查找重叠时间的计数

Question

I have a postgres database that contains:

position (which has values 1-10), starttime , endtime , and status . Status is either "on" or "off". The database lists the start and stop times of when a machine position is "on" or "off", and that machine has 10 "positions".

I am trying to find the best way to count the how much time accrues where all 10 positions have the status off at the same time.

Answer 1

I don't claim that this is the best solution, but it's one of them. The idea is to select only position 1 and join with the table itself where the times overlap and position is not 1. If we find 9 distinct position values, then we found one occurance where all positions are off.

SELECT
  t.machine_id,
  COUNT(*) occurances
FROM (
  SELECT
    t1.machine_id,
    t1.starttime,
    COUNT(DISTINCT t2.position) c
  FROM
    yourTable t1 INNER JOIN
    yourTable t2 ON
    t2.machine_id = t1.machine_id AND
    t2.status = 'off' AND
    t2.position <> 1 AND
    (t2.starttime BETWEEN t1.starttime AND t1.endtime OR  -- those 3 lines check if the
    t2.endtime BETWEEN t1.starttime AND t1.endtime OR  -- times are overlapping
    (t2.starttime < t1.starttime AND t2.endtime > t1.endtime))
  WHERE
    t1.status = 'off' AND
    t1.position = 1
  GROUP BY
    t1.machine_id,
    t1.starttime
  ) t
WHERE
  t.c = 9
GROUP BY
  t.machine_id
;

The results may vary depending on what position you select for t1 . Because if one position takes very short and another a long time then you might count the same overlap several times (also depends on the other position times). So choose the position which takes the longest time and not just 1.