查看以获取具有复杂条件的最小日期

Question

我在 SQL 服务器中有一个 table 是这样的：

+----------+-----------+------------+
| DateFrom | Completed | EmployeeID |
+----------+-----------+------------+
DateFrom: date not null -- unique for each EmployeeID
Completed: bit not null
EmployeeID: bigint not null

每一行都属于一个由开始日期定义的子周期，可以完成也可以不完成。
每个员工可以有多个子期间。
一个周期由一系列有序的子周期定义，直到最后一个子周期结束。

我想创建一个视图，它将 return 每个 EmployeeID 的最后一个周期的开始日期，如下所示：

如果没有Completed为真，则获取最小的DateFrom。 [该员工还有一期未完成]

+----------+-----------+------------+
| DateFrom | Completed | EmployeeID |
+----------+-----------+------------+
|2021-01-01|   false   |     1      |
|2021-01-05|   false   |     1      |
|2021-01-09|   false   |     1      |
|2021-01-10|   false   |     1      |
|2021-01-07|   false   |     2      |
|2021-01-15|   false   |     2      |
+----------+-----------+------------+

Expected Result:
2021-01-01 for EmployeeID = 1
2021-01-07 for EmployeeID = 2

否则，return 最后一个 Completed 之后的最小 DateFrom 为真。 [最后一期还未完成]

+----------+-----------+------------+
| DateFrom | Completed | EmployeeID |
+----------+-----------+------------+
|2021-01-01|   false   |     1      |
|2021-01-05|   true    |     1      |
|2021-01-09|   false   |     1      |
|2021-01-10|   false   |     1      |
|2021-01-07|   true    |     2      |
|2021-01-15|   false   |     2      |
+----------+-----------+------------+

Expected Result:
2021-01-09 for EmployeeID = 1
2021-01-15 for EmployeeID = 2

如果最大 DateFrom has Completed=true，return最后一个 Completed 之前的最小 DateFrom 为 true，在它之前的 true 之后，如果存在。 [最后一期完成多个子期]

+----------+-----------+------------+
| DateFrom | Completed | EmployeeID |
+----------+-----------+------------+
|2021-01-01|   false   |     1      |
|2021-01-05|   true    |     1      |
|2021-01-09|   false   |     1      |
|2021-01-10|   true    |     1      |
|2021-01-07|   false   |     2      |
|2021-01-15|   true    |     2      |
+----------+-----------+------------+

Expected Result:
2021-01-09 for EmployeeID = 1
2021-01-07 for EmployeeID = 2

如果最大 DateFrom 已 Completed=true 并且没有其他行或它之前的行已 Completed=true，则 return 最大 DateFrom。 [最后一期用一个子期完成]

+----------+-----------+------------+
| DateFrom | Completed | EmployeeID |
+----------+-----------+------------+
|2021-01-01|   false   |     1      |
|2021-01-05|   false   |     1      |
|2021-01-09|   true    |     1      |
|2021-01-10|   true    |     1      |
|2021-01-07|   true    |     2      |
+----------+-----------+------------+

Expected Result:
2021-01-10 for EmployeeID = 1
2021-01-07 for EmployeeID = 2

我正在寻找最优化的解决方案。

我试过了，但是我在第三个例子中得到了 NULL 值：

WITH T AS (
    SELECT EmployeeID
        , MAX(CASE WHEN Completed = 0 THEN NULL ELSE DateFrom END) MaxDateFrom 
    FROM TableDates
    GROUP BY EmployeeID
)
SELECT TableDates.EmployeeID, MIN(TableDates.DateFrom) DateFrom
FROM T
LEFT JOIN TableDates ON T.EmployeeID = TableDates.EmployeeID
    AND (T.MaxDateFrom IS NULL OR TableDates.DateFrom > T.MaxDateFrom)
GROUP BY TableDates.EmployeeID

Answer 1

这是一个有效的查询。它可能过于复杂，但我把简化留给你。

处理3种情况，均按要求按EmployeeId分区，如下：

当不存在Completed=1时，使用sum(Completed) over()检测到，则使用first_value(DateFrom)。
当最后一行值为completed=1且前一行为completed=0时，使用last_value(Completed)和lag(Completed)进行检测，然后max(case when Completed = 0 then DateFrom else null end) 被使用。
棘手的情况，当 Completed=1 存在并且不是最后一个时。在这种情况下，找到 Completed=1 的最近行的 DateFrom，然后找到比先前检测到的行更新的所有行的 min(DateFrom)，直到前面的 Completed=1.
如果最后一行有 completed=1 并且倒数第二行有 completed=1 则使用最后一行的 DateFrom。如果所有其他选项都为空，Coalesce 会确保这一点。

insert into @Test (EmployeeId, DateFrom, Completed)
values
-- Scenario 1
(1, '2021-01-01', 0),
(1, '2021-01-02', 0),
(1, '2021-01-03', 0),
-- Scenario 2
(2, '2021-01-01', 0),
(2, '2021-01-02', 1),
(2, '2021-01-03', 0),
(2, '2021-01-04', 0),
-- Scenario 3
(3, '2021-01-01', 0),
(3, '2021-01-02', 1),
(3, '2021-01-03', 0),
(3, '2021-01-04', 1),
-- Special case, single row
(4, '2021-01-01', 1),
-- Scenario 4
(5, '2021-01-01', 0),
(5, '2021-01-02', 0),
(5, '2021-01-03', 1);

with cte as (
  select *
    -- First value of DateFrom over all rows (not the default)
    , first_value (DateFrom) over (partition by EmployeeId order by DateFrom asc rows between unbounded preceding and unbounded following) FirstDateFrom
    -- Last value of Completed over all rows (not the default)
    , last_value (Completed) over (partition by EmployeeId order by DateFrom asc rows between unbounded preceding and unbounded following) LastCompleted
    -- Find the Date of the last row with Completed = 1
    , max (case when Completed = 1 then DateFrom else null end) over (partition by EmployeeId order by DateFrom asc rows between unbounded preceding and unbounded following) LastCompletedNew
    -- Regular row number
    , row_number() over (partition by EmployeeId order by DateFrom desc) RowNumber
    -- Total number of rows with Completed = 1
    , sum(convert(int,Completed)) over (partition by EmployeeId) SumOfCompleted
    -- Max value of DateFrom where Completed = 0
    , max(case when Completed = 0 then DateFrom else null end) over (partition by EmployeeId order by DateFrom asc rows between unbounded preceding and unbounded following) MaxDateFrom
    -- Check the lagged complete to see if the last 2 rows are completed = 1
    , lag(Completed) over (partition by EmployeeId order by DateFrom asc) LaggedComplete
    -- Borrowed from Gordon to check which rows are prior to the last Completed = 1 and before the preceding Completed = 1
    , sum(case when completed = 1 then 1 else 0 end) over (partition by employeeid order by datefrom desc) as completed_seqnum
  from @Test
)
select
  EmployeeId
  -- Use the only DateFrom if there is only one
  , coalesce(case
    -- Scenario 1
    when SumOfCompleted = 0 then FirstDateFrom
    when LastCompleted = 1 then
      case
      -- Scenario 4
      when coalesce(LaggedComplete,0) = 1 then DateFrom
      -- Scenario 3
      else Scenario3
      end
    -- Scenario 2
    else ActualResult
    end, DateFrom) FinalResult
  --, * -- Uncomment for working
from (
  select *
    -- Find the lowest DateFrom which is greater then the DateFrom of the last row where Completed = 1
    , min(case when DateFrom > LastCompletedNew then DateFrom else null end) over (partition by EmployeeId) ActualResult
    -- Find the min DateFrom over the rows between the last Completed=1 and the Completed=1 before it (if it exists)
    , min(case when completed_seqnum = 1 then DateFrom else null end) over (partition by EmployeeId order by DateFrom asc rows between unbounded preceding and unbounded following) Scenario3
  from cte
) x
-- Because we have calculated the same result for every row we just take the first
where RowNumber = 1
order by x.EmployeeId asc, x.DateFrom asc;

注意：这假设每个日期只有一行。

Answer 2

我认为您只需要有条件的聚合——带有一堆逻辑。假设你每天都有行，我想这就是你想要的：

select employeeid,
       (case when -- case 4
                  min(completed) = max(completed) and
                  min(completed) = 'true'
             then max(datefrom) 
             when -- case 1
                  min(completed) = max(completed) and
                  min(completed) = 'false'
             then min(datefrom) 
             when -- case 3
                  max(datefrom) = max(case when completed = 'true' then datefrom end)
             then min(case when completed_seqnum = 1 then datefrom end)
             else dateadd(day, 1, max(case when completed = 'true' then datefrom end))
        end)
from (select t.*,
             sum(case when completed = 'true' then 1 else 0 end) over (partition by employeeid order by datefrom desc) as completed_seqnum
      from t
     ) t
group by employeeid;

每天需要一行实际上只是为了方便——例如，允许代码添加一天以获取特定“true”false 之后的日期。这也可以在子查询中使用 lead() 来完成。

注意：这不会处理所有条件（至少对于非 NULL 日期。例如，它 returns NULL 当末尾有一系列“真”时的数据。

如果这是一个问题 -- 您的问题的这个版本已经被问到。提出一个 new 问题，并提供适当的示例数据和所需的结果。我还认为您可能能够解释您试图解决的问题并简化解释。

编辑：

如果缺少日期，您可以使用：

select employeeid,
       (case when -- case 4
                  min(completed) = max(completed) and
                  min(completed) = 'true'
             then max(datefrom) 
             when -- case 1
                  min(completed) = max(completed) and
                  min(completed) = 'false'
             then min(datefrom) 
             when -- case 3
                  max(datefrom) = max(case when completed = 'true' then datefrom end)
             then min(case when completed_seqnum = 1 then datefrom end)
             else max(case when completed = 'true' then next_datefrom end)
        end)
from (select t.*,
             lead(datefrom) over (partition by employeeid order by datefrom) as next_datefrom,
             sum(case when completed = 'true' then 1 else 0 end) over (partition by employeeid order by datefrom desc) as completed_seqnum
      from t
     ) t
group by employeeid;

查看以获取具有复杂条件的最小日期

View to get the minimum date with a complicated condition

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