我有一个表,其中包含用户在任何给定日期的订阅状态。数据是这样的
+------------+------------+--------------+
| account_id | date | current_plan |
+------------+------------+--------------+
| 1 | 2019-08-01 | free |
| 1 | 2019-08-02 | free |
| 1 | 2019-08-03 | yearly |
| 1 | 2019-08-04 | yearly |
| 1 | 2019-08-05 | yearly |
| ... | | |
| 1 | 2020-08-02 | yearly |
| 1 | 2020-08-03 | free |
| 2 | 2019-08-01 | monthly |
| 2 | 2019-08-02 | monthly |
| ... | | |
| 2 | 2019-08-31 | monthly |
| 2 | 2019-09-01 | free |
| ... | | |
| 2 | 2019-11-26 | free |
| 2 | 2019-11-27 | monthly |
| ... | | |
| 2 | 2019-12-27 | monthly |
| 2 | 2019-12-28 | free |
| 3 | 2020-05-31 | monthly |
| 3 | 2020-06-01 | monthly |
| 4 | 2019-08-01 | yearly |
| ... | | |
| 4 | 2020-06-01 | yearly |
+------------+------------+--------------+
我想有一个表格,提供订阅的开始和结束日期。它看起来像这样。请注意,重要的是
account_ids 3
以及 4
未包含在此表中,因为截至今天(2020年6月1日),它们仍在订阅中。我只想要一份从订阅中搬出来的人的摘要。
+------------+------------+------------+-------------------+
| account_id | start_date | end_date | subscription_type |
+------------+------------+------------+-------------------+
| 1 | 2019-08-03 | 2020-08-02 | yearly |
| 2 | 2019-08-01 | 2019-08-31 | monthly |
| 2 | 2019-11-27 | 2019-12-27 | monthly |
+------------+------------+------------+-------------------+
目前,我有以下这是非常接近,但仍然给我的用户没有退出订阅
select account_id, current_plan, min(date), max(date)
from (select d.*,
row_number() over (partition by account_id order by date) as seqnum,
row_number() over (partition by account_id, current_plan order by date) as seqnum_2
from data d
) d
where current_plan not in ('free', 'trial')
group by account_id, current_plan, (seqnum - seqnum_2);
1条答案
按热度按时间mwecs4sa1#
如果您想对目前已退出的用户进行非常简单的筛选,只需添加:
having max(date)<current_date
但这也将包括以前的余波,如用户\u id=2的第一个余波但是如果你想向前看(比如userid=1)并且只过滤掉最后的余波,你就需要有一个更好的“缺口和孤岛”查询
lag
功能,如果你检查更多的“差距和岛屿”的解决方案,你会发现它。。。一般来说,lag(currrent_plan) over (partition by id order by date)
给你前一天的计划,这样你就可以确定辐射日期,然后在同一个窗口排序,得到每个id的最后一个