Help structure query to list concerts/venues by distance - sql

Question

SQL noob here needing some help. I've got an idea of how to do this in PHP/SQL, but I would really like to condense this into one SELECT statement. OK:
The site I am working on is a list of concerts and venues. Venues have a latitude and longitude, and so do accounts corresponding to that users location.
I have three tables, accounts (users), concerts, I would like to SELECT a list of concerts (and join on venues for that info) that are happening at venues within x miles of the account, using this cheap formula for distance calculation (the site only lists venues in the UK so the error is acceptable):
x = 69.1 * (accountLatitude - venueLatitude);
y = 69.1 * (accountLongitude - venueLongitude) * cos(venueLatitude / 57.3);
distance = sqrt(x * x + y * y);
How can I achieve this in a single query?
Thanks in advance xD

Answer 1

This is done exactly as your formulas suggest.
Just substitute x and y into the distance formula.
If this is for MySQL the below should work (just replace the correct table names / column names).
SELECT concert.name, venue.name,
SQRT(POW(69.1 * (account.Latitude - venue.Latitude), 2) + POW(69.1 * (account.Longitude - venue.Longitude) * cos(venue.Latitude / 57.3), 2)) AS distance
FROM account, venue
LEFT JOIN concert ON conert.ID = venue.concertID
WHERE account.id = UserWhoIsLoggedIn
ORDER BY 3;
This should return all concert names, venue names and distance from user order by the distance.
If you are not using MySQL you may need to change either the POW or SQRT functions.
Also be aware that sin & cos functions take there inputs in Radians.

Answer 2

For everyone elses benefit heres now I did it in the end:
$q = 'SELECT gigs.date, bands.idbands, venues.latitude, venues.longitude, venues.idvenues, bands.name AS band, venues.name AS venue,
SQRT(POW(69.1 * (' . $lat . ' - venues.latitude), 2) + POW(69.1 * (' . $lon . ' - venues.longitude) * cos(venues.latitude / 57.3), 2)) AS distance
FROM gigs
LEFT JOIN bands ON bands.idbands=gigs.bands_idbands
LEFT JOIN venues ON venues.idvenues=gigs.venues_idvenues
WHERE 1
ORDER BY distance';
where $lat and $lon are the latitude and longitude of the user currently logged in!
this selects every single gig thats happening at every venue, and arranges them in order of distance