当前sql代码会将所有最大孤岛大小总和为一。我试图仅找出 连续小时小时的最大岛屿大小，而忽略日期时间内的分钟和秒钟。
select deviceid, grp -- min(readdate) as mn, max(readdate) as mx
from ( select deviceid, readdate, sum(case when prev_rd >= readdate - interval '1 hour' then 1 else 0 end) over (partition by deviceid order by readdate rows between unbounded preceding and current row) as grp
from (
select deviceid, readdate, lag(readdate) over (partition by deviceid order by readdate) as prev_rd
from metersWithExcess
order by readdate)
) tt
group by deviceid, readdate, grp
-- having mx > mn + 1 * interval '1 hour'
是否有一种方法可以返回具有最大大小的设备ID？我试图找到最长的连续小时读数
数据如下：
deviceid readdate
91428 2018-08-27 18:04:42
91428 2018-08-27 17:04:42
91428 2018-08-27 16:04:42
91428 2018-08-27 12:04:42
91428 2018-08-27 11:04:42
91428 2018-08-26 20:04:42
91428 2018-08-26 17:04:40
91428 2018-08-26 16:04:40
91428 2018-08-26 15:04:40
91428 2018-08-26 12:04:40
91489 2018-08-26 06:00:49
91493 2018-08-27 06:58:44
91511 2018-08-27 12:57:01
91511 2018-08-27 10:57:01
91511 2018-08-27 09:57:01
91511 2018-08-26 23:57:01
91511 2018-08-26 22:57:01
91511 2018-08-26 20:57:01
91511 2018-08-26 13:56:59
91511 2018-08-26 12:56:59
91511 2018-08-26 10:56:59
91547 2018-08-26 16:59:07
91547 2018-08-26 15:59:07
91547 2018-08-26 14:59:07
91547 2018-08-26 11:59:07
91547 2018-08-26 04:59:07
91582 2018-08-27 12:01:45
91582 2018-08-26 19:01:45
在此示例中：91428的最大大小为3，因为ID 91428的最大大小为18:04:42 / 17:04:42 / 16:04:42
答案 0 :(得分：0)
您提出问题的方式有点令人困惑，“孤岛”和“最大规模”术语必须特定于您的行业，但是这种蛮力方法会有所帮助吗？格式为：
deviceid，第一个读数，与下一个读数的距离，第二个读数，与下一个读数的距离等
我注释掉了我的标准“ WHERE gap1 = '01：00：00'AND gap2 = '01：00：00'”，但这将找到您连续的每小时读数。
WITH DEVICEID_TEST_DATA AS
( SELECT CAST(91428 AS bigint) AS deviceid , CAST('2018-08-27 18:04:42' AS TIMESTAMP) AS readdate UNION
SELECT 91428 , '2018-08-27 17:04:42' UNION
SELECT 91428 , '2018-08-27 16:04:42' UNION
SELECT 91428 , '2018-08-27 12:04:42' UNION
SELECT 91428 , '2018-08-27 11:04:42' UNION
SELECT 91428 , '2018-08-26 20:04:42' UNION
SELECT 91428 , '2018-08-26 17:04:40' UNION
SELECT 91428 , '2018-08-26 16:04:40' UNION
SELECT 91428 , '2018-08-26 15:04:40' UNION
SELECT 91428 , '2018-08-26 12:04:40' UNION
SELECT 91489 , '2018-08-26 06:00:49' UNION
SELECT 91493 , '2018-08-27 06:58:44' UNION
SELECT 91511 , '2018-08-27 12:57:01' UNION
SELECT 91511 , '2018-08-27 10:57:01' UNION
SELECT 91511 , '2018-08-27 09:57:01' UNION
SELECT 91511 , '2018-08-26 23:57:01' UNION
SELECT 91511 , '2018-08-26 22:57:01' UNION
SELECT 91511 , '2018-08-26 20:57:01' UNION
SELECT 91511 , '2018-08-26 13:56:59' UNION
SELECT 91511 , '2018-08-26 12:56:59' UNION
SELECT 91511 , '2018-08-26 10:56:59' UNION
SELECT 91547 , '2018-08-26 16:59:07' UNION
SELECT 91547 , '2018-08-26 15:59:07' UNION
SELECT 91547 , '2018-08-26 14:59:07' UNION
SELECT 91547 , '2018-08-26 11:59:07' UNION
SELECT 91547 , '2018-08-26 04:59:07' UNION
SELECT 91582 , '2018-08-27 12:01:45' UNION
SELECT 91582 , '2018-08-26 19:01:45'
)
SELECT * FROM (
SELECT deviceid , readdate ,
readdate - (LAG(readdate, 1) OVER (PARTITION BY deviceid ORDER BY readdate ASC) ) AS gap1 ,
LAG(readdate, 1) OVER (PARTITION BY deviceid ORDER BY readdate ASC) AS readdate_1down ,
(LAG(readdate, 1) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) - (LAG(readdate, 2) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) AS gap2 ,
LAG(readdate, 2) OVER (PARTITION BY deviceid ORDER BY readdate ASC) AS readdate_2down ,
(LAG(readdate, 2) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) - (LAG(readdate, 3) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) AS gap3 ,
LAG(readdate, 3) OVER (PARTITION BY deviceid ORDER BY readdate ASC) AS readdate_3down ,
(LAG(readdate, 3) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) - (LAG(readdate, 4) OVER (PARTITION BY deviceid ORDER BY readdate ASC)) AS gap4 ,
LAG(readdate, 4) OVER (PARTITION BY deviceid ORDER BY readdate ASC) AS readdate_4down
FROM DEVICEID_TEST_DATA
-------WHERE deviceid = 91428
)
--------WHERE gap1 = '01:00:00' AND gap2 = '01:00:00'
ORDER BY deviceid , readdate DESC