1

考虑下表

create table EMPLOYEE
(
  empno    NUMBER not null,
  ename    VARCHAR2(100),
  salary   NUMBER,
  hiredate DATE,
  manager  NUMBER
);

alter table EMPLOYEE add constraint PK_EMP primary key (EMPNO);

alter table EMPLOYEE 
   add constraint FK_MGR foreign key (MANAGER)
      references EMPLOYEE (EMPNO);

这是一个自循环表，即每个员工都有一个经理，除了根。

我想在这个表上运行以下查询：

发现所有员工的薪水都高于他们的经理？

附言

结构中只有一个根

考虑以下查询

SELECT LPAD(emp.ename, (LEVEL-1)*5 + LENGTH(emp.ename), ' ') AS "Hierarchy"
    FROM employee emp
    START WITH emp.manager IS NULL
    CONNECT BY manager = PRIOR empno;

结果将是这样的：

Alice
    Alex
    Abbey
Sarah
Jack
    Bill
    Jacob
    Valencia
Bob
    Babak
...

我做了以下查询

SELECT LPAD(emp.ename, (LEVEL-1)*5 + LENGTH(emp.ename), ' ') AS "Hierarchy"
  FROM employee emp
    START WITH empno IN (SELECT empno FROM employee)
    CONNECT BY PRIOR manager = empno;

从下到上为员工表中的每个员工创建一个子树，但我不知道如何导航以获得所需的结果！

4

2 回答 2

1

这是一种方法

with fullemployee (empno, ename, salary, key)
as
(
  select A.empno, A.ename, A.salary, A.empno || '.' from 
      employee A
  where A.manager is null
  union all
  select C.empno, C.ename, C.salary, D.key || '.' || C.empno from 
      employee C
      inner join fullemployee D on C.manager = D.empno
)
select E.ename, F.ename as manager from fullemployee E
inner join fullemployee F on E.key like F.key || '%' and E.key <> F.key
where E.salary > F.salary

或等效地

with fullemployee (empno, ename, salary, key)
as
(
   SELECT empno, ename, salary, SYS_CONNECT_BY_PATH(empno, '.') || '.'
   FROM employee
   START WITH manager is null
   CONNECT BY PRIOR empno = manager
)
select E.ename, F.ename as manager from fullemployee E
inner join fullemployee F on E.key like F.key || '%' and E.key <> F.key
where E.salary > F.salary

SQL 小提琴 - http://sqlfiddle.com/#!4/37f4ae/35

于 2015-06-20T05:18:18.010 回答
0

This should do the work. Remove that or condition if you don't want the 'root' in your list.

select e.empno, e.ename, e.salary from employee e
    inner join employee mgr on mgr.empno = e.manager
where e.salary > mgr.salary
or (e.manager = mgr.empno)
于 2015-06-20T03:56:56.793 回答