具有 Postgres 范围的递归 SQL 查询以查找可用性

Question

我关注了这篇博文： https ://info.crunchydata.com/blog/range-types-recursion-how-to-search-availability-with-postgresql

CREATE TABLE travels (
    id serial PRIMARY KEY,
    travel_dates daterange NOT NULL,
    EXCLUDE USING spgist (travel_dates WITH &&)
);

当我插入持续时间背靠背的行时，发现这个函数有问题

CREATE OR REPLACE FUNCTION travels_get_available_dates(daterange)
RETURNS TABLE(available_dates daterange)
AS $$
    WITH RECURSIVE calendar AS (
        SELECT
            $1 AS left,
             $1 AS center,
             $1 AS right
        UNION
        SELECT
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN daterange(lower(calendar.left), lower(travels.travel_dates * calendar.left))
                ELSE daterange(lower(calendar.right), lower(travels.travel_dates * calendar.right))
            END AS left,
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN travels.travel_dates * calendar.left
                ELSE travels.travel_dates * calendar.right
            END AS center,
            CASE travels.travel_dates && calendar.right
                WHEN TRUE THEN daterange(upper(travels.travel_dates * calendar.right), upper(calendar.right))
                ELSE daterange(upper(travels.travel_dates * calendar.left), upper(calendar.left))
            END AS right
        FROM calendar
        JOIN travels ON
            travels.travel_dates && $1 AND
            travels.travel_dates <> calendar.center AND (
                travels.travel_dates && calendar.left OR
                travels.travel_dates && calendar.right
            )
)
SELECT *
FROM (
    SELECT
        a.left AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.left <> b.left AND
        a.left @> b.left
    GROUP BY a.left
    HAVING NOT bool_or(COALESCE(a.left @> b.left, FALSE))
    UNION
    SELECT
        a.right AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.right <> b.right AND
        a.right @> b.right
    GROUP BY a.right
    HAVING NOT bool_or(COALESCE(a.right @> b.right, FALSE))
) a
$$ LANGUAGE SQL STABLE;

INSERT INTO travels (travel_dates)
VALUES
    (daterange('2018-03-02', '2018-03-02', '[]')),
    (daterange('2018-03-06', '2018-03-09', '[]')),
    (daterange('2018-03-11', '2018-03-12', '[]')),
    (daterange('2018-03-16', '2018-03-17', '[]')),
    (daterange('2018-03-25', '2018-03-27', '[]'));

这在这一点上按预期工作。

SELECT *
FROM travels_get_available_dates(daterange('2018-03-01', '2018-04-01'))
ORDER BY available_dates;

available_dates
-------------------------
[2018-03-01,2018-03-02)
[2018-03-03,2018-03-06)
[2018-03-10,2018-03-11)
[2018-03-13,2018-03-16)
[2018-03-18,2018-03-25)
[2018-03-28,2018-04-01)

但是当添加这一行时：

INSERT INTO travels (travel_dates)
VALUES
(daterange('2018-03-03', '2018-03-05', '[]'));

并重新运行

SELECT *
FROM travels_get_available_dates(daterange('2018-03-01', '2018-04-01'))
ORDER BY available_dates;

我得到

available_dates
-------------------------
empty

Answer 1

我在原始博客文章中添加了一条评论，说明我认为错误的来源，即处理空范围的方式。

如果日期范围是连续的或相邻的，则会导致“左”和“右”列中的任一列或什至两者中的“空”范围。 现在，在递归 CTE 完成后（并假设空范围在“左”列中），在“LEFT OUTER JOIN ... ON ...”子句中，一个免费且有效的 travel_date 与一个 ' 配对来自 B.left 范围的空'范围，因为 A.left <> 'empty' && A.left @> 'empty' 因为所有范围都包含空范围。 理想情况下，它应该与 NULL 配对，因为这是一个左外连接，可以将其包含在最终结果集中，但“空”有点碍事。 'empty' 然后再次出现在 'GROUP BY ... HAVING ...' 子句中，其中 a.left @> 'empty' 评估为 true 并且它被否定，因此所有有效的旅行日期都被丢弃，导致一个空表。 我的解决方案如下，将“空”设为 NULL，并丢弃“中心”中的任何日期范围：

CREATE OR REPLACE FUNCTION travels_get_available_dates(daterange)
RETURNS TABLE(available_dates daterange)
AS $$
    WITH RECURSIVE calendar AS (
        SELECT
            $1 AS left,
             $1 AS center,
             $1 AS right
        UNION
        SELECT
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN daterange(lower(calendar.left), lower(travels.travel_dates * calendar.left))
                ELSE daterange(lower(calendar.right), lower(travels.travel_dates * calendar.right))
            END AS left,
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN travels.travel_dates * calendar.left
                ELSE travels.travel_dates * calendar.right
            END AS center,
            CASE travels.travel_dates && calendar.right
                WHEN TRUE THEN daterange(upper(travels.travel_dates * calendar.right), upper(calendar.right))
                ELSE daterange(upper(travels.travel_dates * calendar.left), upper(calendar.left))
            END AS right
        FROM calendar
        JOIN travels ON
            travels.travel_dates && $1 AND
            travels.travel_dates <> calendar.center AND (
                travels.travel_dates && calendar.left OR
                travels.travel_dates && calendar.right
            )
)
SELECT *
FROM (
    SELECT
        a.left AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.left <> b.left AND
        a.left @> b.left
    GROUP BY a.left
    HAVING NOT bool_or(coalesce(a.left @> case when isempty(b.left) then null else b.left end, FALSE))

    UNION

    SELECT
        a.right AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.right <> b.right AND
        a.right @> b.right
    GROUP BY a.right
    HAVING NOT bool_or(coalesce(a.right @> case when isempty(b.right) then null else b.right end, false))

    EXCEPT

    SELECT a.center AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.center <> b.center AND
        a.center @> b.center
    GROUP BY a.center
    HAVING NOT bool_or(COALESCE(a.center @> b.center, FALSE))
) a
WHERE NOT isempty(a.available_dates)
$$ LANGUAGE SQL STABLE;

Answer 2

我认为你应该采取另一种方法：

CREATE OR REPLACE FUNCTION travels_get_available_dates(daterange)
RETURNS TABLE(
  available_dates daterange
)
AS $$
  WITH RECURSIVE calendar(available_dates) AS
  (
    SELECT 
      CASE 
        WHEN $1 @> travel_dates THEN unnest(array[
          daterange(lower($1),lower(travel_dates)),
          daterange(upper(travel_dates),upper($1)) 
        ])
        WHEN lower($1) < lower(travel_dates) THEN daterange(lower($1),lower(travel_dates)) 
        WHEN upper($1) > upper(travel_dates) THEN daterange(upper(travel_dates),upper($1)) 
      END
    FROM travels 
      WHERE $1 && travel_dates AND NOT travel_dates @> $1
    UNION
    SELECT 
      CASE 
        WHEN available_dates @> travel_dates THEN unnest(array[
          daterange(lower(available_dates),lower(travel_dates)), 
          daterange(upper(travel_dates),upper(available_dates)) 
        ])
        WHEN lower(available_dates) < lower(travel_dates) THEN daterange(lower(available_dates),lower(travel_dates)) 
        WHEN upper(available_dates) > upper(travel_dates) THEN daterange(upper(travel_dates),upper(available_dates)) 
      END
    FROM travels 
      JOIN calendar ON available_dates && travel_dates AND NOT travel_dates @> available_dates
  )

  SELECT $1 AS available_dates 
    WHERE NOT EXISTS(SELECT 1 FROM travels WHERE travel_dates <@ $1)    
  UNION
  SELECT * FROM calendar
    WHERE $1 <> available_dates AND 'empty' <> available_dates
      AND NOT EXISTS(SELECT 1 FROM travels WHERE available_dates && travel_dates)
$$ LANGUAGE SQL STABLE;

我们必须递归地分割左右段上的给定范围，然后只得到那些未被占用的范围。

Answer 3

我最初忘记了“中心”区域的条款。 下面是：

CREATE OR REPLACE FUNCTION travels_get_available_dates(daterange)
RETURNS TABLE(available_dates daterange)
AS $$
    WITH RECURSIVE calendar AS (
        SELECT
            $1 AS left,
             $1 AS center,
             $1 AS right
        UNION
        SELECT
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN daterange(lower(calendar.left), lower(travels.travel_dates * calendar.left))
                ELSE daterange(lower(calendar.right), lower(travels.travel_dates * calendar.right))
            END AS left,
            CASE travels.travel_dates && calendar.left
                WHEN TRUE THEN travels.travel_dates * calendar.left
                ELSE travels.travel_dates * calendar.right
            END AS center,
            CASE travels.travel_dates && calendar.right
                WHEN TRUE THEN daterange(upper(travels.travel_dates * calendar.right), upper(calendar.right))
                ELSE daterange(upper(travels.travel_dates * calendar.left), upper(calendar.left))
            END AS right
        FROM calendar
        JOIN travels ON
            travels.travel_dates && $1 AND
            travels.travel_dates <> calendar.center AND (
                travels.travel_dates && calendar.left OR
                travels.travel_dates && calendar.right
            )
)
SELECT *
FROM (
    SELECT
        a.left AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.left <> b.left AND
        a.left @> b.left
    GROUP BY a.left
    HAVING NOT bool_or(COALESCE(a.left @> b.left, FALSE))
    UNION
    SELECT a.center AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.center <> b.center AND
        a.center @> b.center
    GROUP BY a.center
    HAVING NOT bool_or(COALESCE(a.center @> b.center, FALSE))
    UNION
    SELECT
        a.right AS available_dates
    FROM calendar a
    LEFT OUTER JOIN calendar b ON
        a.right <> b.right AND
        a.right @> b.right
    GROUP BY a.right
    HAVING NOT bool_or(COALESCE(a.right @> b.right, FALSE))
) a
WHERE NOT isempty(a.available_dates)
$$ LANGUAGE SQL STABLE;