Question

#Oracle 我有一个与日期相关的记录，我会在两个日期D1和D2之间按周分组。

我的会议桌是，

id_Meeting | meeting_date 
-----------+--------------
1          | 07/04/2017
2          | 10/06/2017  
3          | 17/04/2017
4          | 08/05/2017
5          | 19/04/2017
6          | 12/05/2017
7          | 06/04/2017
8          | 04/04/2017

现在我打印Date1 = 01/04/2017和Date1 = 01/05 / 2017之间的所有行，

Week |  nb_meeting_by_week
-----+--------------------
w1   |  3
w2   |  0
w3   |  2
w4   |  3

有什么想法吗？

我尝试的是，

Select 
  TRUNC((TO_DATE(m.meeting_date, 'YYYYMMDD') - TO_DATE('01/04/17', 'dd-MM-yy'))/7 +1)
    as week, 
  count(*) as nb_meeting_by_week
from meeting m 
Where TO_DATE(m.meeting_date, 'YYYYMMDD') between '01/04/17' and '01/05/17' 
Group by
  TRUNC((TO_DATE(m.meeting_date, 'YYYYMMDD') - TO_DATE('01/04/17', 'dd-MM-yy'))/7 +1);

但我没有得到第二周[w2 |因为我本周没有开会。但是我在没有会议的所有星期都会有0个。

Answer 1

您可以这样做：

Select to_char(meeting_date, 'w'), count(*)
From your_table
Where date1 = your_date and date2 = your_date2
Group by to_char(meeting_date, 'w');

这是日期功能的使用。日期上有很多功能。

那个人会给你参数日期当月的周数。

Answer 2

您必须首先生成所有星期，因为它们不在您的数据库中（即表中没有每周的记录）。这是通过递归查询完成的。然后外部加入你的桌子。

with weeks(weekno, from_date, max_date) as
(
  select 1 as weekno, date '2017-04-01' as from_date, date '2017-05-01' as max_date
  from dual  
  union all
  select weekno + 1 as weekno, from_date + 7 as from_date, max_date
  from weeks
  where from_date + 7 <= max_date
)
select weeks.weekno, count(mbw.meeting_date)
from weeks
left outer join nb_meeting_by_week mbw
  on  mbw.meeting_date >= weeks.from_date 
  and mbw.meeting_date <  weeks.from_date + 7
  and mbw.meeting_date <= weeks.max_date
group by weeks.weekno
order by weeks.weekno;

Oracle SQL - 在两个日期D1和D2之间按周分组数据

2 个答案: