Question

我想计算一个月中4周的总行数，然后按位置将其分组，结果应类似于：

location week1 week2 week3 week4
Floor A  10    0     3     4
Floor B  5     2     0     0
...

我试图通过给定日期来获得星期数。

CASE WHEN DAY(create_date) >= 1 AND DAY(create_date) < 8 THEN 1 
WHEN DAY(create_date) >= 8 AND DAY(create_date) < 15 THEN 2
WHEN DAY(create_date) >= 15 AND DAY(create_date) < 22 THEN 3
ELSE 4 END AS WeekNumber

这给了我这个结果：

+----+-----------+--------------+------------+
| id | create_by | create_date  | WeekNumber |
+----+-----------+--------------+------------+
|  1 |    555001 | '2018-11-01' |          1 |
|  2 |    555002 | '2018-11-05' |          1 |
|  3 |    555004 | '2018-11-06' |          1 |
|  4 |    555001 | '2018-11-10' |          2 |
|  5 |    555003 | '2018-11-17' |          3 |
|  6 |    555002 | '2018-11-17' |          3 |
|  7 |    555001 | '2018-11-18' |          3 |
|  8 |    555003 | '2018-11-20' |          3 |
|  9 |    555001 | '2018-11-22' |          4 |
| 10 |    555001 | '2018-11-25' |          4 |
+----+-----------+--------------+------------+

我的桌子

tbl_user

+----+----------+----------+
| id | username | location |
+----+----------+----------+
|  1 |   555001 | Floor A  |
|  2 |   555002 | Floor B  |
|  3 |   555003 | Floor C  |
|  4 |   555004 | Floor A  |
|  5 |   555005 | Floor C  |
+----+----------+----------+

tbl_list

+----+-----------+--------------+
| id | create_by | create_date  |
+----+-----------+--------------+
|  1 |    555001 | '2018-11-01' |
|  2 |    555002 | '2018-11-05' |
|  3 |    555004 | '2018-11-06' |
|  4 |    555001 | '2018-11-10' |
|  5 |    555003 | '2018-11-17' |
|  6 |    555002 | '2018-11-17' |
|  7 |    555001 | '2018-11-18' |
|  8 |    555003 | '2018-11-20' |
|  9 |    555001 | '2018-11-22' |
| 10 |    555001 | '2018-11-25' |
+----+-----------+--------------+

Answer 1

您可以尝试使用数据透视

with cte as
(
select *,location,
CASE WHEN DAY(create_date) >= 1 AND DAY(create_date) < 8 THEN 1 
WHEN DAY(create_date) >= 8 AND DAY(create_date) < 15 THEN 2
WHEN DAY(create_date) >= 15 AND DAY(create_date) < 22 THEN 3
ELSE 4 END AS WeekNumber
from tbl_list a left join tbl_user b on a.create_by=b.username
)

select * from
(select * from cte
pivot 
(
count(create_by) for WeekNumber in ([1],[2],[3],[4])
) as pv)A left join 
(select location,count(username) countofUser from tbl_user group by location)B on 
A.location=B.location

Answer 2

我只使用条件聚合：

select u.location,
       sum(case when day(l.create_date) >= 1 and day(l.create_date) < 8 then 1 else 0
           end) as week1,
       sum(case when day(l.create_date) >= 8 and day(l.create_date) < 15 then 1 else 0
           end) as week2,
       sum(case when day(l.create_date) >= 15 and day(l.create_date) < 22 then 1 else 0
           end) as week3,
       sum(case when day(l.create_date) >= 22 and day(l.create_date) < 29 then 1 else 0
           end) as week4
from tbl_list l join
     tbl_user u
     on l.create_by = u.username
group by u.location;

这似乎比尝试使用pivot简单得多。此逻辑完全不需要子查询。

SQL Server计数记录按月份和位置按4周分组

2 个答案: