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计算地图上两坐标点之间的距离

2011-10-08 11:11 309 查看
set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go

ALTER FUNCTION [dbo].[Gps](@gps_1 varchar(80),@gps_2 varchar(80))
returns numeric(10,2)--设置返回值,记住是returns 而不是return
AS

BEGIN

DECLARE @gps1_w float
DECLARE @gps1_s float
DECLARE @gps2_w float
DECLARE @gps2_s float
DECLARE @a1 float
DECLARE @b1 float
DECLARE @a float
DECLARE @b float
declare @s float
DECLARE @return float
IF @gps_1='' and @gps_2=''
BEGIN
SET @return =0.0
END
ELSE
BEGIN

SET @gps1_w =Convert(float,Substring(@gps_1,1,Patindex('%,%',@gps_1)-1))
SET @gps1_s =Convert(float,Substring(@gps_1,Patindex('%,%',@gps_1)+1,20))
SET @gps2_w =Convert(float,Substring(@gps_2,1,Patindex('%,%',@gps_2)-1))
SET @gps2_s =Convert(float,Substring(@gps_2,Patindex('%,%',@gps_2)+1,20))
set @a1=(@gps1_s*Pi()/180.00)
set @b1=(@gps2_s*Pi()/180.00)
set @a=@a1-@b1
set @b=(@gps1_w-@gps2_w)*pi()/180.00
set @s= 2 * Asin(Sqrt(Power(Sin(@a/ 2), 2) +Cos(@a1) * Cos(@b1) * Power(Sin(@b / 2), 2))) * 6378.137
SET @return =@s
END
return @return --返回结果

END
http://www.cnblogs.com/markbum_2008/archive/2011/08/31/sqlserver.html
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