Question

How can I select one, most recent NID, per every 7 days, per UID, starting count back from today.

If today is July 11, the following table

+-----+------------+-----+
| NID |  timestamp | UID |
+-----+------------+-----+
|   1 | 1341719851 |   8 | //July 7
|   2 | 1341115051 |   8 | //July 1
|   3 | 1341547051 |   8 | //July 6
|   4 | 1341719851 |   8 | //July 8
|   5 | 1341979051 |   8 | //July 11
|   6 | 1341806251 |   9 | //July 9
|   7 | 1341460651 |   9 | //July 5
|   8 | 1341892651 |   9 | //July 10
+-----+------------+-----+

Will output this:

+-----+------------+-----+
| NID |  timestamp | UID |
+-----+------------+-----+
|   2 | 1341115051 |   8 | //July 1
|   5 | 1341979051 |   8 | //July 11
|   8 | 1341892651 |   9 | //July 10
+-----+------------+-----+

In the last 7 days, most recent NID for each user is '5' and '8', in the prior 7 days, most recent NID is '2', and so on…

I’m assuming, Group By will do the trick; but I don’t have a clue where to start.

UPDATE

This is the query that worked, based on the top answer:

SELECT nid, timestamp, uid, weeks_ago
FROM (
    SELECT nid, timestamp, uid, FLOOR(
        (UNIX_TIMESTAMP()- timestamp)/604800
    ) weeks_ago
    FROM `table`
    ORDER BY timestamp DESC
) x
GROUP BY uid, weeks_ago 

Answer 1

select nid, max(timestamp), uid, weeks_ago
from (select nid, timestamp, uid, floor(datediff(now(), from_unixtime(timestamp))/7) weeks_ago
      from mytable) x
group by nid, uid, weeks_ago