如何按天统计前 7 天周活？

[killva4624]

比如下表，记录了日期和对应用户:

datetime,username
2023-01-01,UserA
2023-01-01,UserB
2023-01-02,UserA
2023-01-03,UserC
...

需求是按天统计出该天起前七天的周活，当天没有对应用户的话补 0 ，如：

2023-01-08(2023-01-01~2023-01-07), 7
2023-01-09(2023-01-02~2023-01-08), 10

....

[pota]

这种我感觉还是将每天的统计数据存起来直接拿好点

[oRbceGJ]

用偏移类开窗函数？

[Ericcccccccc]

没必要强行用纯 sql 做这个事情吧.

[aw2350]

parttition by ,按照日期汇总

[yuhu96]

能不能按照当前[date ，date - 7 ]作为一个分组字段来进行聚合啊。窗口函数感觉不适用在这里。

[byty18768]

搞一张存储自然日的日期表，excel 生成很快的
`
datetime,start_date,end_date
2023-01-08,2023-01-01,2023-01-07
2023-01-09,2023-01-02,2023-01-08
2023-01-10,2023-01-03,2023-01-09
`
笛卡尔积
`
select a. datetime ,count(distinct username)
from 日期表 a
join 数据表 b on b. datetime between a. start_date and a.end_date
group by 1
`

[ktqFDx9m2Bvfq3y4]

记录日活，主键：
日期_用户 Id 。没活的不记。然后周活就是最近七天的，Group By 用户 Id 即可。

[JKeita]

按天取数据存到一份文本文件中，然后直接用 linux 命令进行去重计数。

[killva4624]

@byty18768 直接用子查询生成这个这个自然日表也可以吧？

[YYYYMMDDHHSS]

with data as (

select '2023-01-01' as dt ,'UserA' as user_name union all
select '2023-01-01' as dt ,'UserB' as user_name union all
select '2023-01-02' as dt ,'UserA' as user_name union all
select '2023-01-03' as dt ,'UserC' as user_name union all
select '2023-01-04' as dt ,'UserA' as user_name union all
select '2023-01-05' as dt ,'UserB' as user_name union all
select '2023-01-06' as dt ,'UserA' as user_name union all
select '2023-01-07' as dt ,'UserC' as user_name union all
select '2023-01-09' as dt ,'UserA' as user_name union all
select '2023-01-10' as dt ,'UserB' as user_name union all
select '2023-01-11' as dt ,'UserA' as user_name union all
select '2023-01-12' as dt ,'UserC' as user_name union all
select '2023-01-13' as dt ,'UserA' as user_name union all
select '2023-01-14' as dt ,'UserB' as user_name union all
select '2023-01-15' as dt ,'UserA' as user_name union all
select '2023-01-16' as dt ,'UserC' as user_name union all
select '2023-01-17' as dt ,'UserA' as user_name union all
select '2023-01-18' as dt ,'UserB' as user_name
)
select
t.dt,
date_add(t.dt, 7) dt_to,
count(distinct t.user_name) as cnt
from
(
select
d1.dt,
d2.user_name
from
data d1
join data d2
where
1 = 1
--and d1.dt > d2.dt
and datediff(d1.dt, d2.dt) <= 7
) t
group by
dt
----

大数据量不建议这样做

[sunny1688]

@Chad0000 老哥的方案是最轻松还解耦的，一条简单的 sql 就行了

[vitoliu]

SQL 挺难想的，我反正比较难硬加出来。通过代码来构建逻辑简单清晰，灵活性更强。
统计某一个时段的周活，可以按天、小时统计，定时任务去跑，还可以无限加条件。比如用户注册时间小于 3 个月，判断新用户留存情况。

[28Sv0ngQfIE7Yloe]

这种不都是定时调度 ETL 做成指标放到数仓里吗？

[byty18768]

@killva4624 主要是你不是说可能会存在整天都没有用户的嘛，所以考虑导入表来补 0 。如果底层每天都是全的肯定没问题