查询按降序排列时如何按事件日期升序排序 MySQL

Question

I have a column name 'dateEvent' in MySQL which stores date in the format 'Ym-d':

+-------------+
| dateEvent   |
|-------------+
| 2018-01-01  |
| 2022-01-02  |
| 2021-01-03  |
| 2019-01-04  |
| 2016-01-05  |
| 2018-01-06  |
| 2020-01-07  |
| 2017-01-08  |
| 2021-01-09  |
| 2015-01-10  |
| 2016-01-11  |
| 2019-01-12  |
| 2018-01-13  |
| 2021-01-14  |
+-------------+

I need to output 3 nearest day right before today (6-jan) regardless of the year (5-jan, 4-jan and 3-jan) and they be sorted in ascending order (3-jan, 4-jan, 5-jan)

My query

SELECT * FROM table WHERE DAYOFYEAR(dateEvent) < DAYOFYEAR(CURDATE()) ORDER BY DAYOFYEAR(dateEvent) DESC LIMIT 3")

Output:

2016-01-05 |                      | 2021-01-03
2019-01-04 | -->I want there are: | 2019-01-04
2021-01-03 |                      | 2016-01-05

When i change ORDER to ASC, output is

2018-01-01 |
2022-01-02 |  incorrect with my request
2021-01-03 |

How can I do? Any help will be appreciated.

Answer 1

Try to order by:

ORDER BY DATE_FORMAT(dateEvent, '%m-%d')

or

ORDER BY MONTH(dateEvent), DAYOFMONTH(dateEvent)

Answer 2

Well, the probably simplest way is to just get the LIMIT ing done in a derived table and apply another ORDER BY in the outer query.

SELECT *
       FROM (SELECT *
                    FROM elbat
                    WHERE dayofyear(dateevent) < dayofyear(curdate())
                    ORDER BY dayofyear(dateevent) DESC
                    LIMIT 3) AS x
       ORDER BY dayofyear(dateevent) ASC;

db<>fiddle