mysql: group by ID, get highest priority per each ID

Question

I have the following mysql table called "pics", with the following fields and sample data:

id   vehicle_id    filename    priority
1    45            a.jpg       4
2    45            b.jpg       1
3    56            f.jpg       4
4    67            cc.jpg      4
5    45            kt.jpg      3
6    67            gg.jpg      1

Is it possible, in a single query, to get one row for each vehicle_id, and the row be the highest priority?

The result I'm looking for:

array (
  [0] => array( [id] => '2', [vehicle_id] => '45', [filename] => 'b.jpg',  [priority] => '1' ),
  [1] => array( [id] => '3', [vehicle_id] => '56', [filename] => 'f.jpg',  [priority] => '4' ),
  [2] => array( [id] => '6', [vehicle_id] => '67', [filename] => 'gg.jpg', [priority] => '1' )
);

If not possible in a single query, what would be the best approach?

Thanks!

Answer 1

This seems to be a typical max per group query. In most databases you could easily do this using ROW_NUMBER:

SELECT id, vehicle_id, filename, priority
FROM (
    SELECT 
        id, vehicle_id, filename, priority,
        ROW_NUMBER() OVER (PARTITION BY vehicle_id
                           ORDER BY priority DESC, id) AS rn
    FROM pics
) AS T1
WHERE rn = 1

Since MySQL doesn't yet support ROW_NUMBER you can emulate it using variables:

SELECT id, vehicle_id, filename, priority
FROM (
    SELECT
        id, vehicle_id, filename, priority,
        @rn := CASE WHEN @prev_vehicle_id = vehicle_id
                    THEN @rn + 1
                    ELSE 1
               END AS rn,
        @prev_vehicle_id := vehicle_id
    FROM (SELECT @prev_vehicle_id := NULL) vars, pics T1
    ORDER BY vehicle_id, priority DESC, id
) T2
WHERE rn = 1

Answer 2

Note that this method was posted before it was made clear that it is required to handle priority ties. I'm leaving it here for reference (see comments below). Check out @Mark's answer for a solution that handles ties as required:

SELECT p.id, p.vehicle_id, p.filename, p.priority
FROM   pics p
JOIN   (
           SELECT   vehicle_id, MAX(priority) max_priority
           FROM     pics
           GROUP BY vehicle_id
       ) sub_p ON (sub_p.vehicle_id = p.vehicle_id AND 
                   sub_p.max_priority = p.priority)
GROUP BY p.vehicle_id;

This assumes that there can be no priority ties for the same vehicle_id.

Test case:

CREATE TABLE pics (id int, vehicle_id int, filename varchar(10), priority int);

INSERT INTO pics VALUES ('1', '45', 'a.jpg', '4');
INSERT INTO pics VALUES ('2', '45', 'b.jpg', '1');
INSERT INTO pics VALUES ('3', '56', 'f.jpg', '4');
INSERT INTO pics VALUES ('4', '67', 'cc.jpg', '4');
INSERT INTO pics VALUES ('5', '45', 'kt.jpg', '3');
INSERT INTO pics VALUES ('6', '67', 'gg.jpg', '1');

Result:

+------+------------+----------+----------+
| id   | vehicle_id | filename | priority |
+------+------------+----------+----------+
|    1 |         45 | a.jpg    |        4 |
|    3 |         56 | f.jpg    |        4 |
|    4 |         67 | cc.jpg   |        4 |
+------+------------+----------+----------+
3 rows in set (0.01 sec)

Answer 3

While this may be the 'accepted' answer, the performance of Mark's solution is under normal circumstances many times better, and equally valid for the question, so by all means, go for his solution in production!

---

SELECT a.id, a.vehicle_id, a.filename, a.priority
FROM pics a
LEFT JOIN pics b               -- JOIN for priority
ON b.vehicle_id = a.vehicle_id 
AND b.priority > a.priority
LEFT JOIN pics c               -- JOIN for priority ties
ON c.vehicle_id = a.vehicle_id 
AND c.priority = a.priority 
AND c.id < a.id
WHERE b.id IS NULL AND c.id IS NULL

Assuming 'id' is a non-nullable column.

[edit]: my bad, need second join, cannot do it with just one.