Haversine formula using SQL server to find closest venue - vb.net

Question

I am grabbing a postcode from a form. I can then convert this postcode to lng,lat coordinates as I have these stored in a table.

SELECT lng, lat from postcodeLngLat WHERE postcode = 'CV1'

I have another table which stores the lng,lat of a selection of venues.

SELECT v.lat, v.lng, v.name, p.lat, p.lng, p.postcode, 'HAVERSINE' AS distance FROM venuepostcodes v, postcodeLngLat p WHERE p.outcode = 'CB6' ORDER BY distance

What I am trying to do is create a datagrid which shows the distance of each venue from the postcode (CV1 in this case). I know that the Haversine formula should do what I am trying to achieve but I'm lost as to where I should start incorporating it into my query. I think the formula needs to go where I've put 'HAVERSINE' in the query above.

Any ideas?

EDIT

SELECT o.outcode AS lead_postcode, v.venue_name, 6371.0E * ( 2.0E *asin(case when 1.0E < (sqrt(square(sin(((RADIANS(CAST(o.lat AS FLOAT)))-(RADIANS(CAST(v.lat AS FLOAT))))/2.0E)) + (cos(RADIANS(CAST(v.lat AS FLOAT))) * cos(RADIANS(CAST(o.lat AS FLOAT))) * square(sin(((RADIANS(CAST(o.lng AS FLOAT)))-(RADIANS(CAST(v.lng AS FLOAT))))/2.0E))))) then 1.0E else (sqrt(square(sin(((RADIANS(CAST(o.lat AS FLOAT)))-(RADIANS(CAST(v.lat AS FLOAT))))/2.0E)) + (cos(RADIANS(CAST(v.lat AS FLOAT))) * cos(RADIANS(CAST(o.lat AS FLOAT))) * square(sin(((RADIANS(CAST(o.lng AS FLOAT)))-(RADIANS(CAST(v.lng AS FLOAT))))/2.0E))))) end )) AS distance FROM venuepostcodes v, outcodepostcodes o WHERE o.outcode = 'CB6' ORDER BY distance

Answer 1

I think you'd do best putting it in a UDF and using that in your query:

SELECT v.lat, v.lng, v.name, p.lat, p.lng, p.postcode, udf_Haversine(v.lat, v.lng, p.lat, p.lng) AS distance FROM venuepostcodes v, postcodeLngLat p WHERE p.outcode = 'CB6' ORDER BY distance

create function dbo.udf_Haversine(@lat1 float, @long1 float, @lat2 float, @long2 float) returns float begin
        declare @dlon float, @dlat float, @rlat1 float, @rlat2 float, @rlong1 float, @rlong2 float, @a float, @c float, @R float, @d float, @DtoR float

        select @DtoR = 0.017453293
        select @R = 3937 --3976

        select 
            @rlat1 = @lat1 * @DtoR,
            @rlong1 = @long1 * @DtoR,
            @rlat2 = @lat2 * @DtoR,
            @rlong2 = @long2 * @DtoR

        select 
            @dlon = @rlong1 - @rlong2,
            @dlat = @rlat1 - @rlat2

        select @a = power(sin(@dlat/2), 2) + cos(@rlat1) * cos(@rlat2) * power(sin(@dlon/2), 2)
        select @c = 2 * atn2(sqrt(@a), sqrt(1-@a))
        select @d = @R * @c

        return @d 
    end

Answer 2

Alternatively uou could also use SQL Server 2008 geography datatypes. If you currently store the longitude/latitide as varchar() in the DB, you will have to store them as geograpghy datatype and then use a function like STIntersects() to get the distance.