Question

有没有办法在SQL Server 2008R2中知道某个点是在另一个点的南，东等等？

例如，我有一个来源point(lat1,lng1)，我想知道point(lat2,lng2)位于该来源的位置：北，西等......

我正在尝试构建风玫瑰图，这可能对我有用。

Answer 1

为了在SQL Server 2008 R2中使用Geography类型计算两个坐标之间的方位，您可以使用此函数：

CREATE FUNCTION [dbo].[CalculateBearing] 
(
    @pointA as geography
    ,@pointB as geography
)

RETURNS decimal(18,12)

AS

    BEGIN

    -- Declare the return variable
    DECLARE @bearing decimal(18,12)

    -- Declare the local variables
    DECLARE @x decimal(18,12)
    DECLARE @y decimal(18,12)
    DECLARE @dLat decimal(18,12)
    DECLARE @dLong decimal(18,12)
    DECLARE @rLat1 decimal(18,12)
    DECLARE @rLat2 decimal(18,12)

    IF(@pointA.STIsEmpty() = 1 OR @pointB.STIsEmpty() = 1)
        set @bearing = null
    ELSE
        BEGIN

        -- Calculate delta between coordinates
        SET @dLat = RADIANS(@pointB.Lat - @pointA.Lat)
        SET @dLong = RADIANS(@pointB.Long - @pointA.Long)

        -- Calculate latitude as radians
        SET @rLat1 = RADIANS(@pointA.Lat)
        SET @rLat2 = RADIANS(@pointB.Lat)

        SET @y = SIN(@dLong)*COS(@rLat2)
        SET @x = COS(@rLat1)*SIN(@rLat2)-SIN(@rLat1)*COS(@rlat2)*COS(@dLong)

        IF (@x = 0 and @y = 0)
            SET @bearing = null
        ELSE
            BEGIN
                SET @bearing = CAST((DEGREES(ATN2(@y,@x)) + 360) as decimal(18,12)) % 360
            END
    END

    -- Return the result of the function
    RETURN @bearing

END

GO

在此之后，您可以像这样使用此功能：

DECLARE @pointA as geography
DECLARE @pointB as geography

SET @pointA = geography::STGeomFromText('POINT(3 45)', 4326)
SET @pointB = geography::STGeomFromText('POINT(4 47)', 4326)

SELECT [dbo].[CalculateBearing](@pointA, @pointB)

更新：添加架构

Bearing explanation

Answer 2

我想出了一种使用标准SQL函数相当简单地计算轴承的方法。 ATAN功能完成了大部分实际工作;两个CASE语句只是特殊情况的更正。 1是源，2是目的地。

atan(([Longitude2]-[Longitude1])/(10e-10+[Latitude2]-[Latitude1]))*360/pi()/2
+case when [Latitude2]<[Latitude1] then 180 else 0 end
+case when [Longitude2]<[Longitude1] and [Latitude2]>[Latitude1] then 360 else 0 end

Answer 3

今天早上，当我们在系统中搜索附近的订单时，我需要此功能为用户提供范围和基本方向。我来到尼古拉斯的答案，它让我大部分都在那里。我创建了第二个函数，它使用Nicolas来为我的UI创建一个缩写的基本方向（N，NE，E等）。

使用此处提供的Nicolas轴承计算结合https://en.wikipedia.org/wiki/Points_of_the_compass的值来确定每个基本方向的范围，

CREATE FUNCTION [dbo].[CalculateCardinalDirection] 
(
    @pointA as geography
    ,@pointB as geography
)

RETURNS varchar(2)

AS
BEGIN
    DECLARE @bearing decimal(18,12)
    -- Bearing calculation provided by http://stackoverflow.com/a/14781032/4142441
    SELECT @bearing = dbo.CalculateBearing(@pointA, @pointB)

    RETURN CASE WHEN @bearing BETWEEN 0 AND 22.5 THEN 'N'
                WHEN @bearing BETWEEN 22.5 AND 67.5 THEN 'NE'
                WHEN @bearing BETWEEN 67.5 AND 112.5 THEN 'E'
                WHEN @bearing BETWEEN 112.5 AND 157.5 THEN 'SE'
                WHEN @bearing BETWEEN 157.5 AND 202.5 THEN 'S'
                WHEN @bearing BETWEEN 202.5 AND 247.5 THEN 'SW'
                WHEN @bearing BETWEEN 247.5 AND 292.5 THEN 'W'
                WHEN @bearing BETWEEN 292.5 AND 337.5 THEN 'NW'
                ELSE 'N' -- Catches NULL bearings and the 337.5 to 360.0 range
           END
END
GO

Answer 4

如果点数据类型为“几何”（例如UTM坐标系），则可以使用以下公式：

DEGREES(ATAN((X2-X1)/(Y2-Y1)))
+case when Y2<Y1 then 180 else 0 end
+case when Y2>Y1 and X2<X1 then 360 else 0 end

以下是用于进一步说明的架构：

Answer 5

X=X2-X1和Y=Y2-Y1.一个公式，用于从0（正Y轴）到360度顺时针方向给出方位角。

  f(X,Y)=180-90*(1+SIGN(Y))*(1-SIGN(X^2))-45*(2+SIGN(Y))*SIGN(X)-180/PI()*SIGN(Y*X)*ATAN((ABS(Y)-ABS(X))/(ABS(Y)+ABS(X)))

确定点之间的基数（罗盘）方向

5 个答案: