检查是否可以在 SQL (Postgresql) 中预订答案

【问题标题】：Checking if booking is possible in SQL (Postgresql)检查是否可以在 SQL (Postgresql) 中预订
【发布时间】：2015-09-26 10:40:46
【问题描述】：

我有这样看起来记录的餐桌预订：

  id  |        from         |         to
------+---------------------+---------------------
  101 | 2015-09-24 08:00:00 | 2015-09-24 09:30:00
 2261 | 2015-09-24 09:00:00 | 2015-09-24 10:00:00
 4061 | 2015-09-24 10:00:00 | 2015-09-24 10:30:00
  204 | 2015-09-24 12:00:00 | 2015-09-24 13:30:00
 2400 | 2015-09-24 13:30:00 | 2015-09-24 14:00:00
 4224 | 2015-09-24 14:00:00 | 2015-09-24 14:30:00
  309 | 2015-09-24 16:00:00 | 2015-09-24 17:30:00
 2541 | 2015-09-24 17:00:00 | 2015-09-24 18:00:00

我正在寻找最佳查询来找到问题的答案：

Is this possible to find a timeslot with duration x (ie. 30 minutes) in above records?

我有使用 postgres 数组或时间范围的想法，但仍在寻找更好的想法......

编辑：我将提供“假”预订作为界限，但如果您有更好的方法，请写信：)

【问题讨论】：

这个问题似乎不完整。如果您没有预订，那么您不能对时间段进行细化。换句话说，您需要提供开始和结束信息。
@GordonLinoff ：你是对的。我目前假设 [ -infinity,+infinity ]
我编辑了问题，谢谢通知！

标签： sql database postgresql gaps-in-data

【解决方案1】：

类似这样的：

select t1.*
from tablename t1
where (select min("from") from tablename t2
       where t2."from" > t1."from") >= t1."to" + interval '30' minute

即如果与下一行的间隔 >= 30 分钟，则返回一行。

注意：from 和 to 是 ANSI SQL 中的保留字，因此它们被分隔为 "from" 和 "to"。

【讨论】：

【解决方案2】：

这是一个使用分析函数的解决方案，它提供了所有窗口都没有预订：

SELECT null as ts_from, min(ts_from) as ts_to
  FROM bookings
 UNION ALL
SELECT ts as ts_from, next_ts as ts_to
  FROM (SELECT ts, lead(ts, 1) over (order by ts) as next_ts, sum(bk) over (order by ts) as bksum
          FROM (SELECT ts_from as ts, 1 as bk
                  FROM bookings
                 UNION ALL
                SELECT ts_to as ts, -1 as bk
                  FROM bookings) as t) as tt
 WHERE bksum = 0
 ORDER BY 1 NULLS FIRST;

SQL 小提琴here.

【讨论】：

【解决方案3】：

你可以使用lag()函数：

select *, book_start- previous_book_end timeslot
from (
    select id, "from" book_start, "to" book_end, 
    lag("to") over (order by "to") previous_book_end
    from test
    ) sub
order by book_end

  id  |     book_start      |      book_end       |  previous_book_end  | timeslot
------+---------------------+---------------------+---------------------+-----------
  101 | 2015-09-24 08:00:00 | 2015-09-24 09:30:00 |                     |
 2261 | 2015-09-24 09:00:00 | 2015-09-24 10:00:00 | 2015-09-24 09:30:00 | -00:30:00
 4061 | 2015-09-24 10:00:00 | 2015-09-24 10:30:00 | 2015-09-24 10:00:00 | 00:00:00
  204 | 2015-09-24 12:00:00 | 2015-09-24 13:30:00 | 2015-09-24 10:30:00 | 01:30:00
 2400 | 2015-09-24 13:30:00 | 2015-09-24 14:00:00 | 2015-09-24 13:30:00 | 00:00:00
 4224 | 2015-09-24 14:00:00 | 2015-09-24 14:30:00 | 2015-09-24 14:00:00 | 00:00:00
  309 | 2015-09-24 16:00:00 | 2015-09-24 17:30:00 | 2015-09-24 14:30:00 | 01:30:00
 2541 | 2015-09-24 17:00:00 | 2015-09-24 18:00:00 | 2015-09-24 17:30:00 | -00:30:00
(8 rows)

选择带有timeslots >= '30m'::interval的行：

select *, book_start- previous_book_end timeslot
from (
    select id, "from" book_start, "to" book_end, 
    lag("to") over (order by "to") previous_book_end
    from test
    ) sub
where book_start- previous_book_end >= '30m'::interval
order by book_end

 id  |     book_start      |      book_end       |  previous_book_end  | timeslot
-----+---------------------+---------------------+---------------------+----------
 204 | 2015-09-24 12:00:00 | 2015-09-24 13:30:00 | 2015-09-24 10:30:00 | 01:30:00
 309 | 2015-09-24 16:00:00 | 2015-09-24 17:30:00 | 2015-09-24 14:30:00 | 01:30:00
(2 rows)

【讨论】：

您不能只查看以前的预订。想象一下第一次预订涵盖了整个时间跨度。您将报告错误的插槽 (example)。
当然，我可以想象一切。也许更好的主意是让 OP 评估这种假设情况是否会发生？
好吧，我弄错了，我们可以假设预订不能重叠（例如它确实重叠，但表中有一些外键）。
这与我对预订本质的理解是一致的。我的回答中的查询有助于避免这种情况。好吧，@Kombajnzbożowy，投反对票有点为时过早，不是吗？
是的，在不重叠预订的约束下，您的解决方案很好。一开始我只是假设了最坏的情况。

【解决方案4】：

非标准自加入：

SELECT
    ll.ts_to AS ts_from
    , hh.ts_from AS ts_to
FROM bookings ll
JOIN bookings hh
    -- enough space 
    ON hh.ts_from >= ll.ts_to + '30 min'::interval
    -- and nothing in between
    AND NOT EXISTS (
        SELECT * FROM bookings nx
        WHERE nx.ts_from >= ll.ts_to
        AND nx.ts_to <= hh.ts_from
        )
UNION ALL   -- before the first
SELECT '-infinity'::timestamp AS ts_from
       , MIN(ts_from) AS ts_to
    FROM bookings
UNION ALL   -- after the last
SELECT MAX(ts_to) AS ts_from
       , 'infinity'::timestamp AS ts_to
    FROM bookings
ORDER BY 1,2
    ;

【讨论】：

【解决方案5】：

生成您的插槽，然后加入它们。 http://sqlfiddle.com/#!15/12bfa

create table t (id integer, "from" timestamp, "to" timestamp);

insert into t values 
(101 , '2015-09-24 08:00:00' , '2015-09-24 09:30:00' ),
(2261 , '2015-09-24 09:00:00' , '2015-09-24 10:00:00' ),
(4061 , '2015-09-24 10:00:00' , '2015-09-24 10:30:00' ),
( 204 , '2015-09-24 12:00:00' , '2015-09-24 13:30:00' ),
(2400 , '2015-09-24 13:30:00' , '2015-09-24 14:00:00' ),
(4224 , '2015-09-24 14:00:00' , '2015-09-24 14:30:00' ),
( 309 , '2015-09-24 16:00:00' , '2015-09-24 17:30:00' ),
(2541 , '2015-09-24 17:00:00' , '2015-09-24 18:00:00' );

SELECT time_slots.t,
       time_slots.t + interval '30 minutes'
FROM generate_series(date'2015-09-24',date'2015-09-25' - interval '30 minutes' ,interval '30 minutes') AS time_slots(t)
LEFT JOIN t ON (time_slots.t BETWEEN t."from" AND t."to")
WHERE t.id IS NULL;


SELECT time_slots.t,
       time_slots.t + interval '30 minutes'
FROM generate_series(date'2015-09-24',date'2015-09-25' - interval '30 minutes',interval '30 minutes') AS time_slots(t)
LEFT JOIN t ON ((time_slots.t,
                 time_slots.t + interval '30 minutes') OVERLAPS (t."from",
                                                                 t."to"))
WHERE t.id IS NULL;

【讨论】：