3

我有这样的记录表预订：

  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:)

4

5 回答 5

1

这是一种使用分析功能的解决方案，它提供了所有没有预订的窗口：

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 Fiddle在这里。

于 2015-09-26T13:50:00.590 回答
1

像这样的东西：

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: fromandto是 ANSI SQL 中的保留字，这就是为什么它们用"from"and分隔的原因"to"。

于 2015-09-26T10:46:21.990 回答
0

您可以使用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)

选择行timeslots >= '30m'::interval：

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)
于 2015-09-26T10:58:32.523 回答
0

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;
于 2015-09-26T20:32:21.373 回答
0

非标自加入：

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
    ;
于 2015-09-26T13:30:55.870 回答