Question

我在尝试解决这个问题时遇到了麻烦，我只想补充一点工资员工的身份是不同的。我以为我可以使用decode（）函数执行此操作，但我无法定义适合的表达式。我的目标是

SUM(DECODE(S.ID,IS DISTINCT,S.SALARY))

但这不起作用！

所以完整查询看起来像

SELECT B.ID, SUM(S.SALARY), COUNT(DISTINCT S.ID), COUNT(DISTINCT RM.MEMBER_ID)
FROM BRANCH B 
INNER JOIN STAFF S ON S.BRANCH_ID = B.ID
INNER JOIN RECRUIT_MEMBER RM ON RM.BRANCH_ID = B.ID
GROUP BY B.ID;

但是问题在于SUM(S.SALARY)它会从重复的ID中增加工资

Answer 1

我不知道DECODE，但这应该有效：

SELECT 
  SUM(S.SALARY)
FROM <table> S
WHERE NOT EXISTS (
  SELECT ID FROM <table> WHERE ID=S.ID GROUP BY ID HAVING COUNT(*)>1
)

Answer 2

也许是这样的......

SELECT E.ID, SUM(E.Salary)
FROM Employers E
WHERE E.ID IN (SELECT DISTINCT E2.ID FROM Employers E2)
GROUP BY E.ID

如果没有，也许您可以发布一些样本数据，以便我能更好地理解

Answer 3

连接引入了重复的行。解决此问题的一种方法是添加行号以顺序识别不同的ID。真正的方法是修复连接，这样就不会发生，但这是第一种方式：

SELECT B.ID, SUM(CASE WHEN SEQNUM = 1 THEN S.SALARY END),
       COUNT(DISTINCT S.ID), COUNT(DISTINCT RM.MEMBER_ID)
FROM (SELECT B.ID, S.ID, RM.MEMBER_ID,
             ROW_NUMBER() OVER (PARTITION BY S.ID ORDER BY S.ID) as seqnum
      FROM BRANCH B 
      INNER JOIN STAFF S ON S.BRANCH_ID = B.ID
      INNER JOIN RECRUIT_MEMBER RM ON RM.BRANCH_ID = B.ID
     ) t
GROUP BY B.ID

Answer 4

你可以像这样创建一个每个ID只有一个薪水的虚拟表......

SELECT
   ...whatever fields you've already got...
   s.Salary
FROM
   ...whatever tables and joins you've already got...

   LEFT JOIN (SELECT ID, MAX(SALARY) as "Salary" FROM SALARY_TABLE GROUP BY ID) s
   ON whatevertable.ID = s.ID

ID不同时的SUM（SALARY）

4 个答案: