Snowflake cloud data platform 如何在不使用它们的情况下，在snowflake中获得按循环连接和按岛连接的结果？

我需要进行分层查询，我需要获得CONNECT_BY_ISCYCLE和CONNECT_BY_ISLEAF的结果，但Oracle不支持Snowflake中的这些功能

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在snowflake中实现CONNECT_BY_ISCYCLE和CONNECT_BY_ISLEAF功能的替代方法有哪些，而不使用它们，因为这些关键字在snowflake中不受支持？

不知道您是否查看了以下snowflake功能

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是的，我看了一下。我还看了一下，它清楚地表明，这些功能在Snowflake中不受支持。 我试图在代码块下面找到一个替代方案，但在《雪花》中遇到了各种各样的错误

person_vertex as (
    select 
        emp_number, 
        user_id 
    from person
),

person_edges as ( 
    select 
        supervisor_emp_number, 
        emp_number 
    from person 
    where supervisor_emp_number is not null
),

select
    pv.emp_number emp_id_pk,
    level,
    CONNECT_BY_ROOT pv.emp_number AS root,
    concat(SYS_CONNECT_BY_PATH(pv.emp_number,':'),':') as path,
    -- CONNECT_BY_ISCYCLE AS iscyclic, ------------------- no idea how to implement this
    -- CONNECT_BY_ISLEAF as isleaf ------------------- i tried below block, but it is not working
    case
        when (pe.supervisor_emp_number in (select emp_number from pv)) then 0
        else 1
    end AS isleaf
from person_vertex pv
    left join person_edges pe on pv.emp_number = pe.emp_number
connect by prior A.emp_number = A.supervisor_emp_number
start with A.supervisor_emp_number is null

非常感谢您对这个街区的任何帮助

谢谢

enter code here

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我想一个代码块/代码片段可以做这件事，但我不能构建这个代码块。