Рекурсивный SQL запрос с Postgres диапазонами для поиска доступности - PullRequest
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Рекурсивный SQL запрос с Postgres диапазонами для поиска доступности

1 голос
/ 18 февраля 2020

Я подписался на этот пост в блоге: 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 &&)
);

и обнаружил, что эта функция глючит, когда я вставляю строки с длительностью назад 

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', '[]'));

На данном этапе это работает, как и ожидалось. 

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)

Но при добавлении этой строки: 

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

И повторный запуск 

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

Я получаю 

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

Ответы [ 2 ]

0 голосов
/ 18 февраля 2020

Я изначально забыл пункт о "центральной" области. Вот оно ниже: 

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;
0 голосов
/ 18 февраля 2020

Я думаю, что вы должны выбрать другой подход: 

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;

Мы должны рекурсивно разделить заданный диапазон на левый и правый сегменты, а затем получить только те, которые не заняты.

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