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所有内容？

当给定'101'作为节点时，您将不知道哪个是根。

当root是给定节点时，查询应该可用。

Answer 1

您需要首先遍历树，然后让所有经理走下去取得所有员工：

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
     )
;

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

编辑：

如果给定节点也可能是根节点，请扩展查询以在父节点列表中包含给定节点：

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
     )
;

在http://www.sqlfiddle.com/#!4/d15e7/32

处小提琴

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

Here是一个小提琴

修改
这是另一个尝试（在了解完整问题之后）：

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)

我使用100，101和1010进行了检查，结果很好 Here是一个小提琴

Answer 3

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 in  ( 
  select employee_id from employees 
  where manager_id is null )

从根节点找到整个树，给出任何节点

3 个答案:

编辑：