sql中经度维度计算距离

Posted 2020-09-19 ly77461

------------------------------------------创建一个方法--------------------------------------------------------------------
--@LatBegin 纬度A(33.8703596)--
--@LngBegin 经度A(-117.9242966)--
--@LatEnd 纬度B(34.0392283)--
--@LngEnd 经度B(-117.8367681)--
CREATE FUNCTION [dbo].[fnGetDistance](@LatBegin REAL, @LngBegin REAL, @LatEnd REAL, @LngEnd REAL) RETURNS FLOAT 
 AS 
 BEGIN 
  --距离(千米) 
  DECLARE @Distance REAL
  --距离(英里)--
  DECLARE @Edistance REAL 
  DECLARE @EARTH_RADIUS REAL 
  DECLARE @MI REAL
  SET @EARTH_RADIUS = 6378.137 --地球半径(千米)--
  Set @MI=0.6213712 --1千米(km)=0.6213712英里(mi)
  DECLARE @RadLatBegin REAL,@RadLatEnd REAL,@RadLatDiff REAL,@RadLngDiff REAL 
  SET @RadLatBegin = @LatBegin *PI()/180.0   
  SET @RadLatEnd = @LatEnd *PI()/180.0   
  SET @RadLatDiff = @RadLatBegin - @RadLatEnd   
  SET @RadLngDiff = @LngBegin *PI()/180.0 - @LngEnd *PI()/180.0  
  SET @Distance = 2 *ASIN(SQRT(POWER(SIN(@RadLatDiff/2), 2)+COS(@RadLatBegin)*COS(@RadLatEnd)*POWER(SIN(@RadLngDiff/2), 2))) 
  SET @Distance = @Distance * @EARTH_RADIUS   
  SET @Distance = Round(@Distance * 10000,1) / 10000  --千米-- 
  SET @Edistance= @Distance * @MI  --英里--
  RETURN @Edistance
 END

 ------------------------------------------删除方法--------------------------------------------------------------------
 drop function dbo.fnGetDistance
 

 ------------------------------------------调用方法--------------------------------------------------------------------
 select dbo.fnGetDistance(33.8703596,-117.9242966,34.0392283,-117.8367681)