计算两点之间的距离（纬度，经度）

Question

我正在尝试计算地图上两个位置之间的距离。我已经在数据中存储了：经度，纬度，X POS和Y POS。

我以前一直在使用以下代码段。

DECLARE @orig_lat DECIMAL

DECLARE @orig_lng DECIMAL

SET @orig_lat=53.381538 set @orig_lng=-1.463526

SELECT *,

    3956 * 2 * ASIN(

          SQRT( POWER(SIN((@orig_lat - abs(dest.Latitude)) * pi()/180 / 2), 2) 

              + COS(@orig_lng * pi()/180 ) * COS(abs(dest.Latitude) * pi()/180)  

              * POWER(SIN((@orig_lng - dest.Longitude) * pi()/180 / 2), 2) )) 

          AS distance

--INTO #includeDistances

FROM #orig dest

但是，我不相信由此产生的数据，它似乎给出了稍微不准确的结果。

一些样本数据，以备不时之需

Latitude        Longitude     Distance 

53.429108       -2.500953     85.2981833133896

任何人都可以帮助我解决我的代码，如果您有一种新的方法来实现这一点，那我就介意您是否要解决已有的问题。

请说明您的结果采用的计量单位。

Answer 1

由于您使用的是SQL Server 2008，因此具有geography可用的数据类型，该数据类型专门用于此类数据：DECLARE @source geography = 'POINT(0 51.5)'DECLARE @target geography = 'POINT(-3 56)'SELECT @source.STDistance(@target)给----------------------538404.100197555(1 row(s) affected)告诉我们，从（伦敦附近）到（爱丁堡附近）大约538公里。当然，首先需要学习很多东西，但是一旦您知道它，就比实现自己的Haversine计算容易得多。再加上您可以获得很多功能。如果您想保留现有的数据结构，则仍然可以使用STDistance，方法是geography使用以下Point方法构造合适的实例：DECLARE @orig_lat DECIMAL(12, 9)DECLARE @orig_lng DECIMAL(12, 9)SET @orig_lat=53.381538 set @orig_lng=-1.463526DECLARE @orig geography = geography::Point(@orig_lat, @orig_lng, 4326);SELECT *,    @orig.STDistance(geography::Point(dest.Latitude, dest.Longitude, 4326))        AS distance--INTO #includeDistancesFROM #orig dest

Answer 2

下面的函数给出两个地理坐标之间的距离（以英里为单位）create function [dbo].[fnCalcDistanceMiles] (@Lat1 decimal(8,4), @Long1 decimal(8,4), @Lat2 decimal(8,4), @Long2 decimal(8,4))returns decimal (8,4) asbegindeclare @d decimal(28,10)-- Convert to radiansset @Lat1 = @Lat1 / 57.2958set @Long1 = @Long1 / 57.2958set @Lat2 = @Lat2 / 57.2958set @Long2 = @Long2 / 57.2958-- Calc distanceset @d = (Sin(@Lat1) * Sin(@Lat2)) + (Cos(@Lat1) * Cos(@Lat2) * Cos(@Long2 - @Long1))-- Convert to milesif @d <> 0beginset @d = 3958.75 * Atan(Sqrt(1 - power(@d, 2)) / @d);endreturn @dend&nbsp;下面的函数给出了两个地理坐标之间的距离（以千米为单位）CREATE FUNCTION dbo.fnCalcDistanceKM(@lat1 FLOAT, @lat2 FLOAT, @lon1 FLOAT, @lon2 FLOAT)RETURNS FLOAT&nbsp;ASBEGIN&nbsp; &nbsp; RETURN ACOS(SIN(PI()*@lat1/180.0)*SIN(PI()*@lat2/180.0)+COS(PI()*@lat1/180.0)*COS(PI()*@lat2/180.0)*COS(PI()*@lon2/180.0-PI()*@lon1/180.0))*6371END下面的函数 使用sql server 2008中引入的Geography数据类型，以公里为单位给出两个地理坐标之间的距离DECLARE @g geography;DECLARE @h geography;SET @g = geography::STGeomFromText('LINESTRING(-122.360 47.656, -122.343 47.656)', 4326);SET @h = geography::STGeomFromText('POINT(-122.34900 47.65100)', 4326);SELECT @g.STDistance(@h);用法：select [dbo].[fnCalcDistanceKM](13.077085,80.262675,13.065701,80.258916)

Answer 3

Create Function [dbo].[DistanceKM]&nbsp;(&nbsp;&nbsp; &nbsp; &nbsp; @Lat1 Float(18),&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; @Lat2 Float(18),&nbsp;&nbsp; &nbsp; &nbsp; @Long1 Float(18),&nbsp;&nbsp; &nbsp; &nbsp; @Long2 Float(18))Returns Float(18)ASBegin&nbsp; &nbsp; &nbsp; Declare @R Float(8);&nbsp;&nbsp; &nbsp; &nbsp; Declare @dLat Float(18);&nbsp;&nbsp; &nbsp; &nbsp; Declare @dLon Float(18);&nbsp;&nbsp; &nbsp; &nbsp; Declare @a Float(18);&nbsp;&nbsp; &nbsp; &nbsp; Declare @c Float(18);&nbsp;&nbsp; &nbsp; &nbsp; Declare @d Float(18);&nbsp; &nbsp; &nbsp; Set @R =&nbsp; 6367.45&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; --Miles 3956.55&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; --Kilometers 6367.45&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; --Feet 20890584&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; --Meters 6367450&nbsp;&nbsp; &nbsp; &nbsp; Set @dLat = Radians(@lat2 - @lat1);&nbsp; &nbsp; &nbsp; Set @dLon = Radians(@long2 - @long1);&nbsp; &nbsp; &nbsp; Set @a = Sin(@dLat / 2)&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;* Sin(@dLat / 2)&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;+ Cos(Radians(@lat1))&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;* Cos(Radians(@lat2))&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;* Sin(@dLon / 2)&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;* Sin(@dLon / 2);&nbsp;&nbsp; &nbsp; &nbsp; Set @c = 2 * Asin(Min(Sqrt(@a)));&nbsp;&nbsp; &nbsp; &nbsp; Set @d = @R * @c;&nbsp;&nbsp; &nbsp; &nbsp; Return @d;&nbsp;EndGO用法：选择dbo.DistanceKM（37.848832506474，37.848732506474，27.83935546875，27.83905546875）输出：0,02849639您可以使用带注释的浮点数更改@R参数。