备注:测试数据库版本为MySQL 8.0

一.问题:

求今年是否为闰年

二.解决方案

根据2月份最后一天来判断

select  day(
        last_day(
        date_add(
        date_add(
        date_add(current_date,
                 interval - dayofyear(current_date) day),
                 interval 1 day),
                 interval 1 month))) dy;

测试记录

mysql> select  day(
    ->         last_day(
    ->         date_add(
    ->         date_add(
    ->         date_add(current_date,
    ->                  interval - dayofyear(current_date) day),
    ->                  interval 1 day),
    ->                  interval 1 month))) dy;
+------+
| dy   |
+------+
|   29 |
+------+
1 row in set (0.01 sec)

这样看起来有点麻烦，分解开
dayofyear(current_date) -- 求出今天是本年的第多少天
date_add(date, interval - n day) -- 今天的天数减去截止当前是今年多少天 得出上年的12月31日
date_add(date, interval + 1 day) -- 得出今年1月的第1天
date_add(date, interval + 1 month) -- 得出今年2月的第一天
last_day(date) -- 得出2月份最后一天
day(date) --求出日

测试记录:

mysql> select dayofyear(current_date);
+-------------------------+
| dayofyear(current_date) |
+-------------------------+
|                     289 |
+-------------------------+
1 row in set (0.00 sec)

mysql> select date_add(current_date,interval -289 day);
+------------------------------------------+
| date_add(current_date,interval -289 day) |
+------------------------------------------+
| 2019-12-31                               |
+------------------------------------------+
1 row in set (0.00 sec)

mysql>
mysql> select date_add('2019-12-31',interval 1 day);
+---------------------------------------+
| date_add('2019-12-31',interval 1 day) |
+---------------------------------------+
| 2020-01-01                            |
+---------------------------------------+
1 row in set (0.00 sec)

mysql>
mysql> select date_add('2020-01-01',interval 1 month);
+-----------------------------------------+
| date_add('2020-01-01',interval 1 month) |
+-----------------------------------------+
| 2020-02-01                              |
+-----------------------------------------+
1 row in set (0.00 sec)

mysql>
mysql> select last_day('2020-02-01');
+------------------------+
| last_day('2020-02-01') |
+------------------------+
| 2020-02-29             |
+------------------------+
1 row in set (0.00 sec)

mysql> select day('2020-02-29');
+-------------------+
| day('2020-02-29') |
+-------------------+
|                29 |
+-------------------+
1 row in set (0.00 sec)