Question

给出Oracle 11gr2中的以下树层次结构示例：

查询：

SELECT Lpad(ename,Length(ename) + LEVEL * 10 - 10,'-') 
FROM   emp 
START WITH mgr IS NULL 
CONNECT BY PRIOR empno = mgr


Result:


KING
----------JONES
--------------------SCOTT
------------------------------ADAMS
--------------------FORD
------------------------------SMITH
----------BLAKE
--------------------ALLEN
--------------------WARD
--------------------MARTIN
--------------------TURNER
--------------------JAMES
----------SAM
--------------------MILLER

我需要修剪我的树查询，这样每当ename以字符“S”开始，i.e. substr(ename,1,1) = 'S'时，显示此值/分支，但忽略下面的每一个，即在此之后不再显示任何内容水平。

因此，基于上述结果示例，新的结果示例将如下所示：

New Result:

KING
----------JONES
--------------------SCOTT
--------------------FORD
------------------------------SMITH
----------BLAKE
--------------------ALLEN
--------------------WARD
--------------------MARTIN
--------------------TURNER
--------------------JAMES
----------SAM

因此，儿童“ADAMS”和“MILLER”已从新结果中删除。

Answer 1

SELECT Lpad(ename,Length(ename) + LEVEL * 10 - 10,'-') 
FROM   emp 
START WITH mgr IS NULL 
CONNECT BY PRIOR empno = mgr and substr(PRIOR ename,1,1) <> 'S';

或者如果您想使用Recursive subquery factoring：

with employees(empno, ename, mgr, emp_level) as
(
    select empno, ename, mgr, 1 emp_level
    from emp
    where mgr is null
    union all
    select emp.empno, emp.ename, emp.mgr, emp_level+1 emp_level
    from emp
    join employees
        on emp.mgr = employees.empno
    where substr(employees.ename,1,1) <> 'S'
) search depth first by empno set order1
select Lpad(ename,Length(ename) + emp_level * 10 - 10,'-')
from employees
order by order1;

SQL Fiddle

如何在Oracle树层次结构中的特定分支/级别停止？

1 个答案: