PostgreSQL：使用相同外键创建另一行之前的持续时间

Question

有table:

id, car_id, status, created_at

有数据：

1, 180, active,   2017-04-02 10:12:00 +0600
2, 190, active,   2017-04-05 17:34:00 +0600
3, 180, inactive, 2017-05-10 16:59:00 +0600
4, 180, active,   2017-06-15 09:23:00 +0600
5, 180, inactive, 2017-07-10 19:12:00 +0600

我想知道180第一次active多长时间，然后inactive多长时间，第二次active多长时间时间.

我不是很擅长SQL，但这就是我最后的结果：

SELECT
  start_log.car_id,
  start_log.status,
  MAX(start_log.updated_at) AS start_time,
  end_log.updated_at AS end_time,
  (end_log.updated_at - start_log.updated_at) AS duration
FROM
  car_availabilities AS start_log
INNER JOIN
  car_availabilities AS end_log ON (
    start_log.car_id = end_log.car_id
    AND
    end_log.updated_at > start_log.updated_at
  )
GROUP BY 
  start_log.status, 
  end_log.updated_at, 
  start_log.updated_at, 
  start_log.car_id
ORDER BY start_time

但是，这会产生每行之间的持续时间，而不仅仅是按时间顺序排列的下一行。

有人可以告诉我如何获得预期的结果吗？即

180, active, 2017-04-02 10:12:00 +0600, 2017-05-10 16:59:00 +0600, 38 days....
180, inactive, 2017-05-10 16:59:00 +0600, 2017-06-15 09:23:00 +0600, 35 days...
etc

谢谢❣️

Answer 1

给你：

WITH t AS (
    SELECT *, lag(created_at, 1) OVER (PARTITION BY car_id ORDER BY created_at) as prev_state_created_at
    FROM car_availabilities
)
SELECT *, (created_at - prev_state_created_at) as duration
FROM t

这是通过使用 PostgreSQL window 函数完成的。

lag returns value evaluated at the row that is offset rows before the current row within the partition;

lag(created_at, 1) OVER (PARTITION BY car_id ORDER BY created_at)

PostgreSQL：使用相同外键创建另一行之前的持续时间

PostgreSQL: Duration before another row is created with the same foreign key

postgresql

time

duration