Question

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

在雪花中无需使用它们即可实现CONNECT_BY_ISCYCLE和CONNECT_BY_ISLEAF功能的替代方法是什么？

Answer 1

是否想看看以下雪花功能？

https://docs.snowflake.net/manuals/user-guide/queries-hierarchical.html#using-connect-by-or-recursive-ctes-to-query-hierarchical-data

Answer 2

是的，我去了那里。我还查看了https://docs.snowflake.net/manuals/sql-reference/constructs/connect-by.html，其中清楚地表明Snowflake不支持这些功能。我正在尝试在代码块下面寻找替代方案，但遇到雪花错误。

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

在此方面提供的任何帮助都非常感谢。

谢谢。

enter code here

如何在不使用雪花的情况下获取CONNECT_BY_ISCYCLE和CONNECT_BY_ISLEAF的结果？

2 个答案: