Calculates the distance between two coordinate points (longitude, latitude) of the Earth SQL function

Go  -Calculates the distance between two coordinate points (longitude, latitude) on the earth SQL function--lordbaby--: www.aspbc.com CREATE function [dbo].[ Fngetdistance] (@LatBegin Real, @LngBegin Real, @LatEnd Real, @LngEnd Real) RETURNS FLOAT  asbegin  -distance (km)  DECLARE @Distance Real  DECLARE @EARTH_RADIUS real  SET @EARTH_RADIUS = 6378.137    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)/ 10000    RETURN @DistanceEND

Here's how to use it:

1	SELECT* FROM商家表名 WHEREdbo.fnGetDistance(121.4625,31.220937,longitude,latitude) < 5

(Mouse over the code, there will be four icons at the top of the code, the first one is to view the source code, the second is to copy the code, the third is to print the code, the fourth is to help)
The longitude,latitude here are the longitude and latitude fields of the hotel, and 121.4625,31.220937 is the longitude of the current customer where the phone is located, and the latter 5 is within 5 km.

Lordbaby also told me that in the SQL2008 database, you can directly use the geography data function to solve, see http://msdn.microsoft.com/zh-cn/library/bb933952 (v=sql.100). aspx.

Calculates the distance between two coordinate points (longitude, latitude) of the Earth SQL function