从如何在PostgreSql的预订表中找到第一次空闲时间的最佳答案中选择

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

用于查找在给定日期和时间（2012年11月17日：在下面的示例中）开始的时间表中的间隔，它还可以查找周六，周日和公共假日。表中定义了公众假期

create table pyha ( pyha date primary key)

如何也排除周末和公共假期？

硬编码空闲时间作为查询之类的保留时间

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 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

对于每个空闲时间范围，都需要在联合中有单独的行。

从给定的日期和时间开始，在工作日和工作时间（8:00 .. 18:00）中返回空闲时间的最佳方法是哪种？

更新

选择答案时，总是在8:00返还空闲时间。如何在指定开始日期的指定开始时间之前返回空闲时间，例如，如果开始时间为9，则不返回2012-11-19
9:00之前的空闲时间？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）也应返回2012-11-19
9:00。仅在2012-11-19的9:00没有空闲时间并且在随后的工作日中第一次空闲8:00时，才应返回8:00。

我试图通过将 2012-11-19 9： 添加到两个地方来解决此问题，如下查询所示，但此查询仍在2012-11-19
8:00返回空闲时间。如何解决此问题，使其在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

更新2

我尝试更新答案，但返回错误数据。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

观察到的第一行：

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

预期的 ：

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

怎么修 ？