如何从 Oracle 表中选择行组？

Question

我在 Oracle 表中的三列（时间戳、状态和值）中有数百万条数据。

从 temp_table rownum

1 2015-04-28T12:41:34.1616834Z 预订 100
2 2015-04-28T12:41:34.1888649Z 删除 100
3 2015-04-28T12:55:59.3789387Z 执行 200
4 2015-04-28T12:55:59.3825833Z 坚持 200
5 2015-04-28T12:55:59.3898336Z 书 200
6 2015-04-28T12:55:59.3903645Z 删除 200
7 2015-04-28T12:57:37.5718992Z 预订 200
8 2015-04-28T12:57:37.5723847Z 删除 200
9 2015-04-28T12:57:37.5725199Z 预订 300
10 2015-04-28T12:57:37.5728888Z 删除 300

我对按 Execute、Persist、Book 和 Delete 分组的四个时间戳列表感兴趣，这些时间戳在当天的每个 orderId 中一起出现。

我对第 7 行和第 8 行不感兴趣，即使 orderId 相同 [因为它们没有按 Execute、Persist、Book 和 Delete 分组]。

我可以通过解析函数来做到这一点吗？

谢谢

Answer 1

不是很优雅，但可以查询：

with t1 as (
  select tstamp ts, status st, orderid oid, 
      decode(status, 'Execute', 1, 'Persist', 2, 'Book', 3, 'Delete', 4, 5) ds
    from temp_table),
t2 as (
  select ts, st, oid, ds
    , decode(ds, 1, 1, lag(ds, ds-1) 
        over (partition by oid, trunc(ts) order by ts, ds)) d1
    , case when ds = 1 then lead(ds) over (partition by oid, trunc(ts) order by ts, ds)
           when ds = 2 then 2
           when ds = 3 then lag(ds) over (partition by oid, trunc(ts) order by ts, ds)
           when ds = 4 then lag(ds, 2) over (partition by oid, trunc(ts) order by ts, ds)
      end d2
    , case when ds = 1 then lead(ds, 2) over (partition by oid, trunc(ts) order by ts, ds)
           when ds = 2 then lead(ds) over (partition by oid, trunc(ts) order by ts, ds)
           when ds = 3 then 3
           when ds = 4 then lag(ds) over (partition by oid, trunc(ts) order by ts, ds)
      end d3
    , decode(ds, 4, 4, lead(ds, 4-ds) 
        over (partition by oid, trunc(ts) order by ts, ds)) d4
  from t1 )
select ts, st, oid from t2
  where (d1, d2, d3, d4) in ((1, 2, 3, 4))
  order by oid, ts, ds