如何从联合声明中删除重复的行

Question

好的 - 我已经看过，看过并找到很多示例，但没有满足我的需要。也许我用错误的词搜索，但我可以使用你的帮助。我会尽可能详细地提供。

我需要生成一个报告，将两个表（或者说一个视图和一个表）的字段合并到一个报表的表中。下面是我使用的语句：

SELECT A.ConfInt, A.Conference, 
     NULL as Ordered, 
     NULL as Approved, 
     NULL as PickedUp, 
     SUM(dbo.Case_Visit_Payments.Qty) AS Qty 
FROM   dbo.Conferences as A INNER JOIN 
         dbo.Case_Table ON A.ConfInt = dbo.Case_Table.Conference_ID INNER JOIN 
         dbo.Case_Visit_Payments ON dbo.Case_Table.Case_ID = dbo.Case_Visit_Payments.Case_ID 
WHERE  (dbo.Case_Visit_Payments.Item_ID = 15 AND A.ProjectCool = 1) 
GROUP BY A.Conference, A.ConfInt 
UNION 
SELECT B.ConfInt, 
     B.Conference, 
     SUM(dbo.Cool_Fan_Order.NumberOfFansRequested) AS Ordered, 
     SUM(dbo.Cool_Fan_Order.Qty_Fans_Approved) AS Approved, 
     SUM(dbo.Cool_Fan_Order.Qty_Fans_PickedUp) AS PickedUp, 
     NULL AS Qty 
FROM   dbo.Conferences as B LEFT OUTER JOIN 
         dbo.Cool_Fan_Order ON B.ConfInt = dbo.Cool_Fan_Order.Conference_ID 
where B.ProjectCool = 1 
GROUP BY B.Conference, B.ConfInt 

，这里是结果：

4 Our Lady  NULL NULL NULL 11 
4 Our Lady  40  40  40  NULL 
7 Holy Rosary  20  20  20  NULL 
11 Little Flower NULL NULL NULL 21 
11 Little Flower 5  5  20  NULL 
19 Perpetual Help NULL NULL NULL 2 
19 Perpetual Help 20  20  20  NULL 

我强烈希望是没有重复的行，如：

4 Our Lady  40  40  40  11 
7 Holy Rosary  20  20  20  NULL 
11 Little Flower 5  5  20  21 
19 Perpetual Help 20  20  20  2 

我希望这个问题很清楚。任何建议将不胜感激。我的确标记为已回答。 :)

格雷戈里

Answer 1

简单的回答是包装您的查询里面一个又一个，

SELECT ConfInt 
    , Conference 
    , SUM(Ordered) AS Ordered 
    , SUM(Approved) As Approved 
    , SUM(PickedUp) AS PickedUp 
    , SUM(Qty) AS Qty 
    FROM (

     <your UNION query here> 

    ) 
GROUP BY ConfInt, Conference 

---

这不是这是实现结果集的唯一方法，但它是满足特定要求的最快速解决方案。

作为替代方案，我相信这些查询将返回相同的结果：

我们可以使用相关子查询在SELECT列表以获得数量：

;WITH q AS 
     (SELECT B.ConfInt 
      , B.Conference 
      , SUM(o.NumberOfFansRequested) AS Ordered 
      , SUM(o.Qty_Fans_Approved) AS Approved 
      , SUM(o.Qty_Fans_PickedUp) AS PickedUp 
      FROM dbo.Conferences as B 
      LEFT 
      JOIN dbo.Cool_Fan_Order o ON o.Conference_ID = B.ConfInt 
     WHERE B.ProjectCool = 1 
     GROUP BY B.ConfInt, B.Conference 
    ) 
SELECT q.ConfInt 
     , q.Conference 
     , q.Ordered 
     , q.Approved 
     , q.PickedUp 
     , (SELECT SUM(v.Qty) 
      FROM dbo.Case_Table t 
      JOIN dbo.Case_Visit_Payments v ON v.Case_ID = t.Case_ID 
      WHERE t.Conference_ID = q.ConfInt 
      AND v.Item_ID = 15 
     ) AS Qty 
    FROM q 
    ORDER BY q.ConfInt, q.Conference 

---

或者，我们可以使用对两个查询进行LEFT JOIN操作，而不是UNION。 （我们知道引用Cool_Fan_Order的查询可以是外连接的左侧，因为我们知道它至少返回与其他查询相同数量的行（基本上，我们知道另一个查询不能返回ConfInt的值和会议不在的Cool_Fan_Order查询。）

;WITH p AS 
     (SELECT A.ConfInt 
      , A.Conference 
      , SUM(v.Qty) AS Qty 
      FROM dbo.Conferences as A 
      JOIN dbo.Case_Table t ON t.Conference_ID = A.ConfInt 
      JOIN dbo.Case_Visit_Payments v ON v.Case_ID = t.Case_ID 
     WHERE A.ProjectCool = 1 
      AND v.Item_ID = 15 
     GROUP BY A.ConfInt, A.Conference 
    ) 
    , q AS 
     (SELECT B.ConfInt 
      , B.Conference 
      , SUM(o.NumberOfFansRequested) AS Ordered 
      , SUM(o.Qty_Fans_Approved) AS Approved 
      , SUM(o.Qty_Fans_PickedUp) AS PickedUp 
      FROM dbo.Conferences as B 
      LEFT 
      JOIN dbo.Cool_Fan_Order o ON B.ConfInt = o.Conference_ID 
     WHERE B.ProjectCool = 1 
     GROUP BY B.ConfInt, B.Conference 
    ) 
SELECT q.ConfInt 
     , q.Conference 
     , q.Ordered 
     , q.Approved 
     , q.PickedUp 
     , p.Qty 
    FROM q 
    LEFT 
    JOIN p ON p.ConfInt = q.ConfInt AND p.Conference = q.Conference 
    ORDER BY q.ConfInt, q.Conference 

这三个之间选择（他们所有所有conditons下返回结果集当量），归结为可读性和可维护性和性能。在足够大的行集，这三条语句之间可能会有一些可观察到的性能差异。

Answer 2

你可以用你的实际查询作为子查询，通过非聚集列使用您的非重复值和组聚合函数（Max或SUM）

SELECT ConfInt, Conference, MAX(Ordered), MAX(Approved), MAX(PickedUp), MAX(Qty) 
FROM (<your actualQuery>) 
GROUP BY ConfInt, Conference. 

Answer 3

我只是在第一次选择并删除联合时将联接添加到了Cool_Fan_Order。

这是否返回相同的结果？

select 
     A.ConfInt, 
     A.Conference, 
     sum(dbo.Cool_Fan_Order.NumberOfFansRequested) as Ordered, 
     sum(dbo.Cool_Fan_Order.Qty_Fans_Approved) as Approved, 
     sum(dbo.Cool_Fan_Order.Qty_Fans_PickedUp) as PickedUp, 
     sum(sub.Qty) as Qty 
    from 
     dbo.Conferences as A 
     left outer join 
     (
      select 
       c.ConfInt, 
       cvp.Qty 
      from dbo.Conferences c 
       inner join dbo.Case_Table ct 
        on a.confInt=ct.Conference_ID 
       inner join dbo.Case_Visit_Payments cvp 
        on ct.Case_ID=cvp.Case_ID 
      where cvp.Item_ID=15 
     ) sub 
      on a.ConfInt=sub.ConfInt 
     left outer join dbo.Cool_Fan_Order 
      on A.ConfInt = dbo.Cool_Fan_Order.Conference_ID 
    where 
     (
     A.ProjectCool = 1 
     ) 
    group by 
     A.Conference, 
     A.ConfInt