Question

我目前有一个sql语句工作，它聚合一些数据以获得具有相同id的表的子集的多行输出（不是简单的最大/最小计算）。以下剪辑根据位置的变化计算特定ID所花费的时间。

WITH id_locations AS (
    SELECT created_at, id, location
        FROM locations
        WHERE id = 1
), travel_locations AS (
    SELECT created_at, id, location, ROW_NUMBER() OVER(ORDER BY created_at) AS row_num
        FROM id_locations
        WHERE 
            created_at IN (
                SELECT min(created_at)
                    FROM id_locations
                    GROUP BY location
            ) OR created_at IN (
                SELECT max(created_at)
                    FROM id_locations
                    GROUP BY location
            )
)
SELECT a.id id, a.created_at departed_at, b.created_at arrived_at, a.location departure_location, b.location arrival_location
    FROM travel_locations AS a
    INNER JOIN travel_locations AS b
        ON a.row_num = b.row_num - 1
    WHERE a.location <> b.location;

Data
id          created_at  location 
1           1           1
1           2           1
1           3           2
1           4           2
1           5           5
1           6           5
1           7           5
1           8           5
2           1           1 
2           2           1 
2           3           1 
2           4           1 
2           5           3 
2           6           3 
2           7           3 
2           8           3 

Desired Output
id          departed_at arrived_at  departure_location  arrival_location
1           2           3           1                   2
1           4           5           2                   5
2           4           5           1                   3

我想创建一个包含此sql语句输出的视图，在按ID分组的表的每个子集上运行

有没有办法用raw sql做这个，或者我需要用一些sql客户端迭代所有可能的id，运行相同的查询，并将结果插入表中以供将来参考？

我使用postgresql 9.6作为参考

Answer 1

SELECT id,
       max_created_at AS departed_at,
       next_created_at AS arrived_at,
       location AS departure_location,
       next_location AS arrival_location
FROM
  (SELECT id,
          created_at,
          location,
          max(created_at) over(partition by id,location) AS max_created_at,
          lead(created_at) over(partition by id order by location,created_at) AS next_created_at,
          lead(location) over(partition by id order by location,created_at) AS next_location
   FROM locations) t
WHERE max_created_at=created_at AND next_created_at IS NOT NULL

<强> Sample Demo

您尝试获取的是max created_at和相应的位置以及下一个created_at，ID的位置。

使用max窗口函数为每个ID，位置获取max created_at。
使用lead从每个ID的下一行（按位置，created_at排序）获取位置，created_at值。
获取max_created_at等于当前行的created_at且next_created_at不为空的行。

通过多次运行子查询来创建PostgreSQL视图

1 个答案: