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我有一个包含工程师、appt_date 和 appt_slot 的 sql 表。

engineer_id, visit_date, visit_slot

1,           20/10/12,   All Day
1,           21/10/12,   AM
2,           20/10/12,   All Day
2,           21/10/12,   All Day
2,           22/10/12,   All Day
3,           20/10/12,   PM
3,           21/10/12,   All Day
3,           22/10/12,   PM

我想展示的是

Engineer_id, start_visit,      end_visit

1,           20/10/12 All Day, 21/10/12 AM
2,           20/10/12 All Day, 22/10/12 All Day
3,           20/10/12 PM,      21/10/12 All Day
3,           22/10/12 PM,      22/10/12 PM

我正在努力使用 sql 来显示预订中的中断，任何帮助将不胜感激。谢谢

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1 回答 1

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您要做的是在数据中查找序列。通常，这是几天的序列。你的情况更难，因为你有这些插槽，大概是半天。

该解决方案通过将时隙转换为时间来工作。我随意使用早上 6:00 和晚上 6 点。我使用 a 来执行此操作cross join，因为新行是为“全天”约会创建的。

使用此数据结构，您可以使用 SQL 技巧来识别属于一起的插槽。这个想法是按日期时间枚举每个工程师的插槽。然后从 datetime 中减去半天 tiems 的枚举值。当值有序时，这是一个常数。有间隙时会出现一个新值。

最后，按engineer_id和group id分组，得到你想要的数据。

以下查询显示了这一点。它将最终结果保留为日期时间，而不是将它们转换回插槽。

with t as (
    select 1 as engineer_id, cast('2012-10-20' as date) as visit_date, 'All Day' as visit_slot union all
    select 1 as engineer_id, cast('2012-10-21' as date) as visit_date, 'AM' as visit_slot union all
    select 2 as engineer_id, cast('2012-10-20' as date) as visit_date, 'All Day' as visit_slot union all
    select 2 as engineer_id, cast('2012-10-21' as date) as visit_date, 'All Day' as visit_slot union all
    select 2 as engineer_id, cast('2012-10-22' as date) as visit_date, 'All Day' as visit_slot union all
    select 3 as engineer_id, cast('2012-10-20' as date) as visit_date, 'PM' as visit_slot union all
    select 3 as engineer_id, cast('2012-10-21' as date) as visit_date, 'All Day' as visit_slot union all
    select 3 as engineer_id, cast('2012-10-22' as date) as visit_date, 'PM' as visit_slot
   ),
   tslots as (
    select t2.*
    from (select t.engineer_id,
                 (CAST(visit_date as datetime) +
                      (case when t.visit_slot in ('All Day', const.slot) then const.hr/24.0 end)
                 ) as visit_datetime
          from t cross join
               (select 'AM' as slot, 6 as hr union all select 'PM', 18) const
         ) t2
    where visit_datetime is not null
    group by engineer_id, visit_datetime
   )
select engineer_id, MIN(visit_datetime) as min_datetime, MAX(visit_datetime) as max_datetime
from (select ts.*,
             ROW_NUMBER() over (partition by engineer_id order by visit_datetime) as seqnum,
             visit_datetime - 0.5*(ROW_NUMBER() over (partition by engineer_id order by visit_datetime)) as groupid
      from tslots ts
     ) ts1
group by engineer_id, groupid
order by 1, 2
于 2012-11-08T14:52:56.670 回答