oracle树形查询

1. WITH AS 示例

with 
e as (select * from employees),
d as (select * from departments)
select * from e,d where e.department_id = d.department_id
and d.department_id = 10

 

2. 树形查询 

    2.1 connect by

    兄弟结点排序： order siblings by 

    PRIOR放在子节点一侧，会从start with开始向下寻找子节点

    PRIOR放在父节点一侧，会从start with开始向上寻找父节点 

SELECT e.employee_id,e.salary,e.last_name,e.manager_id,level from employees e
where level =2  --可直接查询某一层级的数据
start with e.employee_id = 100
connect by  e.employee_id =prior e.manager_id
order siblings by e.salary asc  --兄弟结点间排序

　　2.2  nocycle

　　如果树形结构有嵌套循环的话，会报错ORA-01436，需要在CONNECT BY 后关键字NOCYCLE :

　　CONNECT BY NOCYCLE PRIOR

select last_name "Employee",
       level,
       sys_connect_by_path(last_name, '/') "Path"
  from emp
 where level <= 3
   and department_id = 80
 start with last_name = 'King'
connect by nocycle prior employee_id = manager_id;

　　2.3  CONNECT_BY _ISLEAF，查询所有的叶子节点，含有子节点的为0　

select sys_connect_by_path(e.last_name, '/')
  from employees e
 where connect_by_isleaf = 1
 start with e.employee_id = 100
connect by prior e.employee_id = e.manager_id

　　2.4 CONNECT_BY_ROOT

--如果指定了start_with则获取的是最顶层结点的信息
select e.employee_id,
       e.last_name,
       e.manager_id,
       connect_by_root e.last_name root_last_name,
       connect_by_root e.hire_date root_hire_date
  from employees e
 start with e.employee_id = 100
connect by prior e.employee_id = e.manager_id

--如果为指定start with会依次展示结点的所有的父级结点
with v as  (select e.employee_id,
       e.last_name,
       e.manager_id,
       connect_by_root e.last_name root_last_name,
       connect_by_root e.hire_date root_hire_date
  from employees e
-- start with e.employee_id = 100
connect by prior e.employee_id = e.manager_id)
select * from v where v.employee_id = 206