查看以获取条件复杂的最短日期

Question

我在 SQL Server 中有一个这样的表：

+----------+-----------+------------+
| 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期间，可以完成也可以不完成。
• 每个员工可以有多个子期间。
• 期间由有序子期间列表定义，直到最后一个子期间完成。

我想创建一个视图，该视图将返回每个 EmployeeID 的最后一个期间的开始日期，如下所示：

1. 如果没有 Completed 为 true，则获取最小 DateFrom。 [员工有一个时期尚未完成]

+----------+-----------+------------+
| 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

1. 否则，返回最后一个 Completed 为 true 后的最小 DateFrom。 [最后一期还没有完成]

+----------+-----------+------------+
| 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

1. 如果最大 DateFrom 已 Completed=true，则返回最后一个 Completed 为 true 之前的最小 DateFrom，如果存在，则返回它之前的 true。 [最后一期完成，有多个子期]

+----------+-----------+------------+
| 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

4. 如果最大 DateFrom 的 Completed=true 并且没有其他行或之前的行 Completed=true，则返回最大 DateFrom。 [最后一期以一个子期结束]

+----------+-----------+------------+
| 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，但在第三个例子中我得到了一个 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

我认为您只需要条件聚合——带有一堆逻辑。假设您每天都有行，我认为这可以满足您的要求：

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一种方便——例如，允许代码在特定的“真”假之后添加一天以获取日期。这也可以在子查询中使用lead() 来完成。

注意：这不能处理所有条件（至少对于非 NULL 日期。例如，当数据末尾有一系列“true”时，它会返回 NULL。

如果这是一个问题 - 您的问题的这个版本已被问到。提出一个新问题，并提供适当的样本数据和所需的结果。我还认为您也许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;

Answer 2

这是一个有效的查询。它可能过于复杂，但我把简化留给你。

它处理了3种情况，都按要求按EmployeeId分区，如下：

1. 当不存在Completed=1 时，使用sum(Completed) over() 检测到，然后使用first_value(DateFrom)。

2. 当最后一行值为completed=1，前一行为completed=0时，使用last_value(Completed)和lag(Completed)检测，则使用max(case when Completed = 0 then DateFrom else null end)。

3. 棘手的情况是，Completed=1 存在并且不是最后一个。在这种情况下，找到 Completed=1 的最近行的 DateFrom，然后为所有比先前检测到的行更新的行找到 min(DateFrom)，直到前面的 Completed=1。

4. 如果最后一行有completed=1，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：这里假设每个日期只有一行。