sql - Recursive SQL Query with Postgres Ranges To Find Availability

Question

I followed this blogpost: 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 &&)
);

and found this function to be buggy when I've inserted rows with duration back to back

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', '[]'));

This works as expected at this point.

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)

But when this row is added:

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

And re-run

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

I get

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

Answer 1

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

如果日期范围是连续的，或者更确切地说是相邻的，它会在“左”和“右”列中的任何一个，甚至两个列中产生“空”范围。现在，在递归 CTE 完成后（并假设空范围在“左”列中），在“左外连接 ... ON ...”子句中，一个免费且有效的 travel_date 与一个 ' empty' range from B.left range since 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，以便将其包含在最终结果集中，但“空”有点碍事. 'empty' 然后在 'GROUP BY ... HAVING ...' 子句中再次弹出，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'emptys'设为NULL，并丢弃'center'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;

我们必须递归地在左右段上分割给定的范围，然后只得到那些未被占用的。

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;