如何从必须连接到导致重复的另一个表的表中正确地求和总计

Question

我有两张如下表：

PAY_TABLE

EMPLID  PAY
123     100
123     150
123     150

DEDUCTION_TABLE

EMPLID  DEDUCTION
123     15
123     30

我希望得到如下结果：

TOTAL_PAY
400

我想用一个相当简单的查询得到那个结果，我觉得我错过了一个明显的方法，但我似乎无法弄明白是什么。

例如，此查询返回800，因为PAY_TABLE中的每一行在加入DEDUCTION_TABLE时都会重复：

SELECT SUM(PAY) AS TOTAL_PAY 
FROM PAY_TABLE JOIN DEDUCTION_TABLE USING(EMPLID);

此查询返回250，因为DISTINCT关键字会导致150中的第二个PAY_TABLE值被忽略：

SELECT SUM(DISTINCT PAY) AS TOTAL_PAY 
FROM PAY_TABLE JOIN DEDUCTION_TABLE USING(EMPLID);

可能有几种方法可以做到这一点，但我正在寻找最简单的方法来返回400的结果。

以下是创建示例表以使其更容易的一些代码：

WITH 
PAY_TABLE AS (
  SELECT 123 AS EMPLID, 100 AS PAY FROM DUAL
  UNION ALL 
  SELECT 123, 150 FROM DUAL
  UNION ALL 
  SELECT 123, 150 FROM DUAL
),
DEDUCTION_TABLE AS (
  SELECT 123 AS EMPLID, 15 AS DEDUCTION FROM DUAL 
  UNION ALL
  SELECT 123, 30  FROM DUAL
)

Answer 1

目前还不清楚你需要什么，因为你的例子没有使用DEDUCTION_TABLE表，但我相信你想要的是在你JOIN之前聚合：

;with pay AS (SELECT EmplID,SUM(PAY) AS Pay
              FROM PAY_TABLE
              GROUP BY EmplID
             )
     ,ded AS (SELECT EmplID,SUM(DEDUCTION) AS Ded
              FROM DEDUCTION_TABLE
              GROUP BY EmplID
              )
SELECT *
FROM pay
LEFT JOIN ded
  ON pay.EmplID = ded.EmplID

Answer 2

假设您需要加入DEDUCTION_TABLE才能确保为员工扣除：

SELECT SUM(P.PAY) AS TOTAL_PAY 
FROM PAY_TABLE P
WHERE EXISTS (SELECT NULL FROM DEDUCTION_TABLE D
             WHERE D.EMPLID = P.EMPLID;