SQL按日期和月份选择查询顺序

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
}

And I have 4 records:

id date
1  2013-01-01
2  2013-02-02
3  2013-03-03
4  2013-04-04

I want to build a select query that would return all records in the following order (ordered by date, but upcoming dates first, passed dates appended to the end of the list) :

id date
3  2013-03-03 (upcoming dates first)
4  2013-04-04
1  2013-01-01 (passed dates appended to the end of the list)
2  2013-02-02

All records have the same year. In fact, year is not important, only day and month are. If you can suggest a better structure, you are very welcome.

Answer 1

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

将给出3,4,1,2的顺序

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

如果您使用的是MySQL，我会说只使用RIGHT ，但以下内容应在psql上运行：

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