根据不同的列值过滤日期行

Question

我一直在尝试根据 task_dts 的值过滤具有 code 特定值的行。本质上，我只想要 per id 行，其中 task_dts 时间戳落在当前 code 和 code_dts 的 code_dts 时间戳之间以下 code.

例如；对于 code 等于 'z' 的行，我只想要 task_dts 在值 'z' 和值 [=] 的 code_dts 范围内的行44=]。对于 code 等于 'y' 的行，我只想要 task_dts 在 code_dts 范围内的行，对于值 'y' 和值 'x'等

我的 table 看起来如下：

rowid	id	code	code_dts	task	task_dts
1	a	z	2022-02-01 10:17:08.403000	1	2022-02-01 10:21:27.000000
2	a	z	2022-02-01 10:17:08.403000	2	2022-02-01 10:21:31.000000
3	a	z	2022-02-01 10:17:08.403000	3	2022-02-01 12:41:43.000000
4	a	y	2022-02-01 11:12:13.270000	1	2022-02-01 10:21:27.000000
5	a	y	2022-02-01 11:12:13.270000	3	2022-02-01 12:41:43.000000
6	a	y	2022-02-01 11:12:13.270000	8	2022-02-21 14:57:53.000000
7	a	x	2022-02-21 12:28:50.647000	6	2022-02-21 14:57:53.000000
8	a	x	2022-02-21 12:28:50.647000	7	2022-02-21 14:57:54.000000
9	b	h	2022-04-05 13:44:16.030000	1	2022-04-05 14:03:56.570000
10	b	h	2022-04-05 13:44:16.030000	2	2022-04-05 14:03:56.570000
11	b	i	2022-04-06 13:44:16.030000	1	2022-04-05 14:03:56.570000
12	b	j	2022-04-07 13:44:16.030000	3	2022-04-05 14:03:56.570000

输出将如下所示：

rowid	id	code	code_dts	task	task_dts
1	a	z	2022-02-01 10:17:08.403000	1	2022-02-01 10:21:27.000000
2	a	z	2022-02-01 10:17:08.403000	2	2022-02-01 10:21:31.000000
5	a	y	2022-02-01 11:12:13.270000	3	2022-02-01 12:41:43.000000
7	a	x	2022-02-21 12:28:50.647000	6	2022-02-21 14:57:53.000000
8	a	x	2022-02-21 12:28:50.647000	7	2022-02-21 14:57:54.000000
10	b	h	2022-04-05 13:44:16.030000	2	2022-04-05 14:03:56.570000
11	b	i	2022-04-06 13:44:16.030000	1	2022-04-05 14:03:56.570000
12	b	j	2022-04-07 13:44:16.030000	3	2022-04-05 14:03:56.570000

我试过用 qualify 来解决这个问题，但没有成功。帮助将不胜感激

Answer 1

您可以使用 table 表达式来 pre-compute 时间戳范围。那么过滤就简单了。

例如：

select t.*
from t
join (
  select code, dt, lead(dt) over(order by dt) as next_dt
  from (select code, min(code_dts) as dt from t group by code) x
) y on t.code = y.code
where t.task_dts between y.dt and y.next_dt or y.next_dt is null

Answer 2

看了The Impaler的回答我终于明白你的要求了:-)

这与基于Window函数的相同逻辑：

with cte as
 ( 
   select t.*
     -- next code_dts, i.e. at least one row will return 
     -- the code_dts of the following code 
     ,lead(code_dts,1,task_dts) over (order by code_dts) as next_dts
   from tab as t
 )
select *
from cte
qualify task_dts between code_dts
                     -- assign the next code's dts to all rows within the same code
                     and max(next_dts) over (partition by code)
;

很难说哪个性能更好...

根据不同的列值过滤日期行

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