SQL - Unqiue results based on criteria - mysql

Question

Making a dealer locator where people search for a dealer near them. I want it to work in such a way that if a dealer has multiple branches near the person, only the closest branch shows. So the "name" field should be unique in the results with the result shown having the least distance to the person searching versus other rows that have the same "name" field. I also want only the closest 5 dealers shown ordered by their dealer level, or medal. Right now I have the following:
$query = sprintf("SELECT
name, address, contact, image, medal, phone, email, website, lat, lng,
( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM dealers
HAVING distance < 60
ORDER BY medal, distance
LIMIT 0 , 5",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($radius));
$result = mysql_query($query);
From what I've read, it sounds like I need to do something like:
SELECT * FROM (
SELECT ROW_NUMBER() OVER (PARTITION BY name ORDER BY distance) AS num
FROM dealers)a
WHERE a.num = 1
Or something like that, but I can't get it to work right. Any insights on how I can get this to function would be greatly appreciated.

Answer 1

Get the minimum distance per dealer and of these take the first five. Then select again from the table in order to get complete records, but take only those records already identified by dealer and distance.
A WITH clause would be helpful, but MySQL doesn't support it. Well, ...
SELECT
name, address, contact, image, medal, phone, email, website, lat, lng,
( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM dealers
WHERE (name, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) )) IN
(
SELECT
name, min( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM dealers
GROUP BY name
HAVING distance < 60
ORDER BY distance
LIMIT 5
)
ORDER BY distance, medal;

Answer 2

Final Solution was:
SELECT name, address, contact, image, medal, phone, email, website, lat, lng,
( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM dealers
WHERE (name,
( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) )
IN
(SELECT name, distance
FROM (SELECT name,
MIN( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM dealers
WHERE (3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) < 90
GROUP BY name)
t)
ORDER BY medal, distance
LIMIT 0, 5",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat)
Thank you Thorsten Kettner, for putting me on the right track!