Hvis du kender dit input-år og -måned, kan du altid indstille den første dag, lad os sige, at inputtet er Y = 2012 M=02 , vil den første dag altid være 2012-02-01 og ved at bruge den dato kan du få den sidste dag og derefter datoerne i det interval. Noget som
select a.Date
from (
select last_day('2012-02-01') - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a
where a.Date between '2012-02-01' and last_day('2012-02-01') order by a.Date;
+------------+
| Date |
+------------+
| 2012-02-01 |
| 2012-02-02 |
| 2012-02-03 |
| 2012-02-04 |
| 2012-02-05 |
| 2012-02-06 |
| 2012-02-07 |
| 2012-02-08 |
| 2012-02-09 |
| 2012-02-10 |
| 2012-02-11 |
| 2012-02-12 |
| 2012-02-13 |
| 2012-02-14 |
| 2012-02-15 |
| 2012-02-16 |
| 2012-02-17 |
| 2012-02-18 |
| 2012-02-19 |
| 2012-02-20 |
| 2012-02-21 |
| 2012-02-22 |
| 2012-02-23 |
| 2012-02-24 |
| 2012-02-25 |
| 2012-02-26 |
| 2012-02-27 |
| 2012-02-28 |
| 2012-02-29 |
+------------+
29 rows in set (0.00 sec)