查看获取条件复杂的最小日期

编程问答 2022-05-28

问题描述

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周期，可以完成也可以不完成。
  • 每位员工可以有多个子时段。
  • 一个时期由一系列有序的子时期定义，直到最后一个子时期完成。

我想创建一个视图，该视图将返回每个 EmployeeID 上一期的开始日期，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 为真后的最小 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 之前和之前的 true 之后的最小 DateFrom（如果存在）。 [最后一期由多个子期完成]
+----------+-----------+------------+
| 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
  1. 如果最大 DateFrom 的 Completed=true 并且没有其他行或它之前的行 Completed=true，则返回最大 DateFrom。 [最后一期以一个子期完成]
+----------+-----------+------------+
| 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

解决方法

我认为您只想要条件聚合——带有一堆逻辑。假设您每天都有行，我认为这可以满足您的需求：

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，允许代码添加一天以获取特定“真”假之后的日期。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，当数据末尾有一系列“true”时，它返回 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。我还认为您可能能够解释您正在尝试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;
,

这是一个有效的查询。它可能过于复杂，但我把简化留给你。

处理这3种情况，都按要求按EmployeeId划分，如下：

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

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

  3. 棘手的情况，当 Completed=1 存在并且它不是最后。在这种情况下，找到 Completed=1 的最近行的 DateFrom，然后为比先前检测到的行更近的所有行找到 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，则 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','2021-01-03',-- Scenario 2
(2,(2,1),'2021-01-04',-- Scenario 3
(3,(3,-- Special case,single row
(4,-- Scenario 4
(5,(5,1);

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。

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