简化 sql 查询（日期到期间）

Question

得到这张表： initial table

需要得到这样的表： Target table

我用了什么：

Select ID_CLIENT, BALANCE, HIST_DT as Start_dt, isnull(lead(hist_dt) over (partition by id_client order by hist_dt asc), '2999.12.31') as End_dt
from (
select ID_CLIENT, ID_STATUS, balance, hist_dt, lag(id_status) over (partition by id_client order by id_status) as Prev_ID_status
from Client_History) a
where a.ID_STATUS = a.Prev_ID_status or a.ID_STATUS = 1
order by ID_CLIENT, HIST_DT

我认为它非常复杂。很高兴听到任何简化此查询的建议。

Answer 1

这是一个间隙和孤岛问题，最容易通过行数和聚合的差异来解决：

select id_client, balance, min(hist_dt), max(hist_dt)
from (select ch.*,
             row_number() over (partition by id_client order by balance_hist_dt) as seqnum,
             row_number() over (partition by id_client, balance order by balance_hist_dt) as seqnum_2
      from client_history ch
     ) ch
group by id_client, balance, (seqnum - seqnum_2)
order by id_client, min(hist_dt);

为什么这行得通有点难以描述。但是如果您查看子查询的结果，您会发现差异如何捕获具有相同余额的相邻行。