SQL select 3 小时平均

Question

我明白了 table:

MariaDB [table]> 
select insert_time, host_id, tx 
from host_daily 
where host_id = 2 
order by insert_time desc 
limit 24;

+---------------------+---------+------------+
| insert_time         | host_id | tx         |
+---------------------+---------+------------+
| 2016-02-06 14:00:00 |       2 | 9676875156 |
| 2016-02-06 13:00:00 |       2 | 9671544048 |
| 2016-02-06 12:00:00 |       2 | 9669464371 |
| 2016-02-06 11:00:00 |       2 | 9667087098 |
| 2016-02-06 10:00:00 |       2 | 9665014071 |
| 2016-02-06 09:00:00 |       2 | 9662931956 |
| 2016-02-06 08:00:00 |       2 | 9660874138 |
| 2016-02-06 07:00:00 |       2 | 9658624162 |
| 2016-02-06 06:00:00 |       2 | 9656555329 |
| 2016-02-06 05:00:00 |       2 | 9654443169 |
| 2016-02-06 04:00:00 |       2 | 9651362676 |
| 2016-02-06 03:00:00 |       2 | 9648531733 |
| 2016-02-06 02:00:00 |       2 | 9633368883 |
| 2016-02-05 23:00:00 |       2 | 9464826179 |
| 2016-02-05 22:00:00 |       2 | 9363099844 |
| 2016-02-05 21:00:00 |       2 | 9270841166 |
| 2016-02-05 20:00:00 |       2 | 9140988502 |
| 2016-02-05 19:00:00 |       2 | 9022460285 |
| 2016-02-05 18:00:00 |       2 | 8925920799 |
| 2016-02-05 17:00:00 |       2 | 8825711136 |
| 2016-02-05 16:00:00 |       2 | 8802081092 |
| 2016-02-05 15:00:00 |       2 | 8755784419 |
+---------------------+---------+------------+

这不是很明显，但可能缺少一些时间。例如。 23 点到 2 点之间。如果可能的话，我想用 0 来填补这个缺失的时间。现在我想 select 上周三个小时的平均值，但我似乎没有得到正确的查询。我正在尝试并修改此查询一段时间：

select 
    n.n as id, 
    host_id,
    avg(tx),
    insert_time + interval 3 * n.n hour as 'from',
    insert_time + interval 3 * (n.n + 1) hour as 'to'
from (SELECT  0 AS n
          UNION ALL SELECT  1 UNION ALL SELECT  2 UNION ALL SELECT  3
          UNION ALL SELECT  4 UNION ALL SELECT  5 UNION ALL SELECT  6
          UNION ALL SELECT  7 UNION ALL SELECT  8 UNION ALL SELECT  9
          UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
          UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
          UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18
          UNION ALL SELECT 19 UNION ALL SELECT 20 UNION ALL SELECT 21
          UNION ALL SELECT 22 UNION ALL SELECT 23 
     ) n 
left join host_daily
     on n.n = hour(insert_time) div 3
where host_id = 2 
  and insert_time > NOW() - INTERVAL 24 HOUR group by n.n;

导致：

+----+---------+-----------------+---------------------+---------------------+
| id | host_id | avg(tx)         | from                | to                  |
+----+---------+-----------------+---------------------+---------------------+
|  4 |       2 | 9672627858.3333 | 2016-02-07 00:00:00 | 2016-02-07 03:00:00 |
|  3 |       2 | 9665011041.6667 | 2016-02-06 18:00:00 | 2016-02-06 21:00:00 |
|  2 |       2 | 9658684543.0000 | 2016-02-06 12:00:00 | 2016-02-06 15:00:00 |
|  1 |       2 | 9651445859.3333 | 2016-02-06 06:00:00 | 2016-02-06 09:00:00 |
|  0 |       2 | 9605672722.3333 | 2016-02-06 00:00:00 | 2016-02-06 03:00:00 |
|  7 |       2 | 9366255729.6667 | 2016-02-06 18:00:00 | 2016-02-06 21:00:00 |
|  6 |       2 | 9029789862.0000 | 2016-02-06 12:00:00 | 2016-02-06 15:00:00 |
|  5 |       2 | 9019970953.0000 | 2016-02-06 06:00:00 | 2016-02-06 09:00:00 |
+----+---------+-----------------+---------------------+---------------------+

有人好心为我指出正确的方向吗？

编辑： 通过将查询结果编辑为预期结果，从 24 小时视图返回，保持示例完整性。

Answer 1

您需要将每次四舍五入到三个小时。假设你每三个小时至少有一个记录，你可以直接使用四舍五入：

select from_unixtime(floor(unix_timestamp(insert_time)/(60*60*3))*60*60*3) as timestart,
       host_id,
       sum(tx) / 3
from host_daily hd
group by from_unixtime(floor(unix_timestamp(insert_time)/(60*60*3))*60*60*3), hostid
order by 1, 2;

这种假设您希望将缺失值视为 0。如果不是，则使用 avg(tx)。

如果这仍然缺少行，您可以在缺少所有数据期间使用您想要的任何值添加它们。您的问题对于在这种情况下应包括什么内容含糊不清。

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