Oracle 層次化查詢

create table employees(

employee_id integer constraint employees_pk primary key,
manager_id integer constraint employees_fk_employees references employees(employee_id),
first_name varchar2(10) not null,
last_name varchar2(10) not null,
title varchar2(20),
salary number(6,0)
);insert into employees(employee_id,first_name,last_name,title,salary) values(1,'james','smith','ceo',800000);
insert into employees values(2,1,'ron','johnson','sales manager',600000);
insert into employees values(3,2,'fred','hobbs','sales person',200000);
insert into employees values(4,1,'susan','jones','support manager',500000);
insert into employees values(5,2,'rob','green','sales person',40000);
insert into employees values(6,4,'jane','brown','support person',45000);
insert into employees values(7,4,'jone','grey','support person',30000);
insert into employees values(8,7,'jean','blue','support person',29000);
insert into employees values(9,6,'henry','heyson','support person',30000);
insert into employees values(10,1,'kenvin','black','ops manager',100000);
insert into employees values(11,10,'keith','long','ops person',50000);
insert into employees values(12,10,'frank','howard','ops person',50000);
insert into employees values(13,10,'doreen','penn','ops person',47000);

員工的關係式乙個樹形結構，要提取不同員工之間的關係可以使用select 語句的connect by 和start with 子句執行層次化查詢。

語法：select [level],column,...

from table

[where where_condition]

start with start_condition connect by prior prior_condition

level 是乙個偽列，代表樹的第幾層。根節點的值是1，根節點的子節點是2，以此類推。

start_condition 定義了層次化查詢的起點。當編寫層次化查詢時必須指定start with 子句。例如：將start_condition 定義為 employee_id=1,表示從員工id=1開始查詢。

prior_condition 定義了父行和子行之間的關係。當編寫層次化查詢時必須指定prior_codition子句。例如將prior_condition 定義為 employee_id=manager_id,表示父節點的employee_id和子節點的manager_id之間存在相等關係。

select employee_id,manager_id,first_name,last_name 

from employees 
start with employee_id=1 connect by prior employee_id=manager_id;

1		james	smith

2 1 ron johnson
3 2 fred hobbs
5 2 rob green
4 1 susan jones
6 4 jane brown
9 6 henry heyson
7 4 jone grey
8 7 jean blue
10 1 kenvin black
11 10 keith long
12 10 frank howard
13 10 doreen penn

select level,employee_id,manager_id,first_name,last_name from employees start with employee_id=1 connect by prior employee_id=manager_id order by level;

1	1		james	smith

2 10 1 kenvin black
2 2 1 ron johnson
2 4 1 susan jones
3 13 10 doreen penn
3 7 4 jone grey
3 11 10 keith long
3 5 2 rob green
3 3 2 fred hobbs
3 12 10 frank howard
3 6 4 jane brown
4 8 7 jean blue
4 9 6 henry heyson

select level,lpad('-',2*(level-1))||first_name||' '||last_name as employee from employees start with employee_id=1 connect by prior employee_id=manager_id;

1	james smith

2 -ron johnson
3 -fred hobbs
3 -rob green
2 -susan jones
3 -jane brown
4 -henry heyson
3 -jone grey
4 -jean blue
2 -kenvin black
3 -keith long
3 -frank howard
3 -doreen penn

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