Access 中的 Min() 子查询

Min() Subquery in Access

我的任务是显示该经理的最低薪员工的 MGR 和薪水。

我需要排除 MGR 未知的任何人,并排除最低工资低于 1000 美元的任何群体。结果应按薪水降序排列。

这里是 table:

    +-------+--------+-----------+------+------------+-----------+-----------+------+
    | Empno | Ename  |    Job    | Mgr  |  Hiredate  |    Sal    |   Comm    | Dept |
    +-------+--------+-----------+------+------------+-----------+-----------+------+
    |  7839 | KING   | PRESIDENT |      | 11/17/1981 | ,000.00 | [=10=].00     |   10 |
    |  7782 | CLARK  | MANAGER   | 7839 | 6/9/1981   | ,450.00 | [=10=].00     |   10 |
    |  7934 | MILLER | CLERK     | 7782 | 1/23/1982  | ,300.00 | [=10=].00     |   10 |
    |  7902 | FORD   | ANALYST   | 7566 | 12/3/1981  | ,000.00 | [=10=].00     |   20 |
    |  7788 | SCOTT  | ANALYST   | 7566 | 12/9/1982  | ,000.00 | [=10=].00     |   20 |
    |  7876 | ADAMS  | CLERK     | 7788 | 1/12/1983  | ,100.00 | [=10=].00     |   20 |
    |  7369 | SMITH  | CLERK     | 7902 | 12/17/1980 | 0.00   | [=10=].00     |   20 |
    |  7566 | JONES  | MANAGER   | 7839 | 4/2/1981   | [=10=].00     | [=10=].00     |   20 |
    |  7698 | BLAKE  | MANAGER   | 7839 | 5/1/1981   | ,850.00 | [=10=].00     |   30 |
    |  7499 | ALLEN  | SALESMAN  | 7698 | 2/20/1981  | ,600.00 | 0.00   |   30 |
    |  7844 | TURNER | SALESMAN  | 7698 | 9/8/1981   | ,500.00 | [=10=].00     |   30 |
    |  7521 | WARD   | SALESMAN  | 7698 | 2/22/1981  | ,250.00 | 0.00   |   30 |
    |  7654 | MARTIN | SALESMAN  | 7698 | 9/28/1981  | ,250.00 | ,400.00 |   30 |
    |  7900 | JAMES  | CLERK     | 7698 | 12/3/1981  | 0.00   | [=10=].00     |   30 |
    +-------+--------+-----------+------+------------+-----------+-----------+------+

到目前为止,这是我的代码:

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;

我当前代码的问题是它没有考虑最低工资参数。我相信这需要通过使用子查询来完成,尽管我完全确定如何继续......

有人可以帮忙吗?

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

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;

以下内容与杰拉德的回答密切相关,但解决了我在评论中提出的问题:

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,returns 与 Jerrad 相同的行,只是省略了经理 7839 或 7698 的行。除此之外,这两位管理人员的薪水为 0 美元和 950 美元。按照我解释原始问题 ("exclude any groups where the minimum salary is less than 00") 的方式,这些经理 应该 被排除在结果之外。