Question

从How to find first free time in reservations table in PostgreSql

中选择最佳答案

create table reservation (during tsrange,
 EXCLUDE USING gist (during WITH &&)
 );

用于查找从给定日期和时间开始的时间表中的差距（2012-11-17 8：下面的示例）它也发现了周六，周日和公众假期。公共假期在表

中定义

create table pyha ( pyha date primary key)

如何排除周末和公众假期？

硬编码空闲时间作为查询保留，如

with gaps as (
  select
    upper(during) as start,
    lead(lower(during),1,upper(during)) over (ORDER BY during) - upper(during) as gap
  from (
    select during
    from reservation
   union all values
     ('(,2012-11-17 8:)'::tsrange), -- given date and hour from which to find free work time
     ('[2012-11-17 0:,2012-11-18 24:)'::tsrange), -- exclude saturday
     ('[2012-11-18 0:,2012-11-19 8:)'::tsrange),  -- exclude sunday
     ('[2012-11-19 18:,2012-11-20 8:)'::tsrange),
     ('[2012-11-20 18:,2012-11-21 8:)'::tsrange),
     ('[2012-11-21 18:,2012-11-22 8:)'::tsrange),
     ('[2012-11-22 18:,2012-11-23 8:)'::tsrange),
     ('[2012-11-23 18:,2012-11-24 24:)'::tsrange),
     ('[2012-11-24 0:,2012-11-25 24:)'::tsrange), -- exclude saturday
     ('[2012-11-25 0:,2012-11-26 8:)'::tsrange)  -- exclude sunday
 ) as x
)
select *
  from gaps
where gap > '0'::interval
order by start

每个空闲时间范围都需要单独的联合行。

从给定的日期和时间开始，哪个是在工作日和工作时间（8：00-18：00）返回空闲时间的最佳方式？

更新

选择回答始终在8:00返回空闲时间。如果在指定的开始日期之前不是在指定的开始时间之前返回空闲时间，例如在2012-11-19 9:00之前如果开始时间是9则返回？开始时间可能只有值8,9,10,11,12,13,14,15,16或17

即使2012-11-19 8:00如果有空，也应该返回2012-11-19 9:00。只有在2012-11-19 9:00没有空闲时间且8:00在接下来的工作日内首次免费时，才应返回8:00。

我尝试通过将 2012-11-19 9： 添加到两个地方来解决此问题，如下面的查询所示，但此查询仍会返回2012-11的空闲时间-19 8:00 如何解决这个问题，以便在2012-11-19 9:00返回空闲时间？

create table reservation (during tsrange,
 EXCLUDE USING gist (during WITH &&)
 );
create table pyha ( pyha date primary key);
with gaps as (
    select
        upper(during) as start,
        lead(lower(during),1,upper(during)) over (ORDER BY during) - upper(during) as gap
    from (
        select during
          from reservation
             where upper(during)>= '2012-11-19 9:'
       union all values
         ('(,2012-11-19 9:)'::tsrange)
        union all
        select
            unnest(case
                when pyha is not null then array[tsrange(d, d + interval '1 day')]
                when date_part('dow', d) in (0, 6) then array[tsrange(d, d + interval '1 day')]
                else array[tsrange(d, d + interval '8 hours'),
                           tsrange(d + interval '18 hours', d + interval '1 day')]
            end)
        from generate_series(
            '2012-11-19'::timestamp without time zone,
            '2012-11-19'::timestamp without time zone+ interval '3 month',
            interval '1 day'
        ) as s(d)
        left join pyha on pyha = d::date
    ) as x
)

select start,
   date_part('epoch', gap) / (60*60) as hours
  from gaps
where gap > '0'::interval
order by start

UPDATE2

我尝试了更新的答案，但它返回了错误的数据。完整的测试用例是：

create temp table reservation  ( during tsrange ) on commit drop;
insert into reservation values(
'[2012-11-19 11:00:00,2012-11-19 11:30:00)'::tsrange );

with gaps as (
    select
        upper(during) as start,
        lead(lower(during),1,upper(during)) over (ORDER BY during) - upper(during) as gap
    from (
        select during
          from reservation
        union all
        select
            unnest(case
                when pyha is not null then array[tsrange(d, d + interval '1 day')]
                when date_part('dow', d) in (0, 6) then array[tsrange(d, d + interval '1 day')]
                when d::date =  DATE'2012-11-19' then array[
                            tsrange(d, '2012-11-19 12:'),  -- must return starting at 12:00
                            tsrange(d + interval '18 hours', d + interval '1 day')]
                else array[tsrange(d, d + interval '8 hours'), 
                           tsrange(d + interval '18 hours', d + interval '1 day')]
            end)
        from generate_series(
            DATE'2012-11-19'::timestamp without time zone,
            DATE'2012-11-19'::timestamp without time zone+ interval '3 month',
            interval '1 day'
        ) as s(d) 
        left join pyha on pyha = d::date
    ) as x 
)

select start,
   date_part('epoch', gap) / (60*60) as tunde
  from gaps 
where gap > '0'::interval
order by start

观察第一行：

"2012-11-19 11:30:00"

预期：

"2012-11-19 12:00:00"

如何解决？

Answer 1

您可以使用generate_series（）函数来屏蔽非工作时间：

with gaps as (
    select
        upper(during) as start,
        lead(lower(during),1,upper(during)) over (ORDER BY during) - upper(during) as gap
    from (
        select during
        from reservation
        union all
        select
            unnest(case
                when pyha is not null then array[tsrange(d, d + interval '1 day')]
                when date_part('dow', d) in (0, 6) then array[tsrange(d, d + interval '1 day')]
                when d::date = '2012-11-14' then array[tsrange(d, d + interval '9 hours'), tsrange(d + interval '18 hours', d + interval '1 day')]
                else array[tsrange(d, d + interval '8 hours'), tsrange(d + interval '18 hours', d + interval '1 day')]
            end)
        from generate_series(
            '2012-11-14'::timestamp without time zone, 
            '2012-11-14'::timestamp without time zone + interval '2 week', 
            interval '1 day'
        ) as s(d) 
        left join pyha on pyha = d::date
    ) as x 
)
select *
    from gaps
where gap > '0'::interval
order by start

让我解释一些棘手的部分：

您不必将sat / sun的日期插入pyha表，因为您可以使用date_part('dow', d)函数。仅将pyha表用于公众假期。对于太阳或周六，'dow'分别返回0或6。
公共假期和sat / sun可以表示为单个间隔（0..24）。工作日必须由两个间隔（0..8）和（18..24）表示，因此unfst（）和array []
您可以在generate_series（）函数

根据您对问题的更新，我向when添加了另一个case：

when d::date = '2012-11-14' then array[tsrange(d, d + interval '9 hours'), tsrange(d + interval '18 hours', d + interval '1 day')]

想法是为开始日期（d::date = '2012-11-14'）生成不同的间隔：（0..9）和（18..24）

如何从PostgreSql中的预约返回工作时间？

1 个答案: