Question

I have read that grouping happens before ordering, is there any way that I can order first before grouping without having to wrap my whole query around another query just to do this?

Let’s say I have this data:

id  | user_id | date_recorded
1   | 1       | 2011-11-07
2   | 1       | 2011-11-05    
3   | 1       | 2011-11-06
4   | 2       | 2011-11-03
5   | 2       | 2011-11-06

Normally, I’d have to do this query in order to get what I want:

SELECT
*
FROM (
    SELECT * FROM table ORDER BY date_recorded DESC
) t1
GROUP BY t1.user_id

But I’m wondering if there’s a better solution.

Answer 1

Your question is somewhat unclear but I have a suspicion what you really want is not any GROUP aggregates at all, but rather ordering by date first, then user ID:

SELECT
  id, 
  user_id, 
  date_recorded
FROM tbl
ORDER BY date_recorded DESC, user_id ASC

Here would be the result. Note reordering by date_recorded from your original example

id  | user_id | date_recorded
1   | 1       | 2011-11-07
3   | 1       | 2011-11-06
2   | 1       | 2011-11-05    
5   | 2       | 2011-11-06
4   | 2       | 2011-11-03

Update

To retrieve the full latest record per user_id, a JOIN is needed. The subquery (mx) locates the latest date_recorded per user_id, and that result is joined to the full table to retrieve the remaining columns.

SELECT 
  mx.user_id,
  mx.maxdate,
  t.id
FROM (
  SELECT
    user_id,
    MAX(date_recorded) AS maxdate
  FROM tbl
  GROUP BY user_id
) mx JOIN tbl t ON mx.user_id = t.user_id AND mx.date_recorded = t.date_recorded