我使用的是Oracle 11g，我遇到了这个问题。我还没有想出任何想法来解决它。
我有一张带有教室的桌子。我需要找到的是日期时间范围之间的可用时间。例如，我有A，B和C房间，被占用教室的桌子如下所示：
Classroom start end
A 10/10/2013 10:00 10/10/2013 11:30
B 10/10/2013 09:15 10/10/2013 10:45
B 10/10/2013 14:30 10/10/2013 16:00
我需要得到的是这样的：
with date time range between '10/10/2013 07:00' and '10/10/2013 21:15'
Classroom avalailable_from available_to
A 10/10/2013 07:00 10/10/2013 10:00
A 10/10/2013 11:30 10/10/2013 21:15
B 10/10/2013 07:00 10/10/2013 09:15
B 10/10/2013 10:45 10/10/2013 14:30
B 10/10/2013 16:00 10/10/2013 21:15
C 10/10/2013 07:00 10/10/2013 21:15
有没有办法用sql或pl / sql来实现？
答案 0 :(得分：5)
我正在寻找一个至少与Wernfried相似的概念解决方案，但我认为它的发布也不同。开始是相同的想法，首先生成可能的时间段，并假设您正在查看15分钟的窗口：我正在使用CTE，因为我认为它们比嵌套选择更清晰，特别是在这么多级别。
with date_time_range as (
select to_date('10/10/2013 07:00', 'DD/MM/YYYY HH24:MI') as date_start,
to_date('10/10/2013 21:15', 'DD/MM/YYYY HH24:MI') as date_end
from dual
),
time_slots as (
select level as slot_num,
dtr.date_start + (level - 1) * interval '15' minute as slot_start,
dtr.date_start + level * interval '15' minute as slot_end
from date_time_range dtr
connect by level <= (dtr.date_end - dtr.date_start) * (24 * 4) -- 15-minutes
)
select * from time_slots;
这为您提供了指定的开始日期和结束日期之间的57个15分钟的插槽。 date_time_range的CTE并非绝对必要，您可以将日期直接放入time_slots条件，但是您必须重复它们，然后引入可能的失败点（并且意味着绑定从JDBC或其他任何地方多次使用相同的值。
然后可以将这些插槽交叉连接到教室列表，我假设它们已经在另一个表中，这样可以提供171（3x57）种组合;这些可以与现有的预订进行比较 - 一旦被淘汰，你将被留下153个15分钟的无预订位置。
with date_time_range as (...),
time_slots as (...),
free_slots as (
select c.classroom, ts.slot_num, ts.slot_start, ts.slot_end,
lag(ts.slot_end) over (partition by c.classroom order by ts.slot_num)
as lag_end,
lead(ts.slot_start) over (partition by c.classroom order by ts.slot_num)
as lead_start
from time_slots ts
cross join classrooms c
left join occupied_classrooms oc on oc.classroom = c.classroom
and not (oc.occupied_end <= ts.slot_start
or oc.occupied_start >= ts.slot_end)
where oc.classroom is null
)
select * from free_slots;
但是你必须将它们折叠成连续的范围。有各种方法可以做到这一点;在这里，我正在查看上一行和下一行，以确定特定值是否是范围的边缘：
with date_time_range as (...),
time_slots as (...),
free_slots as (...),
free_slots_extended as (
select fs.classroom, fs.slot_num,
case when fs.lag_end is null or fs.lag_end != fs.slot_start
then fs.slot_start end as slot_start,
case when fs.lead_start is null or fs.lead_start != fs.slot_end
then fs.slot_end end as slot_end
from free_slots fs
)
select * from free_slots_extended
where (fse.slot_start is not null or fse.slot_end is not null);
现在我们已经减少了12行。 （外部where子句消除了上一步中153个插槽中的所有141个中间范围，因为我们只关心边缘）：
CLASSROOM SLOT_NUM SLOT_START SLOT_END
--------- ---------- ---------------- ----------------
A 1 2013-10-10 07:00
A 12 2013-10-10 10:00
A 19 2013-10-10 11:30
A 57 2013-10-10 21:15
B 1 2013-10-10 07:00
B 9 2013-10-10 09:15
B 16 2013-10-10 10:45
B 30 2013-10-10 14:30
B 37 2013-10-10 16:00
B 57 2013-10-10 21:15
C 1 2013-10-10 07:00
C 57 2013-10-10 21:15
所以那些代表边缘，但是在不同的行上，最后一步结合它们：
...
select distinct fse.classroom,
nvl(fse.slot_start, lag(fse.slot_start)
over (partition by fse.classroom order by fse.slot_num)) as slot_start,
nvl(fse.slot_end, lead(fse.slot_end)
over (partition by fse.classroom order by fse.slot_num)) as slot_end
from free_slots_extended fse
where (fse.slot_start is not null or fse.slot_end is not null)
或者把所有这些放在一起：
with date_time_range as (
select to_date('10/10/2013 07:00', 'DD/MM/YYYY HH24:MI') as date_start,
to_date('10/10/2013 21:15', 'DD/MM/YYYY HH24:MI') as date_end
from dual
),
time_slots as (
select level as slot_num,
dtr.date_start + (level - 1) * interval '15' minute as slot_start,
dtr.date_start + level * interval '15' minute as slot_end
from date_time_range dtr
connect by level <= (dtr.date_end - dtr.date_start) * (24 * 4) -- 15-minutes
),
free_slots as (
select c.classroom, ts.slot_num, ts.slot_start, ts.slot_end,
lag(ts.slot_end) over (partition by c.classroom order by ts.slot_num)
as lag_end,
lead(ts.slot_start) over (partition by c.classroom order by ts.slot_num)
as lead_start
from time_slots ts
cross join classrooms c
left join occupied_classrooms oc on oc.classroom = c.classroom
and not (oc.occupied_end <= ts.slot_start
or oc.occupied_start >= ts.slot_end)
where oc.classroom is null
),
free_slots_extended as (
select fs.classroom, fs.slot_num,
case when fs.lag_end is null or fs.lag_end != fs.slot_start
then fs.slot_start end as slot_start,
case when fs.lead_start is null or fs.lead_start != fs.slot_end
then fs.slot_end end as slot_end
from free_slots fs
)
select distinct fse.classroom,
nvl(fse.slot_start, lag(fse.slot_start)
over (partition by fse.classroom order by fse.slot_num)) as slot_start,
nvl(fse.slot_end, lead(fse.slot_end)
over (partition by fse.classroom order by fse.slot_num)) as slot_end
from free_slots_extended fse
where (fse.slot_start is not null or fse.slot_end is not null)
order by 1, 2;
给出了：
CLASSROOM SLOT_START SLOT_END
--------- ---------------- ----------------
A 2013-10-10 07:00 2013-10-10 10:00
A 2013-10-10 11:30 2013-10-10 21:15
B 2013-10-10 07:00 2013-10-10 09:15
B 2013-10-10 10:45 2013-10-10 14:30
B 2013-10-10 16:00 2013-10-10 21:15
C 2013-10-10 07:00 2013-10-10 21:15
答案 1 :(得分：2)
当你喜欢“选择不存在的东西”时，这总是一个挑战。首先，您需要所有可用教室和时间的列表（间隔15分钟）。然后您可以通过跳过占用的项目来选择它们。
我设法在没有任何PL / SQL的情况下进行查询：
CREATE TABLE Table1
(Classroom VARCHAR2(10), start_ts DATE, end_ts DATE);
INSERT INTO Table1 VALUES ('A', TIMESTAMP '2013-01-10 10:00:00', TIMESTAMP '2013-01-10 11:30:00');
INSERT INTO Table1 VALUES ('B', TIMESTAMP '2013-01-10 09:15:00', TIMESTAMP '2013-01-10 10:45:00');
INSERT INTO Table1 VALUES ('B', TIMESTAMP '2013-01-10 14:30:00', TIMESTAMP '2013-01-10 16:00:00');
WITH all_rooms AS
(SELECT CHR(64+LEVEL) AS ROOM FROM dual CONNECT BY LEVEL <= 3),
all_times AS
(SELECT CAST(TIMESTAMP '2013-01-10 07:00:00' + (LEVEL-1) * INTERVAL '15' MINUTE AS DATE) AS TIMES, LEVEL AS SLOT
FROM DUAL
CONNECT BY TIMESTAMP '2013-01-10 07:00:00' + (LEVEL-1) * INTERVAL '15' MINUTE <= TIMESTAMP '2013-01-10 21:15:00'),
all_free_slots AS
(SELECT ROOM, TIMES, SLOT,
CASE SLOT-LAG(SLOT, 1, 0) OVER (PARTITION BY ROOM ORDER BY SLOT)
WHEN 1 THEN 0
ELSE 1
END AS NEW_WINDOW
FROM all_times
CROSS JOIN all_rooms
WHERE NOT EXISTS
(SELECT 1 FROM TABLE1 WHERE ROOM = CLASSROOM AND TIMES BETWEEN START_TS + INTERVAL '1' MINUTE AND END_TS - INTERVAL '1' MINUTE)),
free_time_windows AS
(SELECT ROOM, TIMES, SLOT,
SUM(NEW_WINDOW) OVER (PARTITION BY ROOM ORDER BY SLOT) AS WINDOW_ID
FROM all_free_slots)
SELECT ROOM,
TO_CHAR(MIN(TIMES), 'yyyy-mm-dd hh24:mi') AS free_time_start,
TO_CHAR(MAX(TIMES), 'yyyy-mm-dd hh24:mi') AS free_time_end
FROM free_time_windows
GROUP BY ROOM, WINDOW_ID
HAVING MAX(TIMES) - MIN(TIMES) > 0
ORDER BY ROOM, 2;
ROOM FREE_TIME_START FREE_TIME_END
---- ----------------------------------
A 2013-01-10 07:00 2013-01-10 10:00
A 2013-01-10 11:30 2013-01-10 21:15
B 2013-01-10 07:00 2013-01-10 09:15
B 2013-01-10 10:45 2013-01-10 14:30
B 2013-01-10 16:00 2013-01-10 21:15
C 2013-01-10 07:00 2013-01-10 21:15
为了理解查询，您可以从顶部拆分子查询，例如
WITH all_rooms AS
(SELECT CHR(64+LEVEL) AS ROOM FROM dual CONNECT BY LEVEL <= 3),
all_times AS
(SELECT CAST(TIMESTAMP '2013-01-10 07:00:00' + (LEVEL-1) * INTERVAL '15' MINUTE AS DATE) AS TIMES, LEVEL AS SLOT
FROM DUAL
CONNECT BY TIMESTAMP '2013-01-10 07:00:00' + (LEVEL-1) * INTERVAL '15' MINUTE <= TIMESTAMP '2013-01-10 21:15:00')
SELECT ROOM, TIMES, SLOT,
CASE SLOT-LAG(SLOT, 1, 0) OVER (PARTITION BY ROOM ORDER BY SLOT)
WHEN 1 THEN 0
ELSE 1
END AS NEW_WINDOW
FROM all_times
CROSS JOIN all_rooms
WHERE NOT EXISTS (SELECT 1 FROM TABLE1 WHERE ROOM = CLASSROOM AND TIMES BETWEEN START_TS + INTERVAL '1' MINUTE AND END_TS - INTERVAL '1' MINUTE)
ORDER BY ROOM, SLOT