为每个表行创建 n 分钟间隔

Question

我正在尝试根据表格获取时间间隔。

我的源表是这样的：

ID	OTHER_DATA	TIME_BEG	TIME_END	DURATION
1	abcd	10:00	11:00	15
2	xyzt	16:00	17:00	30

期望的输出：

ID	OTHER_DATA	ITVL_BEG	ITVL_END
1	abcd	10:00	10:15
1	abcd	10:15	10:30
1	abcd	10:30	10:45
1	abcd	10:45	11:00
2	xyzt	16:00	16:30
2	xyzt	16:30	17:00

TIME_BEG 和 TIME_END 是 VARCHAR 列，但我也将它们设置为 DAY TO SECOND INTERVAL，此处未显示（分别为 TIME_BEG_INT 和 TIME_END_INT）。

基本上我需要在一个 SQL 中复制每一行 TRUNC (EXTRACT (DAY FROM 24 * 60 * (TIME_END_INT - TIME_BEG_INT)) / DURATION) 次并将 this*DURATION 添加到我的日期中。

感谢任何帮助。

Answer 1

如果您使用间隔：

CREATE TABLE table_name (ID, OTHER_DATA, TIME_BEG, TIME_END, DURATION) AS
SELECT 1, 'abcd', INTERVAL '10:00' HOUR TO MINUTE, INTERVAL '11:00' HOUR TO MINUTE, INTERVAL '15' MINUTE FROM DUAL UNION ALL
SELECT 2, 'xyzt', INTERVAL '16:00' HOUR TO MINUTE, INTERVAL '17:00' HOUR TO MINUTE, INTERVAL '30' MINUTE FROM DUAL;

那么你可以使用：

WITH range (ID, OTHER_DATA, TIME_BEG, TIME_INT_END, TIME_END, DURATION) AS (
  SELECT ID,
         OTHER_DATA,
         TIME_BEG,
         LEAST(time_beg + duration, time_end),
         TIME_END,
         DURATION
  FROM   table_name
UNION ALL
  SELECT ID,
         OTHER_DATA,
         TIME_INT_END,
         LEAST(time_int_end + duration, time_end),
         TIME_END,
         DURATION
  FROM   range
  WHERE  time_int_end < time_end
)
SEARCH DEPTH FIRST BY id SET id_order
SELECT ID,
       OTHER_DATA,
       TIME_BEG AS itvl_beg,
       TIME_INT_END AS itvl_end
FROM   range;

哪些输出：

ID	OTHER_DATA	ITVL_BEG	ITVL_END
1	abcd	+000000000 10:00:00.000000000	+000000000 10:15:00.000000000
1	abcd	+000000000 10:15:00.000000000	+000000000 10:30:00.000000000
1	abcd	+000000000 10:30:00.000000000	+000000000 10:45:00.000000000
1	abcd	+000000000 10:45:00.000000000	+000000000 11:00:00.000000000
2	xyzt	+000000000 16:00:00.000000000	+000000000 16:30:00.000000000
2	xyzt	+000000000 16:30:00.000000000	+000000000 17:00:00.000000000

---

如果您将值作为字符串，那么您可以先将它们转换为间隔：

CREATE TABLE table_name (ID, OTHER_DATA, TIME_BEG, TIME_END, DURATION) AS
SELECT 1, 'abcd', '10:00', '11:00', 15 FROM DUAL UNION ALL
SELECT 2, 'xyzt', '16:00', '17:00', 30 FROM DUAL;

WITH data(ID, OTHER_DATA, TIME_BEG, TIME_END, DURATION) AS (
  SELECT ID,
         OTHER_DATA,
         TO_DSINTERVAL('0 '||TIME_BEG||':00'),
         TO_DSINTERVAL('0 '||TIME_END||':00'),
         NUMTODSINTERVAL(DURATION, 'MINUTE')
  FROM   table_name
),
range (ID, OTHER_DATA, TIME_BEG, TIME_INT_END, TIME_END, DURATION) AS (
  SELECT ID,
         OTHER_DATA,
         TIME_BEG,
         LEAST(time_beg + duration, time_end),
         TIME_END,
         DURATION
  FROM   data
UNION ALL
  SELECT ID,
         OTHER_DATA,
         TIME_INT_END,
         LEAST(time_int_end + duration, time_end),
         TIME_END,
         DURATION
  FROM   range
  WHERE  time_int_end < time_end
)
SEARCH DEPTH FIRST BY id SET id_order
SELECT ID,
       OTHER_DATA,
       TIME_BEG AS itvl_beg,
       TIME_INT_END AS itvl_end
FROM   range;

db小提琴here

Answer 2

这是基于pre-intervalDATE计算的经典解决方案

with time_int(ID, OTHER_DATA, ITVL_BEG, ITVL_END, DURATION, TIME_END) as (
select 
 ID, OTHER_DATA, TIME_BEG, 
 to_char(to_date(time_beg,'HH24:mi')+duration/(24*60),'HH24:mi'), 
 DURATION, TIME_END 
from tab
union all
select 
 ID, OTHER_DATA, ITVL_END, 
 to_char(to_date(ITVL_END,'HH24:mi')+duration/(24*60),'HH24:mi'), 
 DURATION, TIME_END 
from time_int
where ITVL_END <= TIME_END
)
select 
   ID, OTHER_DATA, ITVL_BEG, ITVL_END
from time_int
order by 1,3;

        ID OTHE ITVL_ ITVL_
---------- ---- ----- -----
         1 abcd 10:00 10:15
         1 abcd 10:15 10:30
         1 abcd 10:30 10:45
         1 abcd 10:45 11:00
         1 abcd 11:00 11:15
         2 xyzt 16:00 16:30
         2 xyzt 16:30 17:00
         2 xyzt 17:00 17:30

递归 CTE 与 step 一起使用，基于以下计算得到下一个区间边界（假设您的时间存储为 VARCHAR2）

to_char(to_date(time_beg,'HH24:mi')+duration/(24*60),'HH24:mi')

请注意，您可以将10:00 转换为日期，Oracle 会提供您不需要的默认缺失日、月和年，因为在增加duration 之后，您会转换回@ 987654326@字符串。