mySQL select zipcodes within x km/miles within range of y

Question

Note: Although I use a zipcode database with Dutch zipcodes, this question is country independent.

I have a database with every zipcode in the Netherlands +

Answer 1

You want to do something like this:

SELECT zipcode FROM zipcodes WHERE DistanceFormula(lat, long, 4.808855, 52.406332) < $range

It may be slow if your table of zip codes is large. You may also want to check out the geospatial extensions for MySQL.

Answer 2

You have to use something called the Haversine formula:

$sql = "
    SELECT zipcode
    FROM zipcodes
    WHERE ".mysqlHaversine($lat, $lon, $distance)."
";

And the formula:

function mysqlHaversine($lat = 0, $lon = 0, $distance = 0)
{
    if($distance > 0)
    {
        return ('
        ((6372.797 * (2 *
        ATAN2(
            SQRT(
                SIN(('.($lat*1).' * (PI()/180)-latitude*(PI()/180))/2) *
                SIN(('.($lat*1).' * (PI()/180)-latitude*(PI()/180))/2) +
                COS(latitude * (PI()/180)) *
                COS('.($lat*1).' * (PI()/180)) *
                SIN(('.($lon*1).' * (PI()/180)-longitude*(PI()/180))/2) *
                SIN(('.($lon*1).' * (PI()/180)-longitude*(PI()/180))/2)
                ),
            SQRT(1-(
                SIN(('.($lat*1).' * (PI()/180)-latitude*(PI()/180))/2) *
                SIN(('.($lat*1).' * (PI()/180)-latitude*(PI()/180))/2) +
                COS(latitude * (PI()/180)) *
                COS('.($lat*1).' * (PI()/180)) *
                SIN(('.($lon*1).' * (PI()/180)-longitude*(PI()/180))/2) *
                SIN(('.($lon*1).' * (PI()/180)-longitude*(PI()/180))/2)
            ))
        )
        )) <= '.($distance/1000). ')');
    }

    return '';
}

Usually I do not use code without understanding the way it works first, but I must confess this function is a little bit over my head...

Answer 3

While Captaintokyo's method is accurate, it's also fairly slow. I can't help but think it'd be more advantageous to use a temporary table of all zipcodes whose boundaries are within the range, then to refine those results by distance.