从根节点找到整个树,给出任何节点

Question

如何在给定树的节点的情况下找到整个树？

树的例子:

       100
  101        102
1010 1011   1020  1021


select level, employee_id, last_name, manager_id ,
       connect_by_root employee_id as root_id
  from employees
 connect by prior employee_id = manager_id
 start with employee_id = 101
;

表中的根是(父,子)示例(100,101)表中没有(null,100)行.

上面的查询只给出了从101开始的孩子.但是,让我说我想要从根开始的一切？

当'101'作为节点时,您将不知道哪个是根.

当根是给定节点时,查询应该可用.

Answer 1

您需要首先遍历树以使所有经理然后遍历以获取所有员工:

select level, employee_id, last_name, manager_id ,
       connect_by_root employee_id as root_id
   from employees
connect by prior employee_id = manager_id -- down the tree
start with manager_id in ( -- list up the tree
     select manager_id 
       from employees
     connect by employee_id = prior manager_id -- up the tree
     start with employee_id = 101
     )
;

见http://www.sqlfiddle.com/#!4/d15e7/18

编辑:

如果给定节点也可能是根节点,则扩展查询以将父节点列表中的给定节点包括在内:

非根节点的示例:

select distinct employee_id, last_name, manager_id 
   from employees
connect by prior employee_id = manager_id -- down the tree
start with manager_id in ( -- list up the tree
     select manager_id 
       from employees
     connect by employee_id = prior manager_id -- up the tree
     start with employee_id = 101
     union 
     select manager_id -- in case we are the root node
       from employees
     where manager_id = 101
     )
;

根节点示例:

select distinct employee_id, last_name, manager_id 
   from employees
connect by prior employee_id = manager_id -- down the tree
start with manager_id in ( -- list up the tree
     select manager_id 
       from employees
     connect by employee_id = prior manager_id -- up the tree
     start with employee_id = 100
     union 
     select manager_id -- in case we are the root node
       from employees
     where manager_id = 100
     )
;

小提琴在http://www.sqlfiddle.com/#!4/d15e7/32

Answer 2

为什么不呢:

select level, employee_id, last_name, manager_id ,
connect_by_root manager_id as root_id
from employees
connect by prior employee_id = manager_id
start with manager_id = 100

这是一个小提琴

编辑
这是另一个尝试(在了解完整问题后):

with t as (
select case when mgr.employee_id is null then
1 else 0 end is_root, emp.employee_id employee, emp.manager_id manager, emp.last_name last_name

from employees mgr right outer join employees emp
on mgr.employee_id = emp.manager_id
),
tmp as (

select level, employee, last_name, manager ,
connect_by_root manager as root_id,
manager||sys_connect_by_path(employee,
',') cbp

from t
connect by prior employee = manager
start with t.is_root =
1 )
select * from tmp
where tmp.root_id in (select root_id from tmp where employee= 101 or manager = 101)

我检查过它100,101并且1010它运作良好
这是一个小提琴