以天为单位的平均打开错误寿命

Question

我希望根据严重性确定未解决问题的平均生命周期（以天为单位）。

bug	severity	status	date_assigned
1	A	open	2021-9-14
1	A	in progress	2021-9-15
1	A	fixed	2021-9-16
1	A	verified	2021-9-17
1	A	closed	2021-9-18
2	B	opened	2021-10-18
2	B	in progress	2021-10-19
2	B	closed with fix	2021-10-20
3	C	open	2021-11-20
3	C	review	2021-11-21
3	C	close	2021-11-22
4	A	open	2021-12-24
4	A	closed	2021-12-26

我想跟踪每个严重​​性（A、B、C、D）从它们的第一个状态（如（打开/打开）到其状态更改为关闭（关闭/关闭/关闭）所需的天数）修复）

这是我迄今为止尝试过的，

select severity, count(distinct date_assigned) as no_of_days 
from Table 1 as a
join Table 1 as b
where status = 'open' | 'opened' and status = 'close' | 'closed' | 'closed with fix'

但这并没有给我预期的输出

Answer 1

假设每个 bug 有一个“打开”和“关闭”，那么您可以使用两个级别的聚合：

select severity,
       avg( close_date - open_date ) as avg_days
from (select bug, severity, min(date_assigned) as open_date,
             max(date_assigned) as close_date
      from table1
      where status like '%open%' or
            status like '%closed%'
      group by bug, severity
     ) t1
group by severity
order by severity;

请注意，日期函数因数据库而异。数据差异的确切语法可能与上述不同。

编辑：

在 Databricks 中，我认为您想要：

select severity,
       avg( datediff(close_date, open_date) ) as avg_days