4

十年的大部分时间里，我一直在使用 SQL Server，而这种分组（或分区，或排名......我不确定答案是什么！）让我难过。感觉它也应该是一件容易的事。我将概括我的问题：

假设我有 3 名员工（不用担心他们辞职或其他任何事情......总是有 3 名），我会按月分配他们的薪水。

Month   Employee  PercentOfTotal
--------------------------------
1       Alice     25%
1       Barbara   65%
1       Claire    10%

2       Alice     25%
2       Barbara   50%
2       Claire    25%

3       Alice     25%
3       Barbara   65%
3       Claire    10%

如您所见，我在第 1 个月和第 3 个月支付了相同的百分比，但在第 2 个月，我给了 Alice 相同的 25%，但 Barbara 得到了 50%，而 Claire 得到了 25%。

我想知道的是我曾经给出的所有不同的分布。在这种情况下，将有两个——一个用于第 1 个月和第 3 个月，一个用于第 2 个月。

我希望结果看起来像这样（注意：ID，或定序器，或任何东西，无关紧要）

ID      Employee  PercentOfTotal
--------------------------------
X       Alice     25%
X       Barbara   65%
X       Claire    10%

Y       Alice     25%
Y       Barbara   50%
Y       Claire    25%

看起来很容易，对吧？我难住了！任何人都有一个优雅的解决方案？我只是在写这个问题时把这个解决方案放在一起，这似乎有效，但我想知道是否有更好的方法。或者也许是一种不同的方式，我可以从中学到一些东西。

WITH temp_ids (Month)
AS
(
  SELECT DISTINCT MIN(Month)
    FROM employees_paid
  GROUP BY PercentOfTotal
)
SELECT EMP.Month, EMP.Employee, EMP.PercentOfTotal
  FROM employees_paid EMP
         JOIN temp_ids IDS ON EMP.Month = IDS.Month
GROUP BY EMP.Month, EMP.Employee, EMP.PercentOfTotal

谢谢大家！-瑞奇

4

5 回答 5

4

这会以与您要求的格式略有不同的格式为您提供答案：

SELECT DISTINCT
    T1.PercentOfTotal AS Alice,
    T2.PercentOfTotal AS Barbara,
    T3.PercentOfTotal AS Claire
FROM employees_paid T1
JOIN employees_paid T2
  ON T1.Month = T2.Month AND T1.Employee = 'Alice' AND T2.Employee = 'Barbara'
JOIN employees_paid T3
  ON T2.Month = T3.Month AND T3.Employee = 'Claire'

结果：

Alice   Barbara  Claire
25%     50%      25%
25%     65%      10%

如果您愿意，可以使用UNPIVOT将此结果集转换为您要求的形式。

SELECT rn AS ID, Employee, PercentOfTotal
FROM (
    SELECT *, ROW_NUMBER() OVER (ORDER BY Alice) AS rn
    FROM (
        SELECT DISTINCT
            T1.PercentOfTotal AS Alice,
            T2.PercentOfTotal AS Barbara,
            T3.PercentOfTotal AS Claire
        FROM employees_paid T1
        JOIN employees_paid T2 ON T1.Month = T2.Month AND T1.Employee = 'Alice'
                                                      AND T2.Employee = 'Barbara'
        JOIN employees_paid T3 ON T2.Month = T3.Month AND T3.Employee = 'Claire'
    ) T1
) p UNPIVOT (PercentOfTotal FOR Employee IN (Alice, Barbara, Claire)) AS unpvt

结果：

ID  Employee  PercentOfTotal  
1   Alice     25%
1   Barbara   50%      
1   Claire    25%             
2   Alice     25%             
2   Barbara   65%              
2   Claire    10%
于 2010-06-14T22:31:00.943 回答
3

您想要的是让每个月的分布充当您希望在其他月份找到的值的签名或模式。尚不清楚的是，获得价值的员工是否与百分比分解一样重要。例如，Alice=65%、Barbara=25%、Claire=10% 是否与您示例中的第 3 个月相同？在我的示例中，我假设它不会相同。与 Martin Smith 的解决方案类似，我通过将每个百分比乘以 10 来找到签名。这假定所有百分比值都小于 1。例如，如果某人可能有 110% 的百分比，那么这会给这个解决方案带来问题。

With Employees As
    (
    Select 1 As Month, 'Alice' As Employee, .25 As PercentOfTotal
    Union All Select 1, 'Barbara', .65
    Union All Select 1, 'Claire', .10
    Union All Select 2, 'Alice', .25
    Union All Select 2, 'Barbara', .50
    Union All Select 2, 'Claire', .25
    Union All Select 3, 'Alice', .25
    Union All Select 3, 'Barbara', .65
    Union All Select 3, 'Claire', .10
    )
    , EmployeeRanks As
    (
    Select Month, Employee, PercentOfTotal
        , Row_Number() Over ( Partition By Month Order By Employee, PercentOfTotal ) As ItemRank
    From Employees
    )
    , Signatures As
    (
    Select Month
        , Sum( PercentOfTotal * Cast( Power( 10, ItemRank ) As bigint) ) As SignatureValue
    From EmployeeRanks
    Group By Month
    )
    , DistinctSignatures As
    (
    Select Min(Month) As MinMonth, SignatureValue
    From Signatures
    Group By SignatureValue
    )
Select E.Month, E.Employee, E.PercentOfTotal
From Employees As E
    Join DistinctSignatures As D
        On D.MinMonth = E.Month
于 2010-06-14T23:55:11.127 回答
2

如果我对您的理解正确，那么对于一般解决方案，我认为您需要将整个组连接在一起 - 例如生产Alice:0.25, Barbara:0.50, Claire:0.25. 然后选择不同的组，这样就可以做到（相当笨拙）。

WITH EmpSalaries
AS
(

SELECT 1 AS Month, 'Alice' AS Employee, 0.25 AS PercentOfTotal UNION ALL
SELECT 1 AS Month, 'Barbara' AS Employee, 0.65 UNION ALL
SELECT 1 AS Month, 'Claire' AS Employee, 0.10 UNION ALL

SELECT 2 AS Month, 'Alice' AS Employee, 0.25 UNION ALL
SELECT 2 AS Month, 'Barbara' AS Employee, 0.50 UNION ALL
SELECT 2 AS Month, 'Claire' AS Employee, 0.25 UNION ALL

SELECT 3 AS Month,  'Alice' AS Employee, 0.25 UNION ALL
SELECT 3 AS Month,  'Barbara' AS Employee, 0.65 UNION ALL
SELECT 3 AS Month,  'Claire' AS Employee, 0.10 
),
Months AS 
(
SELECT DISTINCT Month FROM EmpSalaries
),
MonthlySummary AS
(
SELECT Month,
Stuff(
            (
            Select ', ' + S1.Employee + ':' + cast(PercentOfTotal as varchar(20))
            From EmpSalaries As S1
            Where S1.Month = Months.Month
            Order By S1.Employee
            For Xml Path('')
            ), 1, 2, '') As Summary
FROM Months
)
SELECT * FROM EmpSalaries
WHERE Month IN (SELECT MIN(Month)
                FROM MonthlySummary
                GROUP BY Summary)
于 2010-06-14T23:31:49.047 回答
2

我假设性能不会很好（子查询的原因）

SELECT * FROM employees_paid where Month not in (
     SELECT
          a.Month
     FROM
          employees_paid a
          INNER JOIN employees_paid b ON 
               (a.employee = B.employee AND 
               a.PercentOfTotal = b.PercentOfTotal AND 
               a.Month > b.Month)
     GROUP BY
          a.Month,
          b.Month
     HAVING
          Count(*) = (SELECT COUNT(*) FROM employees_paid c 
               where c.Month = a.Month)
     )
  1. 内部 SELECT 执行自联接以识别匹配的员工和百分比组合（同月除外）。JOIN 中的 > 确保只获取一组匹配项，即如果 Month1 条目 = Month3 条目，我们将仅得到 Month3-Month1 条目组合，而不是 Month1-Month3、Month3-Month1 和 Month3-Month3。
  2. 然后，我们对每个月-月组合的匹配条目进行 COUNT 分组
  3. 然后 HAVING 排除了没有与月份条目一样多的匹配的月份
  4. 外部 SELECT 获取除内部查询返回的条目之外的所有条目（具有完整匹配的条目）
于 2010-06-15T06:32:19.710 回答
2

我只是在写这个问题时把这个解决方案放在一起，这似乎有效

我认为它不起作用。在这里，我添加了另外两组（分别为月 = 4 和 5），我认为它们是不同的，但结果是相同的，即仅月 = 1 和 2：

WITH employees_paid (Month, Employee, PercentOfTotal)
AS 
(
 SELECT 1, 'Alice', 0.25
 UNION ALL
 SELECT 1, 'Barbara', 0.65
 UNION ALL
 SELECT 1, 'Claire', 0.1
 UNION ALL
 SELECT 2, 'Alice', 0.25
 UNION ALL
 SELECT 2, 'Barbara', 0.5
 UNION ALL
 SELECT 2, 'Claire', 0.25
 UNION ALL
 SELECT 3, 'Alice', 0.25
 UNION ALL
 SELECT 3, 'Barbara', 0.65
 UNION ALL
 SELECT 3, 'Claire', 0.1
 UNION ALL
 SELECT 4, 'Barbara', 0.25
 UNION ALL
 SELECT 4, 'Claire', 0.65
 UNION ALL
 SELECT 4, 'Alice', 0.1
 UNION ALL
 SELECT 5, 'Diana', 0.25
 UNION ALL
 SELECT 5, 'Emma', 0.65
 UNION ALL
 SELECT 5, 'Fiona', 0.1
), 
temp_ids (Month)
AS
(
 SELECT DISTINCT MIN(Month)
   FROM employees_paid
  GROUP 
     BY PercentOfTotal
)
SELECT EMP.Month, EMP.Employee, EMP.PercentOfTotal
  FROM employees_paid AS EMP
       INNER JOIN temp_ids AS IDS 
          ON EMP.Month = IDS.Month
 GROUP 
    BY EMP.Month, EMP.Employee, EMP.PercentOfTotal;
于 2010-06-15T08:02:41.510 回答