Postgres 查找两列数据集之间的缺失日期

Question

我正在尝试创建一个返回两列和多行之间缺失日期的查询。

示例：租约

move_in          move_out       hotel_id
2021-04-01    2021-04-14    1
2021-04-17   2021-04-30     1
2021-04-01    2021-04-14    2
2021-04-17   2021-04-30     2

结果应该是

date         hotel_id
2021-04-15   1
2021-04-16   1
2021-04-15   2
2021-04-16   2

Answer 1

如果您使用的是 postgresql 14+，则可以使用 multiranges 来执行此操作：

CREATE TEMP TABLE t (
    "move_in"  DATE,
    "move_out" DATE,
    "hotel_id" INTEGER
);

INSERT INTO t
    ("move_in", "move_out", "hotel_id")
VALUES ('2021-04-01', '2021-04-14', '1')
     , ('2021-04-17', '2021-04-30', '1')
     , ('2021-05-03', '2021-05-30', '1')  -- added this as a test case
     , ('2021-04-01', '2021-04-14', '2')
     , ('2021-04-17', '2021-04-30', '2');

SELECT hotel_id, datemultirange(DATERANGE(MIN(move_in), MAX(move_out))) - range_agg(DATERANGE(move_in, move_out, '[]')) AS r
FROM t
GROUP BY hotel_id

returns

+--------+-------------------------------------------------+
|hotel_id|r                                                |
+--------+-------------------------------------------------+
|2       |{[2021-04-14,2021-04-17)}                        |
|1       |{[2021-04-14,2021-04-17),[2021-04-30,2021-05-03)}|
+--------+-------------------------------------------------+

如果你想每天有 1 行，你可以使用 unnest 和 generate_series 来扩展 multiranges:

WITH available_ranges AS(
    SELECT hotel_id, unnest(datemultirange(DATERANGE(MIN(move_in), MAX(move_out), '[]')) - range_agg(DATERANGE(move_in, move_out, '[]'))) AS r
    FROM t
    GROUP BY hotel_id
)
SELECT hotel_id, generate_series(lower(r), upper(r) - 1, '1 day'::interval)
FROM available_ranges
ORDER BY 1, 2
;

returns

+--------+---------------------------------+
|hotel_id|generate_series                  |
+--------+---------------------------------+
|1       |2021-04-15 00:00:00.000000 +00:00|
|1       |2021-04-16 00:00:00.000000 +00:00|
|1       |2021-05-01 00:00:00.000000 +00:00|
|1       |2021-05-02 00:00:00.000000 +00:00|
|2       |2021-04-15 00:00:00.000000 +00:00|
|2       |2021-04-16 00:00:00.000000 +00:00|
+--------+---------------------------------+

Answer 2

您正在找出两组之间的差异。一种是租赁酒店的日子。另一个是四月份的所有日子。你正在为所有酒店这样做。

我们可以为所有酒店制作一组四月的所有日子。首先我们需要构建四月份所有日期的集合：generate_series('2022-04-01'::date, '2022-04-30'::date, '1 day').

然后我们需要将它与所有酒店 ID 交叉连接。

select *
from generate_series('2021-04-01'::date, '2021-04-30'::date, '1 day') as dates(day)
cross join (
  select distinct hotel_id as id
  from leases
) as hotels(id)

现在，对于每一天，我们都可以加入当天的租约。

select *
from generate_series('2021-04-01'::date, '2021-04-30'::date, '1 day') as dates(day)
cross join (
  select distinct hotel_id as id
  from leases
) as hotels(id)

没有租约的任何天数都不会 lease.id，因此请对此进行过滤。

select day, hotels.id
from generate_series('2021-04-01'::date, '2021-04-30'::date, '1 day') as dates(day)
cross join (
  select distinct hotel_id as id
  from leases
) as hotels(id)
left join leases on day between move_in and leases.move_out and hotel_id = hotels.id
where leases.id is null
order by hotels.id, day

Demonstration.

Postgres 查找两列数据集之间的缺失日期

Postgres Find missing dates between a dataset of two columns

postgresql