我的表中有4列,rii、uii、rdi和udi。如下所示:

+----------+------+----------+------+
|       rdi|   rii|       udi|   uii|
+----------+------+----------+------+
|2002-02-06|1376.Q|2002-02-06|1376.Q|
|2002-02-28|1376.Q|2002-02-28|1376.Q|
|2002-03-06|1376.Q|2002-03-06|1376.Q|
|2002-02-01|1792.T|2002-02-01|1792.T|
|2002-03-07|1802.T|2002-03-07|1802.T|
|2002-03-08|1802.T|2002-03-08|1802.T|
|2002-04-03|1802.T|2002-04-03|1802.T|
|2002-03-07|1805.T|2002-03-07|1805.T|
|2002-02-18|1810.T|2002-02-18|1810.T|
|2002-03-22|1821.T|2002-03-22|1821.T|
|2002-02-27|1862.T|2002-02-27|1862.T|
|2002-04-11|1878.T|2002-04-11|1878.T|
|2002-04-18|1884.T|2002-04-18|1884.T|
|2002-02-27|1899.T|2002-02-27|1899.T|
|2002-03-11|1924.T|2002-03-11|1924.T|
|2002-02-05|1925.T|2002-02-05|1925.T|
|2002-01-23|1926.T|2002-01-23|1926.T|
|2002-03-19|1926.T|2002-03-19|1926.T|
|2002-01-25|1942.T|2002-01-25|1942.T|
|2002-01-31|1942.T|2002-01-31|1942.T|
+----------+------+----------+------+

我只想得到一个逻辑上的唯一rii的数量,比如如果我把lookback设为2,那么它应该在一组天内给出唯一的rii的数量(在那个特定的rdi上,在rdi的前两天)
因此,我将回首作为2,我的结果应该是这样的(对于rdi=2002-02-06,它应该在rdi in(2002-02-062002-02-052002-02-04)中找到唯一的rii)
+----------+-------------+----------+------+
|       rdi|          rii|       udi|   uii|
+----------+-------------+----------+------+
|2002-02-06|1376.Q,1925.T|2002-02-06|1376.Q|

我试过下面的查询,但没有得到所需的o/p
select count(distinct uii) as u,
  rdi,
  (select count(distinct rii) from `mytable` where rdi between DATE_SUB(rdi, INTERVAL 2 DAY) AND rdi) as r
  from `mytable`
  group by rdi
  order by rdi;

检查我的小提琴here

最佳答案

您可以使用LEFT JOIN将每个记录与前几天的记录关联:

select t1.rdi,
       group_concat(t2.rii) as rii,
       t1.udi,
       count(distinct t2.uii)
from `mytable` as t1
left join `mytable` as t2
   on t2.rdi between DATE_SUB(t1.rdi, INTERVAL 2 DAY) AND t1.rdi
group by rdi
order by rdi;

输出:
mysql - 获取一组日期(特定日期和前2天)中ID的不同计数-LMLPHP
Demo here

09-20 01:48