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  • SQL

latlngHaving a store locator on my to do list, I started months back with storing the latitude and longitude for the stores in a database. I forked this repository in 2012 https://github.com/bjorn2404/jQuery-Store-Locator-Plugin in order to have something like a store locator on our website. The downside to that store locator is that it is all client side and puts a copy of all your customer data on the internet. It is great data for your competition. So to disclose as little as possible and still serve your valued customers, I decided to do the calculation on the backend. Closest to the data. So that would be on the SQL server.

SQL Server has datatypes like points and geometry since 2008. With this SQL statement you can get the distance between my location

lat 51.69917 lng 5.30417 and the locations in the stores table.
SELECT top 10 StoreID
      ,round((geography::Point(StoreLatitude, StoreLongitude, 4326).STDistance(geography::Point(51.69917, 5.30417, 4326))) / 1000, 1) as km
  where StoreLongitude < 150 and StoreLongitude > -150 and StoreLatitude < 90 and StoreLatitude > -90
order by (geography::Point(StoreLatitude, StoreLongitude, 4326).STDistance(geography::Point(51.69917, 5.30417, 4326)))

The third parameter for the point data type is 4326 which is an enumeration:

And about 4326 SRID. It also called WSG84(http://en.wikipedia.org/wiki/World_Geodetic_System) system. It's the most common system to represent point on spherical(not flat) earth. It uses degree,minute,second notation and it's x and y coordinates are usually called latitude and longitude.

source http://stackoverflow.com/a/405256/169714

SRID (Spatial Reference System Identifier) : https://en.wikipedia.org/wiki/SRID

This post is more of a reference for me to lookup how to calculate distances on the Microsoft SQL server in SQL. But maybe it helps others too.

Good luck!

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