以下は、私の CREATE TABLE スクリプトです。
create table EMPLOYEES
(EmpID char(4) unique Not null,
Ename varchar(10),
Job varchar(9),
MGR char(4),
Hiredate date,
Salary decimal(7,2),
Comm decimal(7,2),
DeptNo char(2) not null,
Primary key(EmpID),
Foreign key(DeptNo) REFERENCES DEPARTMENTS(DeptNo));
以下は私のINSERTスクリプトです:
insert into EMPLOYEES values (7839,'King','President',null,'17-Nov-11',5000,null,10);
insert into EMPLOYEES values (7698,'Blake','Manager',7839,'01-May-11',2850,null,30);
insert into EMPLOYEES values (7782,'Clark','Manager',7839,'02-Jun-11',2450,null,10);
insert into EMPLOYEES values (7566,'Jones','Manager',7839,'02-Apr-11',2975,null,20);
insert into EMPLOYEES values (7654,'Martin','Salesman',7698,'28-Feb-12',1250,1400,30);
insert into EMPLOYEES values (7499,'Allen','Salesman',7698,'20-Feb-11',1600,300,30);
insert into EMPLOYEES values (7844,'Turner','Salesman',7698,'08-Sep-11',1500,0,30);
insert into EMPLOYEES values (7900,'James','Clerk',7698,'22-Feb-12',950,null,30);
insert into EMPLOYEES values (7521,'Ward','Salesman',7698,'22-Feb-12',1250,500,30);
insert into EMPLOYEES values (7902,'Ford','Analyst',7566,'03-Dec-11',3000,null,20);
insert into EMPLOYEES values (7369,'Smith','Clerk',7902,'17-Dec-10',800,null,20);
insert into EMPLOYEES values (7788,'Scott','Analyst',7566,'09-Dec-12',3000,null,20);
insert into EMPLOYEES values (7876,'Adams','Clerk',7788,'12-Jan-10',1100,null,20);
insert into EMPLOYEES values (7934,'Miller','Clerk',7782,'23-Jan-12',1300,null,10);
以下は私のSELECTスクリプトです:
select distinct e.Ename as Employee, m.mgr as reports_to
from EMPLOYEES e
inner join Employees m on e.mgr = m.mgr;
対応するマネージャーの ID を持つ従業員を取得しています。
Ford 7566
Scott 7566
Allen 7698
James 7698
Martin 7698
Turner 7698
Ward 7698
Miller 7782
Adams 7788
Blake 7839
Clark 7839
Jones 7839
Smith 7902
マネージャーの名前も記載するにはどうすればよいですか? *正しい内部結合を行っていますか?*