根据给定日期的最大状态计数，使用分组数据

Question

我的示例是一个票务系统，保存状态更新条目和创建票证。

Fiddle: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=a5ff4600adbab185eb14b08586f1bd29

ID	TICKETID	STATUS	TICKET_CREATED	STATUS_CHANGED
1	1	other_error	01-JAN-20	01-JAN-20 08.00.00
2	2	tech_error	01-JAN-20	01-JAN-20 09.00.00
3	3	unknown	01-JAN-20	01-JAN-20 09.10.00
4	4	unknown	01-JAN-20	01-JAN-20 09.20.00
5	4	tech_error	01-JAN-20	02-JAN-20 09.30.00
6	1	solved	01-JAN-20	02-JAN-20 10.00.00
7	2	solved	01-JAN-20	02-JAN-20 07.00.00
8	5	tech_error	02-JAN-20	02-JAN-20 08.00.00
9	6	unknown	02-JAN-20	02-JAN-20 08.30.00
10	6	solved	02-JAN-20	02-JAN-20 09.30.00
11	5	solved	02-JAN-20	03-JAN-20 08.00.00
12	4	unknown	01-JAN-20	03-JAN-20 09.00.00

我想根据工单创建日期来评估数据，获取特定日期的三件事：

（完成） 在给定日期总共创建了多少票
（完成） 在给定日期
（未完成） 在给定日期有多少票处于 'unknown' 状态？棘手！因为重要的是给定日期午夜以下最大 STATUS_CHANGED 的状态。

2021 年 1 月 1 日的预期结果：

TICKET_CREATED	Total Created	Tickets created in Unknown status	Total tickets in Unknown status
01-JAN-20	4	2	2

解释：20 年 1 月 1 日，工单 3 和 4 在当天结束时处于 'unknown' 状态

2021 年 1 月 2 日的预期结果：

TICKET_CREATED	Total Created	Tickets created in Unknown status	Total tickets in Unknown status
02-JAN-20	2	1	1

解释：在 2020 年 1 月 2 日，只有工单 3 在当天结束时处于 'unknown' 状态

第 1 + 2 部分的当前解决方案：

select ticket_created, 
count(*) as "Total Created",
sum(case when status = 'unknown' then 1 else 0 end) as "Unknown tickets created",
'?' as "Total tickets in Unknown status"
from myTable
where id in
    (select min(id) as id
    from myTable
    where ticket_created = to_date('01.01.2020', 'DD.MM.YYYY')
    group by ticketid)
group by ticket_created

你能给我一些关于如何处理第 3 点的提示吗？

Answer 1

假设我正确理解了您的逻辑，这就是我实现您的目标的方式：

with ticket_info as (select id,
                            ticketid,
                            status,
                            ticket_created,
                            status_changed,
                            row_number() over (partition by ticketid, trunc(status_changed) order by status_changed desc) rn_per_id_day_desc,
                            row_number() over (partition by ticketid order by status_changed) rn_per_id_asc
                     from   mytable)
select ticket_created,
       count(distinct case when trunc(ticket_created) = to_date('01/01/2020', 'dd/mm/yyyy') then ticketid end) as "Total Created",
       count(case when rn_per_id_asc = 1 and status = 'unknown' then 1 end) as "Unknown tickets created",
       count(case when rn_per_id_day_desc = 1 and status = 'unknown' then 1 end) as "Total tickets in Unknown status"
from   ticket_info
where  status_changed >= to_timestamp('01/01/2020', 'dd/mm/yyyy')
and    status_changed < to_timestamp('01/01/2020', 'dd/mm/yyyy') + interval '1' day
group by ticket_created;

db<>fiddle

你可以看到，首先，我使用了几个 row_number() 分析函数来为行提供标签 - 一个按照行的更改顺序为每个 id 标记行（这允许我们识别每个 id 的第一行，即票证创建的行），另一个按降序标记每个 id 和天的行（这允许我们识别每个 id 当天的最后一行）。

使用该信息，我们可以计算出您的所有三个案例：

一天创建的工单 - 此处我使用了不同的计数，但您可以将其更改为 count(case when rn_per_id_asc = 1 then 1 end)，这可能更有效且更容易理解。
当天创建的工单为“未知”- 此处我使用了条件计数：如果它是第一行且状态未知，则对其进行计数
在一天结束时处于“未知”状态的工单 - 这里我使用了另一个条件计数：如果它是当天的最后一行并且状态未知，请计算它。

预计到达时间：第三部分的逻辑被修改为计算当天结束时状态未知的有效工单，我认为这应该可以解决问题：

with date_of_interest as (select start_date + level -1 dt,
                                 start_date + level next_dt
                          from   (select to_date('01/01/2020', 'dd/mm/yyyy') start_date,
                                         to_date('03/01/2020', 'dd/mm/yyyy') end_date
                                  from   dual)
                          connect by level <= (end_date - start_date) + 1),
          ticket_info as (select mt.id,
                                 mt.ticketid,
                                 mt.status,
                                 mt.ticket_created,
                                 mt.status_changed,
                                 row_number() over (partition by mt.ticketid, doi.dt order by mt.status_changed) rn_per_id_asc,
                                 row_number() over (partition by mt.ticketid, doi.dt order by mt.status_changed desc) rn_per_id_desc,
                                 doi.dt,
                                 doi.next_dt
                          from   mytable mt
                                 inner join date_of_interest doi on mt.status_changed < doi.next_dt
                          )
select dt,
       count(case when ticket_created = dt and rn_per_id_asc = 1 then 1 end) as "Total Created",
       count(case when ticket_created = dt and rn_per_id_asc = 1 and status = 'unknown' then 1 end) as "Unknown tickets created",
       count(case when rn_per_id_desc = 1 and status = 'unknown' then 1 end) as "Total tickets in Unknown status"
from   ticket_info
group by dt
order by dt;

您会注意到我已经将查询更新为运行多天 - 如果查询一次只针对一个日期运行，您可以替换date_of_interest 像这样的子查询：

with date_of_interest as (select dt,
                                 dt + 1 next_dt
                          from   (select to_date('03/01/2020', 'dd/mm/yyyy') dt
                                  from   dual)),

已更新db<>fiddle

N.B。这不会是最有效的做事方式；随着时间的推移，随着越来越多的记录出现，查询会变慢。如果你能想出一种方法来轻松识别活动工单，尤其是如果你能在索引中获取该信息，那就更好了。

Answer 2

这是一个单独计算第 3 个指标的解决方案。
然后将它们加入您已经知道的指标。

with cte_ranges as (
  select id, status, ticketid, ticket_created
  , status_changed as started
  , coalesce(
     lead(status_changed) over (partition by ticketid order by status_changed)
    , current_timestamp) as ended
  from myTable
  where trunc(ticket_created) between DATE'2020-01-01' and DATE'2020-01-02'
)
select q.ticket_date   as "Ticket Created"
     , q.total_tickets as "Total Created"
     , q.total_unknown as "Unknown tickets created"
     , endofday.total_unknown "Total tickets in Unknown status"
from
(
  select trunc(t.ticket_created) as ticket_date
  , count(distinct t.ticketid) as total_tickets
  , count(distinct case when t.status = 'unknown' then t.ticketid end) as total_unknown
  from cte_ranges t
  group by trunc(t.ticket_created) 
) q
left join (
  select trunc(cast(dt as date)) as ticket_date
  , count(distinct case when status = 'unknown' then ticketid end) as total_unknown
  from cte_ranges
  join (
    select distinct 
     cast(trunc(ticket_created)+1 as timestamp) - interval '1' second as dt 
    from cte_ranges
  ) cutoff on dt between started and ended
  group by cast(dt as date)
) endofday 
on endofday.ticket_date = q.ticket_date;

Ticket Created	Total Created	Unknown tickets created	Total tickets in Unknown status
01-JAN-20	4	2	2
02-JAN-20	2	1	1

db<>fiddle here

诀窍是首先使用 LEAD 来计算状态处于活动状态的范围。

然后将截止时间（一天的最后一分钟）加入这些范围。
这样你就可以得到状态仍然有效的天数。

两个子查询都使用了 CTE。这样您只需更改 CTE 中的日期标准。

根据给定日期的最大状态计数，使用分组数据

Count based on the max status on a given date, with grouped data

sql

oracle

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oracle18c