SQL select query order by day and month

Question

Let\'s pretend today is the 3rd of February.

And I have a table:

CREATE TABLE devotion
(
  id serial NOT NULL,
  date timestamp without time zone
}


        

Answer 1

If you were on MySQL i'd say just use RIGHT, but the below should work on psql:

SELECT * FROM devotion ORDER BY substring(to_char( date, 'YYYY-MM-DD') from char_length(to_char( date, 'YYYY-MM-DD')) - 5)

Answer 2

Simpler:

ORDER BY (date1 < now()), date1

You can just order by the boolean value of the expression (date1 < now()).
FALSE sorts before TRUE sorts before NULL.

Answer 3

Here is another solution. Ofcourse above answer is good enough. You can't assume today is Feb 3rd since we are taking Now() which stand for Feb 2nd. So I used demo data 2013-02-01. And this answer is mainly depending on your ID. To say that ID is sequential so is the date. Anyone is free to comment on the dodgy part of this logic..

MYSQL DEMO

select id, `date`,
case when `date` > Date_Format(Now(),'%Y-%m-%d')
then id-3
else id+2 end as x
from demo
order by x asc
;

| ID |                            DATE | X |
--------------------------------------------
|  3 |    March, 03 2013 00:00:00+0000 | 0 |
|  4 |    April, 04 2013 00:00:00+0000 | 1 |
|  1 |  January, 01 2013 00:00:00+0000 | 3 |
|  2 | February, 01 2013 00:00:00+0000 | 4 |

Answer 4

order by case when date1 > now() then 0 else 1 end case, date1

will give order of 3,4,1,2