Question

我正在尝试编写一个复杂的查询，我是Postgresql的初学者。

以下是我的架构，其中包含一些示例数据，目前已完成进度：

在架构no列中，只是主键。

我的函数接受参数course_no，执行查询并返回一个json。

我的查询应该执行以下操作：

时间段按isodow，开始时间和结束时间分组。
连续工作日分组仅显示第一个上周工作日（如周一至周五或周一至周五）
Json按start_timestamp排序，因此min（start_timestamp）位于json的第一个组中。

在星期一11：00-11：45：

"[{"schedule":"Mon 11:00-11:45","count_timeslots":47}]"

课程_在星期二09：00-09：30有15个时段，在星期四10：00-10：45有20个时段：

"[{"schedule":"Tue 09:00-09:30","count_timeslots":15},{"schedule":"Thu 10:00-10:45","count_timeslots":20}]"

在星期一，星期二和星期三09：00-09：30以及星期三17：00-18：00的23个时段有46个时段的航线_no。

通知“周一至周三......”

"[{"schedule":"Mon-Wed 09:00-09:45","count_timeslots":46},{"schedule":"Wed 17:00-18:00","count_timeslots":23}]"

我知道如何通过start_time和end_time从时间戳和分组中提取时间，但我不知道如何连续分组？

更新：进度

我写了一个查询，其中没有插槽，GROUP BY start_timestamp和end_timestamp以及ORDER BY min（start_timestamp）。

我只需要帮助就可以将它们分组为isodow（星期几），只有当它们是连续的时候。

SELECT COUNT(*) AS 
count_timeslot, 
(EXTRACT(hour FROM start_timestamp) || ':' || 
EXTRACT(minute FROM start_timestamp)) AS start_time,     
(EXTRACT(hour FROM end_timestamp) || ':' || EXTRACT(minute FROM end_timestamp)) 
AS end_time FROM timeslot GROUP BY start_time, end_time ORDER BY MIN(start_timestamp);

更新2：进度

在Postgresql的窗口函数的帮助下，几乎完成了这个查询。

首先，我根据start_time，end_time，day_of_week对其进行分组。
然后我通过计算grp来创建day_of_week - ROW NO() over partition of start_time，它在连续时给我一个常数值。
第三，我计算了count_timeslot OVER partition of start_time and grp的总和。

我的查询

SELECT *, SUM(count_timeslot) OVER (PARTITION BY start_time, grp) 
AS n_count_time 
FROM (
   SELECT *, day_of_week - ROW_NUMBER() 
   OVER (PARTITION BY start_time ORDER BY day_of_week) AS grp 
   FROM( 
     SELECT COUNT(*) AS count_timeslot, 
     (EXTRACT(hour FROM start_timestamp) || ':' || EXTRACT(minute FROM 
     start_timestamp)) AS start_time, 
     (EXTRACT(hour FROM end_timestamp) || ':' || EXTRACT(minute FROM 
     end_timestamp)) AS end_time, 
     EXTRACT(ISODOW FROM start_timestamp) AS day_of_week FROM 
     timeslot GROUP BY start_time, end_time, day_of_week 
     ORDER BY MIN(start_timestamp)
     )foo
   )foo1;

Answer 1

以下是解决方案：

WITH temp(k, v) 
AS(VALUES('Mon', 1), ('Tue', 2), ('Wed', 3), ('Thu', 4), ('Fri', 5), ('Sat', 6), ('Sun', 7))

SELECT array_to_json(array_agg(row_to_json(foo4)))
FROM(
    SELECT grouped || ' ' || start_time || '-' || end_time AS schedule
    , n_count_time AS count_timeslots 
    FROM(
        SELECT * 
        , ROW_NUMBER() OVER (PARTITION BY start_time, grp) AS row_no 
        FROM(
            SELECT * 
            , CASE 
            WHEN COUNT( * ) OVER (PARTITION BY start_time, grp) > 1 
            THEN first_value(name_of_day) OVER (PARTITION BY start_time, grp) 
                 || '-' || last_value(name_of_day) OVER (PARTITION BY start_time, grp) 
            ELSE first_value(name_of_day) OVER (PARTITION BY start_time, grp) 
            END AS grouped
            , SUM(count_timeslot) OVER (PARTITION BY start_time, grp) AS n_count_time 
            FROM(
                SELECT * 
                , day_of_week - ROW_NUMBER() OVER (PARTITION BY start_time ORDER BY day_of_week) AS grp
                , k AS name_of_day FROM(SELECT COUNT( * ) AS count_timeslot
                , to_char(start_timestamp, 'HH24:MI') AS start_time
                , to_char(end_timestamp, 'HH24:MI') AS end_time
                , EXTRACT(ISODOW FROM start_timestamp) AS day_of_week 
                FROM timeslot 
                GROUP BY start_time, end_time, day_of_week 
                ORDER BY MIN(start_timestamp)
                ) foo, temp WHERE v = day_of_week) foo1
            ) foo2
        ) foo3 WHERE row_no = 1
    ) foo4;

工作小提琴：http://sqlfiddle.com/#!15/e8b13/47

PostgreSQL查询将连续几天分组

1 个答案: