用户权限查询上的redshift listagg

Question

我正在使用 listagg 对具有相同权限的用户进行分组，根据来自以下堆栈问题的查询，根据我的需要对其进行了一些调整。 How do I view grants on Redshift

这失败了，说 listagg 是计算节点函数，应该在用户创建的表上使用。有什么方法可以在目录表上使用 listagg 和 has_*_privilege 函数，这两个函数都在领导节点上运行？

with cte1 as (
SELECT *
FROM 
    (    SELECT 
        fullobj
        ,usename
        ,case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'select') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'S' else '' end ||  
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'insert') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'I' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'update') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'U' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'delete') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'D' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'references') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'R' else '' end AS perm
    FROM
        (
        SELECT schemaname, 't' AS obj_type, tablename AS objectname, schemaname + '.' + tablename AS fullobj FROM pg_tables
        WHERE schemaname not in ('pg_internal')
        UNION
        SELECT schemaname, 'v' AS obj_type, viewname AS objectname, schemaname + '.' + viewname AS fullobj FROM pg_views
        WHERE schemaname not in ('pg_internal')
        ) AS objs
        ,(SELECT usename::text FROM pg_user) AS usrs
    ORDER BY fullobj
    ) WHERE perm != '')
select fullobj, perm, listagg(usename,',') from cte1 group by fullobj, perm;
SQL Error [500310] [XX000]: [Amazon](500310) Invalid operation: One or more of the used functions must be applied on at least one user created tables. Examples of user table only functions are LISTAGG, MEDIAN, PERCENTILE_CONT, etc;

我试图将查询输出存储在一个表中，但它不允许存储该表，因为 has_table_privilege 函数是leader-only函数

create  table user_perms as
SELECT *
FROM 
    (    SELECT 
        fullobj
        ,usename
        ,case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'select') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'S' else '' end ||  
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'insert') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'I' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'update') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'U' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'delete') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'D' else '' end ||
        case when HAS_TABLE_PRIVILEGE(usrs.usename, fullobj, 'references') AND has_schema_privilege(usrs.usename, schemaname, 'usage') then 'R' else '' end AS perm
    FROM
        (
        SELECT schemaname, 't' AS obj_type, tablename AS objectname, schemaname + '.' + tablename AS fullobj FROM pg_tables
        WHERE schemaname not in ('pg_internal')
        UNION
        SELECT schemaname, 'v' AS obj_type, viewname AS objectname, schemaname + '.' + viewname AS fullobj FROM pg_views
        WHERE schemaname not in ('pg_internal')
        ) AS objs
        ,(SELECT usename::text FROM pg_user) AS usrs
    ORDER BY fullobj
    ) WHERE perm != '';
SQL Error [500310] [0A000]: [Amazon](500310) Invalid operation: Specified types or functions (one per INFO message) not supported on Redshift tables.;

对于能够在领导节点函数/表上运行 listagg 的问题是否有任何解决方案？

Answer 1

没有。

如您正确理解，listagg 是 Redshift 实现的函数，而不是继承自 Postgres/Paraccel，并且仅在工作节点上实现。

has 函数来自 Postgres，仅在领导节点上实现。

查询计划器不允许使用仅领导节点功能的查询来招募工作节点，因此您不能调用listagg。

（顺便说一句，如果我没记错的话，reltype 的 'v' 也会获取物化视图。）

顺便说一句，您实际上可以直接从系统表中获取您要查找的信息，但这是一项漫长而复杂的工作。我是 Redshift 专家，第一个版本花了我两个月的时间，虽然我当时正在工作。