gpt4 book ai didi

sql - CTE 未按预期表现的每周汇总

转载 作者:行者123 更新时间:2023-12-01 11:46:42 25 4
gpt4 key购买 nike

我有这个 USERS 表,其中的用户可以是两种不同类型(A 和 B)。我需要显示一份报告,其中包含每周每种类型的汇总。到目前为止,我的查询运行良好,只是有几周没有正确分组。在下面的示例中,从 1 月 28 日开始的一周应该有一行,而不是两行。

Week Starts |Week| Type A | Type B
------------+----+--------+------
2013-02-04 | 14 | 2 | 26
2013-01-28 | 13 | 5 | 191
2013-01-28 | 13 | 0 | 24
2013-01-21 | 12 | 1 | 134
2013-01-21 | 12 | 0 | 20
2013-01-14 | 11 | 1 | 143
2013-01-14 | 11 | 0 | 2
2013-01-07 | 10 | 0 | 233
2013-01-07 | 10 | 0 | 23
2012-12-31 | 9 | 0 | 12
2012-12-31 | 9 | 4 | 164
2012-12-31 | 9 | 0 | 20

SQL

;with cte as
(
select DATEADD(m,-3,GETDATE()) firstday, DATEADD(m,-3,GETDATE()) + 6 - DATEDIFF(day, 0, DATEADD(m,-3,GETDATE())) %7 lastday, 1 week
union all
select lastday + 1, case when GETDATE() < lastday + 7 then GETDATE() else lastday + 7 end, week + 1
from cte
where lastday < GETDATE()
)
SELECT
cast(firstday as date) 'Week Starts',
cte.week as 'Week',
Sum(CASE WHEN USR_TYPE = 'A' THEN 1 ELSE 0 END) As 'Type A',
Sum(CASE WHEN USR_TYPE = 'B' THEN 1 ELSE 0 END) As 'Type B'
FROM cte left join USERS
ON cte.firstday <= USERS.CREATED
AND cte.lastday > USERS.CREATED
GROUP BY cte.week, cte.firstday, cte.lastday, DATEPART(YEAR,USERS.CREATED), DATEPART(wk,USERS.CREATED)
ORDER BY week desc

我做错了什么?

最佳答案

在没有看到 users 表中的任何数据的情况下,我将进行猜测。

您在 CTE 中生成的日期列表包括时间

您可能需要 cast() 您的 firstdaylastday 值作为 date 或生成没有时间列出。

查看 SQL Fiddle Demo

来自您的 CTE 和新日期的样本:

| CASTFIRSTDAY | CASTLASTDAY | WEEK |                        FIRSTDAY |                         LASTDAY |
---------------------------------------------------------------------------------------------------------
| 2012-11-05 | 2012-11-11 | 1 | November, 05 2012 20:08:10+0000 | November, 11 2012 20:08:10+0000 |
| 2012-11-12 | 2012-11-18 | 2 | November, 12 2012 20:08:10+0000 | November, 18 2012 20:08:10+0000 |
| 2012-11-19 | 2012-11-25 | 3 | November, 19 2012 20:08:10+0000 | November, 25 2012 20:08:10+0000 |
| 2012-11-26 | 2012-12-02 | 4 | November, 26 2012 20:08:10+0000 | December, 02 2012 20:08:10+0000 |
| 2012-12-03 | 2012-12-09 | 5 | December, 03 2012 20:08:10+0000 | December, 09 2012 20:08:10+0000 |
| 2012-12-10 | 2012-12-16 | 6 | December, 10 2012 20:08:10+0000 | December, 16 2012 20:08:10+0000 |

您可能想要编辑您的 CTE 以仅返回 date 值:

;with cte as
(
select
cast(DATEADD(m,-3,GETDATE()) as date) firstday,
cast(DATEADD(m,-3,GETDATE()) + 6 - DATEDIFF(day, 0, DATEADD(m,-3,GETDATE())) %7 as DATE) lastday,
1 week
union all
select
cast(DATEADD(DAY, 1, lastday) as date),
case
when cast(GETDATE() as date) < cast(DATEADD(DAY, 7, lastday) as date)
then cast(GETDATE() as date)
else cast(DATEADD(DAY, 7, lastday) as date)
end,
week + 1
from cte
where cast(lastday as date) < cast(GETDATE() as date)
)
select *
from cte

参见 SQL Fiddle with Demo

关于sql - CTE 未按预期表现的每周汇总,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14715911/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com