2017-05-25 46 views
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我的任务是显示该经理的最低薪酬员工的MGR和薪水。最小()Access中的子查询

我需要排除MGR未知的任何人,并且排除最低工资低于1000美元的任何群体。结果应按工资降序排列。

下面是表:

+-------+--------+-----------+------+------------+-----------+-----------+------+ 
    | Empno | Ename | Job | Mgr | Hiredate | Sal | Comm | Dept | 
    +-------+--------+-----------+------+------------+-----------+-----------+------+ 
    | 7839 | KING | PRESIDENT |  | 11/17/1981 | $5,000.00 | $0.00  | 10 | 
    | 7782 | CLARK | MANAGER | 7839 | 6/9/1981 | $2,450.00 | $0.00  | 10 | 
    | 7934 | MILLER | CLERK  | 7782 | 1/23/1982 | $1,300.00 | $0.00  | 10 | 
    | 7902 | FORD | ANALYST | 7566 | 12/3/1981 | $3,000.00 | $0.00  | 20 | 
    | 7788 | SCOTT | ANALYST | 7566 | 12/9/1982 | $3,000.00 | $0.00  | 20 | 
    | 7876 | ADAMS | CLERK  | 7788 | 1/12/1983 | $1,100.00 | $0.00  | 20 | 
    | 7369 | SMITH | CLERK  | 7902 | 12/17/1980 | $800.00 | $0.00  | 20 | 
    | 7566 | JONES | MANAGER | 7839 | 4/2/1981 | $0.00  | $0.00  | 20 | 
    | 7698 | BLAKE | MANAGER | 7839 | 5/1/1981 | $2,850.00 | $0.00  | 30 | 
    | 7499 | ALLEN | SALESMAN | 7698 | 2/20/1981 | $1,600.00 | $300.00 | 30 | 
    | 7844 | TURNER | SALESMAN | 7698 | 9/8/1981 | $1,500.00 | $0.00  | 30 | 
    | 7521 | WARD | SALESMAN | 7698 | 2/22/1981 | $1,250.00 | $500.00 | 30 | 
    | 7654 | MARTIN | SALESMAN | 7698 | 9/28/1981 | $1,250.00 | $1,400.00 | 30 | 
    | 7900 | JAMES | CLERK  | 7698 | 12/3/1981 | $950.00 | $0.00  | 30 | 
    +-------+--------+-----------+------+------------+-----------+-----------+------+ 

这里是我到目前为止的代码:

SELECT EMp.Mgr, EMp.Ename, EMp.Sal AS Sal 
FROM EMp 
GROUP BY EMp.Mgr, EMp.Ename, EMp.Sal 
HAVING (((EMp.Mgr) Is Not Null) AND ((EMp.Sal)>1000)) 
ORDER BY EMp.Sal DESC; 

我目前的代码的问题是,它没有考虑到最低工资的参数。我相信这需要通过使用子查询完成,但我完全相信如何继续...

任何人都可以请协助吗?

+0

什么是sal列的数据类型? –

+0

这是一个货币数据类型 – Darren

+0

你可以试试emp.sal> CCur(1000)? –

回答

1

试试这个:

SELECT EMp.Mgr, EMp.Ename, EMp.Sal AS Sal 
FROM EMp 
WHERE emp.Sal = (select MIN(sal) from emp as emp2 where emp2.MGr = emp.Mgr and emp2.sal > 1000) 
GROUP BY EMp.Mgr, EMp.Ename, EMp.Sal 
HAVING EMp.Mgr Is Not Null 
ORDER BY EMp.Sal DESC; 
+0

感谢Jerrad!我相信这正是我正在寻找的东西 – Darren

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我并不完全相信这个答案涉及到1000美元的部分。你是否只想不列出任何经理的最低薪酬雇员的薪水低于此值?此代码将显示所有雇员支付超过1000美元的经理的所有经理。 –

0

请尝试

with one as 
(
SELECT EMp.Mgr,min(EMp.Sal) MinSlary 
FROM EMp 
GROUP BY EMp.Mgr 
) 
select a.Mgr,b.EName,b.Sal from one a 
inner join Emp b on a.Mgr=b.Mgr and a.MinSlary=b.Sal 
where a.Mgr is not null and a.MinSlary>1000 
+0

嗯我复制/粘贴你的代码准确,并给出了错误:“无效的SQL语句;预期的'删除','插入','程序','选择'或'更新' – Darren

+0

访问不支持'与'statement。 – Jerrad

+0

感谢Jerrad,对于Info,我并没有意识到它 –

0

以下是紧密依托Jerrad的回答,但解决有关我在评论那里提出的关切:

SELECT Emp.Mgr, Emp.Ename, Emp.Sal AS Sal 
FROM Emp 
WHERE Emp.Sal=(SELECT MIN(sal) 
       FROM Emp as Emp2 
       WHERE Emp2.MGr = Emp.Mgr 
       HAVING min(Emp2.sal) >= 1000) 
GROUP BY Emp.Mgr, Emp.Ename, Emp.Sal 
HAVING Emp.Mgr Is Not Null 
ORDER BY Emp.Sal DESC; 

使用您的样本数据,返回与Jerrad相同的行,但省略与经理7839或7698一起。其中,这两个管理人员的工资为0美元和950美元。我解释原始问题的方式(“排除最低工资低于1000美元的任何团体”),这些经理应该被排除在结果之外。