4,这意味着最远的距离是从3到4。当然,这是错误的。现在我不知道下一步该怎么办。创建从一个点到具有相同ID的其他点的迭代,然后选择ID 1的最大值?嗯,也许这是一个正确的方式:/但是把你的尝试放在问题上。使用该功能。@Roman您可以添加数据和表格脚本吗?好的,我已经编辑了我的问题,您可以查看它。谢谢。那正是我想要的!非常感谢@hsn!!谢谢你的回复。价值观因为第一个ID是正确的,但是第二个ID是不正确的。正确的ID_点是5和8。谢谢你的解决方案,但我会用@hsn这个答案!
;WITH CTE AS
(
SELECT ID FROM #points_on_lin GROUP BY ID
)
SELECT CTE.ID, Point1, Point2, Distance FROM CTE
CROSS APPLY
(
SELECT TOP 1
T1.id_point AS Point1, T2.id_point AS Point2, T1.Point.STDistance(T2.Point) AS Distance
FROM
#points_on_lin T1 join #points_on_lin T2 on T1.ID = T2.ID AND T1.id_point < T2.id_point
WHERE
T1.ID = CTE.ID
ORDER BY
Distance DESC
) A
ID Point1 Point2 Distance
1 2 3 56.228317019116
2 5 8 105.177655821281