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# How to Use Google Map API From Stored Procedure

Posted on 2016-10-05
Medium Priority
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Hi Guys,

Are there any tutorials on how to use google map API and connect it with the stored procedure to make the marker points?

regards
0
Question by:shah36
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Author Comment

ID: 41829568
Hi Terry,

Sorry i should have been more specific. So the requirement is that if a user types an address and the radius. The google maps should create the markers on the suppliers within that radius. Now the address of the suppliers are saved within our SQL Server database with Lat ad Long.

regards
0

LVL 27

Accepted Solution

Zberteoc earned 2000 total points
ID: 41830341
You can't use Google APIs from SQL server and they were never intended for that. SO your problem, if I understand correctly is this:

From an application you get a location in the form of lat-long coordinates, a radius and then you have to find all the suppliers within that radius by querying the supplier table in the database.

This can be done in 2 ways:

1. Using the lat long given by the user and select all the rows from the table where the distance between supplier lat long and the user lat long <= radius

2. You can use the spatial datatypes to store the location of the suppliers and then buffer the user location by the radius and select from the table all the suppliers where location intersects the user buffered location.

I would recommend the first approach because is simpler and doesn't need spatial datatype which are not familiar for everyone. For this you need a distance function to be created in teh same database where the suppliers table is:
``````CREATE FUNCTION [dbo].[fnDistance]
(
@Lat1 decimal(11, 8),
@Lng1 decimal(11, 8),
@Lat2 decimal(11, 8),
@Lng2 decimal(11, 8)
)
RETURNS decimal(10, 6) -- max 10km
AS
BEGIN

if abs(@Lat2-@Lat1) > 1 or abs(@Lng2-@Lng1) > 1
return 10
declare	@distance decimal(30, 8)

declare @R as int
select @R = 6371000

declare @dLat as decimal(21, 17)
declare @dLng as decimal(21, 17)
select @dLat = (@Lat2-@Lat1) * pi() / 180
select @dLng = (@Lng2-@Lng1) * pi() / 180

declare @a as decimal(18, 18)
declare @c as decimal(18, 18)
select @a = Sin(@dLat / 2) * Sin(@dLat / 2) +
Cos(@Lat1 * pi() / 180) * Cos(@Lat2 * pi() / 180) * Sin(@dLng /2) * Sin(@dLng / 2);
select @c = 2 * Atn2(Sqrt(@a), Sqrt(1-@a))

select @distance = (@R * @c )/1000-- % 10000

return @distance

END
``````
Use this function to find the suppliers like this:
``````declare
@UserLat decimal(11,8)=42.998282,
@UserLong decimal (11,8)=-81.243822,

select
*
from
Suppliers
where
``````
0

Author Closing Comment

ID: 41846614
Thanks a lot it did help.
Kindest regards
0

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