Question

我需要计算存储在数据库中的所有操作的平均时间。我存储操作的表格如下：

 creation time       | operation_type | operation_id
 2017-01-03 11:14:25 | START          | 1
 2017-01-03 11:14:26 | START          | 2
 2017-01-03 11:14:28 | END            | 2
 2017-01-03 11:14:30 | END            | 1

在这种情况下，操作1需要5秒钟，操作2需要2秒才能完成。

如何计算MySQL中这些操作的平均值？

编辑：似乎operation_id不需要是唯一的 - 给定的操作可能会执行多次，因此表可能如下所示：

 creation time       | operation_type | operation_id
 2017-01-03 11:14:25 | START          | 1
 2017-01-03 11:14:26 | START          | 2
 2017-01-03 11:14:28 | END            | 2
 2017-01-03 11:14:30 | END            | 1
 2017-01-03 11:15:00 | START          | 1
 2017-01-03 11:15:10 | END            | 1

我应该在查询中添加什么来正确计算所有这些操作的平均时间？

Answer 1

由于操作的END始终在START之后，因此可以使用MIN和MAX

select avg(diff)
from
(
      select operation_id, 
             TIME_TO_SEC(TIMEDIFF(max(creation_time), min(creation_time))) as diff
      from your_table
      group by operation_id
) tmp

Answer 2

我不确定子查询是否必要......

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Answer 3

select avg(diff)
from
(
select a1.operation_id, timediff(a2.operation_time, a1.operation_time) as diff
from oper a1 -- No table name provided, went with 'oper' because it made sense in my head
inner join oper a2
  on a1.operation_id = a2.operation_id
where a1.operation_type = 'START'
and a2.operation_type = 'END'
)

存储在数据库中的平均操作时间

3 个答案: