Fast nearest-location finder for SQL (MySQL, PostgreSQL, SQL Server)

fullquery

SELECT zip, primary_city,
       latitude, longitude, distance
  FROM (
 SELECT z.zip,
        z.primary_city,
        z.latitude, z.longitude,
        p.radius,
        p.distance_unit
                 * DEGREES(ACOS(LEAST(1.0, COS(RADIANS(p.latpoint))
                 * COS(RADIANS(z.latitude))
                 * COS(RADIANS(p.longpoint - z.longitude))
                 + SIN(RADIANS(p.latpoint))
                 * SIN(RADIANS(z.latitude))))) AS distance
  FROM zip AS z
  JOIN (   /* these are the query parameters */
        SELECT  42.81  AS latpoint,  -70.81 AS longpoint,
                50.0 AS radius,      111.045 AS distance_unit
    ) AS p ON 1=1
  WHERE z.latitude
     BETWEEN p.latpoint  - (p.radius / p.distance_unit)
         AND p.latpoint  + (p.radius / p.distance_unit)
    AND z.longitude
     BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
         AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
 ) AS d
 WHERE distance <= radius
 ORDER BY distance
 LIMIT 15

partialquery

 SELECT z.zip,
        z.primary_city,
        z.latitude, z.longitude,
        p.distance_unit
                 * DEGREES(ACOS(LEAST(1.0, COS(RADIANS(p.latpoint))
                 * COS(RADIANS(z.latitude))
                 * COS(RADIANS(p.longpoint) - RADIANS(z.longitude))
                 + SIN(RADIANS(p.latpoint))
                 * SIN(RADIANS(z.latitude))))) AS distance_in_km
  FROM zip AS z
  JOIN (   /* these are the query parameters */
        SELECT  42.81  AS latpoint,  -70.81 AS longpoint,
                50.0 AS radius,      111.045 AS distance_unit
    ) AS p ON 1=1
  WHERE z.latitude
     BETWEEN p.latpoint  - (p.radius / p.distance_unit)
         AND p.latpoint  + (p.radius / p.distance_unit)
    AND z.longitude
     BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
         AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
  ORDER BY distance_in_km
  LIMIT 15

Thank you very much for sharing! I have a question: Could I also use a HAVING clause instead of the second SELECT or is it much slower? I mean something like this:

 SELECT z.zip,
        z.primary_city,
        z.latitude, z.longitude,
        p.radius,
        p.distance_unit
                 * DEGREES(ACOS(LEAST(1.0, COS(RADIANS(p.latpoint))
                 * COS(RADIANS(z.latitude))
                 * COS(RADIANS(p.longpoint - z.longitude))
                 + SIN(RADIANS(p.latpoint))
                 * SIN(RADIANS(z.latitude))))) AS distance
  FROM zip AS z
  JOIN (   /* these are the query parameters */
        SELECT  42.81  AS latpoint,  -70.81 AS longpoint,
                50.0 AS radius,      111.045 AS distance_unit
    ) AS p ON 1=1
  WHERE z.latitude
     BETWEEN p.latpoint  - (p.radius / p.distance_unit)
         AND p.latpoint  + (p.radius / p.distance_unit)
    AND z.longitude
     BETWEEN p.longpoint - (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
         AND p.longpoint + (p.radius / (p.distance_unit * COS(RADIANS(p.latpoint))))
 HAVING distance <= radius
 ORDER BY distance
 LIMIT 15

What is the better approach?

Edit: This StackOverflow answer says there shouldn't be any difference regarding perfomance.

@LostInDarkMath in my case I had to use HAVING to take advantage of the indices.