Oracle SQL完整路径层次结构

Question

我正在寻找解决方案如何使用sys_connect_by_path仅选择完整的层次结构＆＃39;分支＆＃39;。我在互联网上找到了这样的例子，但结果显示它没有我感兴趣的第二个查询。以下示例。有谁知道我怎么能实现第二种形式的输出 - 完全层次结构跳过较短的输出？

  SQL>
    SQL> column  full_path format a40
    SQL>
    SQL> select  ename
      2  ,       connect_by_root ename            as Designer
      3  ,       sys_connect_by_path(ename,' > ') as full_path
      4  from    emp
      5  start   with job = 'Designer'
      6  connect by prior empno = mgr;

    ENAME    DESIGNER FULL_PATH
    -------- -------- ----------------------------------------
    Jane     Jane      > Jane
    Smart    Jane      > Jane > Smart
    Ana      Jane      > Jane > Smart > Ana
    Fake     Jane      > Jane > Fake
    Tom      Jane      > Jane > Fake > Tom
    Black    Black     > Black
    Jack     Black     > Black > Jack
    Wil      Black     > Black > Wil
    Mary     Black     > Black > Mary
    Take     Black     > Black > Take
    Jane     Black     > Black > Jane
    Chris    Chris     > Chris
    Mike     Chris     > Chris > Mike
    Peter    Peter     > Peter
    Jane     Peter     > Peter > Jane
    Smart    Peter     > Peter > Jane > Smart
    Ana      Peter     > Peter > Jane > Smart > Ana
    Fake     Peter     > Peter > Jane > Fake
    Tom      Peter     > Peter > Jane > Fake > Tom

    ENAME    DESIGNER FULL_PATH
    -------- -------- ----------------------------------------
    Black    Peter     > Peter > Black
    Jack     Peter     > Peter > Black > Jack
    Wil      Peter     > Peter > Black > Wil
    Mary     Peter     > Peter > Black > Mary
    Take     Peter     > Peter > Black > Take
    Jane     Peter     > Peter > Black > Jane
    Chris    Peter     > Peter > Chris
    Mike     Peter     > Peter > Chris > Mike

Answer 1

使用CONNECT_BY_ISLEAF pseudocolumn仅过滤层次结构树叶子上的那些行：

select  ename
,       connect_by_root ename            as Designer
,       sys_connect_by_path(ename,' > ') as full_path
from    emp
WHERE   CONNECT_BY_ISLEAF = 1
start   with job = 'Designer'
connect by prior empno = mgr;