Question

我想在SELECT上运行table查询，该查询会获取year_end字段中存储的日期，并自动将其视为过去12个月内该日期的发生

即：今天是5月14日所以：

如果我有2012-01-01，那么我希望保持不变

如果我有2010-05-05，那么我想要2012-05-05

如果我有2012-10-10，那么我想要2011-10-10

我需要SELECT的任何想法吗？

Answer 1

以下是您的需求：

select
concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',
right(@mydate,5)),right(@mydate,5))) dttm
from
(
    select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    (select substr(date(@mydate),6) my_md) AA,
    (select substr(date(now()),6) md,year(now()) yr) BB
) A;

以下是您选择的三个日期作为示例+ 2个闰年日期

set @mydate = '2012-01-01';
set @mydate = '2010-05-05';
set @mydate = '2012-10-10';
set @mydate = '2008-02-29';
set @mydate = '2012-02-29';

这里他们被执行

的MySQL＆GT;设置@mydate ='2012-01-01'; 查询正常，0行受影响（0.00秒）

mysql> select
    -> concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',right(@mydate,5)),right(@mydate,5))) dttm
    -> from
    -> (
    ->     select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    ->     IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    ->     (select substr(date(@mydate),6) my_md) AA,
    ->     (select substr(date(now()),6) md,year(now()) yr) BB
    -> ) A;
+------------+
| dttm       |
+------------+
| 2012-01-01 |
+------------+
1 row in set (0.00 sec)

mysql> set @mydate = '2010-05-05';
Query OK, 0 rows affected (0.00 sec)

mysql> select
    -> concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',right(@mydate,5)),right(@mydate,5))) dttm
    -> from
    -> (
    ->     select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    ->     IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    ->     (select substr(date(@mydate),6) my_md) AA,
    ->     (select substr(date(now()),6) md,year(now()) yr) BB
    -> ) A;
+------------+
| dttm       |
+------------+
| 2012-05-05 |
+------------+
1 row in set (0.00 sec)

mysql> set @mydate = '2012-10-10';
Query OK, 0 rows affected (0.00 sec)

mysql> select
    -> concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',right(@mydate,5)),right(@mydate,5))) dttm
    -> from
    -> (
    ->     select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    ->     IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    ->     (select substr(date(@mydate),6) my_md) AA,
    ->     (select substr(date(now()),6) md,year(now()) yr) BB
    -> ) A;
+------------+
| dttm       |
+------------+
| 2012-10-10 |
+------------+
1 row in set (0.00 sec)

mysql> set @mydate = '2008-02-29';
Query OK, 0 rows affected (0.00 sec)

mysql> select
    -> concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',right(@mydate,5)),right(@mydate,5))) dttm
    -> from
    -> (
    ->     select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    ->     IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    ->     (select substr(date(@mydate),6) my_md) AA,
    ->     (select substr(date(now()),6) md,year(now()) yr) BB
    -> ) A;
+------------+
| dttm       |
+------------+
| 2012-02-29 |
+------------+
1 row in set (0.00 sec)

mysql> set @mydate = '2012-02-29';
Query OK, 0 rows affected (0.00 sec)

mysql> select
    -> concat(yr,'-',if(leapyear=0,if(right(dt,5)='02-29','03-01',right(@mydate,5)),right(@mydate,5))) dttm
    -> from
    -> (
    ->     select concat(if(my_md > md,yr -1,yr),'-',my_md) dt,yr,my_md,
    ->     IF(MOD(yr,4)>0,0,IF(MOD(yr,100),1,(MOD(yr,400)=0))) leapyear FROM
    ->     (select substr(date(@mydate),6) my_md) AA,
    ->     (select substr(date(now()),6) md,year(now()) yr) BB
    -> ) A;
+------------+
| dttm       |
+------------+
| 2012-02-29 |
+------------+
1 row in set (0.00 sec)

mysql>

试一试!!!

Answer 2

Probbaly效率低下的查询，关于性能，但它有效：

SELECT 
    dateX + INTERVAL ( YEAR(CURDATE())
                     - YEAR(dateX) 
                     - ( ( MONTH(dateX), DAY(dateX) ) 
                       > ( MONTH(CURDATE()), DAY(CURDATE()) )
                       )
                     ) YEAR 
      AS calc_date
FROM 
    tableX ;

您可以对其进行测试SQL-Fiddle。另一个版本，稍微更具可读性：

SELECT 
    d + INTERVAL ( now.yy - t.yy - ((t.mm, t.dd) > (now.mm, now.dd))
                 ) YEAR 
      AS calc_date
FROM
    ( SELECT 
          dateX        AS d 
        , YEAR(dateX)  AS yy
        , MONTH(dateX) AS mm
        , DAY(dateX)   AS dd           
      FROM tableX
    ) AS t
  CROSS JOIN
    ( SELECT 
          YEAR(CURDATE())  AS yy
        , MONTH(CURDATE()) AS mm
        , DAY(CURDATE())   AS dd            
    ) AS now ;

将MySQL中的日期视为过去12个月，无论存储的实际年份如何

2 个答案: