postgresql - 查看模式权限

Question

我是否可以运行查询以显示特定架构当前分配的权限？

即。如此分配的权限：

GRANT USAGE ON SCHEMA dbo TO MyUser

我试过了

SELECT *
FROM information_schema.usage_privileges;

但这仅返回内置PUBLIC角色的授权。相反，我想看看哪些用户已被授予各种架构的权限。

注意：我实际上使用的是Amazon Redshift而不是纯粹的PostgreSQL，但如果在Amazon Redshift中无法实现，我会接受纯粹的PostgreSQL答案。 （虽然我怀疑它是）

Answer 1

在console util psql中：

\dn+

会告诉你

     Name      |  Owner   |   Access privileges   |      Description   

Answer 2

权限存储在pg_namespace的nspacl字段中。由于它是一个数组字段，你必须做一些奇特的编码来解析它。此查询将为您提供用于用户和组的授权语句：

select 
'grant ' || substring(
          case when charindex('U',split_part(split_part(array_to_string(nspacl, '|'),pu.usename,2 ) ,'/',1)) > 0 then ',usage ' else '' end 
          ||case when charindex('C',split_part(split_part(array_to_string(nspacl, '|'),pu.usename,2 ) ,'/',1)) > 0 then ',create ' else '' end 
       , 2,10000)
|| ' on schema '||nspname||' to "'||pu.usename||'";' 
from pg_namespace pn,pg_user pu
 where  array_to_string(nspacl,',') like '%'||pu.usename||'%' --and pu.usename='<username>' 
and nspowner > 1 
union
select 
'grant ' || substring(
          case when charindex('U',split_part(split_part(array_to_string(nspacl, '|'),pg.groname,2 ) ,'/',1)) > 0 then ',usage ' else '' end 
          ||case when charindex('C',split_part(split_part(array_to_string(nspacl, '|'),pg.groname,2 ) ,'/',1)) > 0 then ',create ' else '' end 
       , 2,10000)
|| ' on schema '||nspname||' to group "'||pg.groname||'";' 
from pg_namespace pn,pg_group pg
 where array_to_string(nspacl,',') like '%'||pg.groname||'%' --and pg.groname='<username>' 
 and nspowner > 1 

Answer 3

列出所有模式及其当前用户的权益：

WITH "names"("name") AS (
  SELECT n.nspname AS "name"
    FROM pg_catalog.pg_namespace n
      WHERE n.nspname !~ '^pg_'
        AND n.nspname <> 'information_schema'
) SELECT "name",
  pg_catalog.has_schema_privilege(current_user, "name", 'CREATE') AS "create",
  pg_catalog.has_schema_privilege(current_user, "name", 'USAGE') AS "usage"
    FROM "names";

响应将是例如：

  name   | create | usage 
---------+--------+-------
 public  | t      | t
 test    | t      | t
 awesome | f      | f
(3 rows)

在此示例中，当前用户不是awesome架构的所有者。

正如您所猜测的，对特定架构的类似请求：

SELECT
  pg_catalog.has_schema_privilege(
    current_user, 'awesome', 'CREATE') AS "create",
  pg_catalog.has_schema_privilege(
    current_user, 'awesome', 'USAGE') AS "usage";

并回复：

 create | usage 
--------+-------
 f      | f

如您所知，可以将pg_catalog.current_schema()用于当前架构。

所有可能的特权

-- SELECT
-- INSERT
-- UPDATE
-- DELETE
-- TRUNCATE
-- REFERENCES
-- TRIGGER
-- CREATE
-- CONNECT
-- TEMP
-- EXECUTE
-- USAGE

唯一的CREATE和USAGE允许使用架构。

与current_schema()一样，current_user可以替换为特定角色。

带有current列的

BONUS

WITH "names"("name") AS (
  SELECT n.nspname AS "name"
    FROM pg_catalog.pg_namespace n
      WHERE n.nspname !~ '^pg_'
        AND n.nspname <> 'information_schema'
) SELECT "name",
  pg_catalog.has_schema_privilege(current_user, "name", 'CREATE') AS "create",
  pg_catalog.has_schema_privilege(current_user, "name", 'USAGE')  AS "usage",
  "name" = pg_catalog.current_schema() AS "current"
    FROM "names";

--   name   | create | usage | current
-- ---------+--------+-------+---------
--  public  | t      | t     | t
--  test    | t      | t     | f
--  awesome | f      | f     | f
-- (3 rows)

WITH | System Information Functions | GRANT (privileges)

Answer 4

试试这个（适用于PUBLIC角色）：

SELECT nspname,
       coalesce(nullif(role.name,''), 'PUBLIC') AS name,
       substring(
          CASE WHEN position('U' in split_part(split_part((','||array_to_string(nspacl,',')), ','||role.name||'=',2 ) ,'/',1)) > 0 THEN ', USAGE' ELSE '' END 
          || CASE WHEN position('C' in split_part(split_part((','||array_to_string(nspacl,',')), ','||role.name||'=',2 ) ,'/',1)) > 0 THEN ', CREATE' ELSE '' END 
       , 3,10000) AS privileges
FROM pg_namespace pn, (SELECT pg_roles.rolname AS name
   FROM pg_roles UNION ALL SELECT '' AS name) AS role
 WHERE (','||array_to_string(nspacl,',')) LIKE '%,'||role.name||'=%'
 AND nspowner > 1;

Answer 5

适用于AWS Redshift的组合版本（组，用户，PUBLIC）：

    SELECT *
FROM (SELECT CASE
               WHEN charindex ('U',SPLIT_PART(SPLIT_PART(ARRAY_TO_STRING(nspacl,'|'),pu.usename,2),'/',1)) > 0 THEN ' USAGE'
               ELSE ''
             END ||case WHEN charindex('C',SPLIT_PART(SPLIT_PART(ARRAY_TO_STRING(nspacl,'|'),pu.usename,2),'/',1)) > 0 THEN ' CREATE' ELSE '' END AS rights,
             nspname AS schema,
             '' AS role,
             pu.usename AS user
      FROM pg_namespace pn,
           pg_user pu
      WHERE ARRAY_TO_STRING(nspacl,',') LIKE '%' ||pu.usename|| '%'
      --and pu.usename='<username>' 
      AND   nspowner > 1

  UNION

      SELECT CASE
               WHEN charindex ('U',SPLIT_PART(SPLIT_PART(ARRAY_TO_STRING(nspacl,'|'),pg.groname,2),'/',1)) > 0 THEN ' USAGE '
               ELSE ''
             END ||case WHEN charindex('C',SPLIT_PART(SPLIT_PART(ARRAY_TO_STRING(nspacl,'|'),pg.groname,2),'/',1)) > 0 THEN ' CREATE' ELSE '' END as rights,
             nspname AS schema,
             pg.groname AS role,
             '' AS user
      FROM pg_namespace pn,
           pg_group pg
      WHERE ARRAY_TO_STRING(nspacl,',') LIKE '%' ||pg.groname|| '%'
      --and pg.groname='<username>' 
      AND   nspowner > 1

  UNION

      SELECT CASE
               WHEN POSITION('U' IN SPLIT_PART(SPLIT_PART((',' ||array_to_string (nspacl,',')),',' ||roles.name|| '=',2),'/',1)) > 0 THEN ' USAGE'
               ELSE ''
             END 
      || CASE
               WHEN POSITION('C' IN SPLIT_PART(SPLIT_PART((',' ||array_to_string (nspacl,',')),',' ||roles.name|| '=',2),'/',1)) > 0 THEN ' CREATE'
               ELSE ''
             END AS rights,
             nspname AS schema,
             COALESCE(NULLIF(roles.name,''),'PUBLIC') AS role,
             '' AS user
      FROM pg_namespace pn,
           (SELECT pg_group.groname AS name
            FROM pg_group
            UNION ALL
            SELECT '' AS name) AS roles
      WHERE (',' ||array_to_string (nspacl,',')) LIKE '%,' ||roles.name|| '=%'
      AND   nspowner > 1) privs

ORDER BY schema,rights

Answer 6

这是psql内部使用的：）

SELECT n.nspname AS "Name",
  pg_catalog.pg_get_userbyid(n.nspowner) AS "Owner",
  pg_catalog.array_to_string(n.nspacl, E'\n') AS "Access privileges",
  pg_catalog.obj_description(n.oid, 'pg_namespace') AS "Description"
FROM pg_catalog.pg_namespace n
WHERE n.nspname !~ '^pg_' AND n.nspname <> 'information_schema'
ORDER BY 1;

Answer 7

更简洁地说，一个人可以做：

  SELECT 
    n.nspname AS schema_name
   FROM pg_namespace n
  WHERE  has_schema_privilege('my_user',n.nspname, 'CREATE, USAGE');

Answer 8

我知道这篇文章很旧，但是我根据不同的答案又做了一个查询，以使查询简短且易于使用：

select
    nspname as schema_name
    , r.rolname as role_name
    , pg_catalog.has_schema_privilege(r.rolname, nspname, 'CREATE') as create_grant
    , pg_catalog.has_schema_privilege(r.rolname, nspname, 'USAGE') as usage_grant
from pg_namespace pn,pg_catalog.pg_roles r
where array_to_string(nspacl,',') like '%'||r.rolname||'%' 
    and nspowner > 1 

我一直在想，有一天我将进行查询以仅在一个视图中拥有所有权利。 ;）

Answer 9

对于当前问题，可以尝试以下问题：

SELECT r.rolname AS role_name,
       n.nspname AS schema_name,
       p.perm AS privilege
FROM pg_catalog.pg_namespace AS n
    CROSS JOIN pg_catalog.pg_roles AS r
    CROSS JOIN (VALUES ('USAGE'), ('CREATE')) AS p(perm)
WHERE has_schema_privilege(r.oid, n.oid, p.perm)
--      AND n.nspname <> 'information_schema'
--      AND n.nspname !~~ 'pg\_%'
--      AND NOT r.rolsuper

在我遇到过许多对象和用户的数据库中，性能可能会很低。因此，我可以使用aclexplode()这样的默认函数来解决此问题：

SELECT  oid_to_rolname(a.grantee) AS role_name,
        n.nspname AS schema_name,
        a.privilege_type AS privilege_type
FROM pg_catalog.pg_namespace AS n,
        aclexplode(nspacl) a
WHERE n.nspacl IS NOT NULL 
        AND oid_to_rolname(a.grantee) IS NOT NULL 
--      AND n.nspname <> 'information_schema'
--      AND n.nspname !~~ 'pg\_%'

但是，注意，最后一个不包括用户从PUBLIC角色获得的特权。 其中oid_to_rolname()是简单的自定义函数SELECT rolname FROM pg_roles WHERE oid = $1。

而且，像@Jaisus一样，我的任务要求拥有所有用户都具有的所有特权。因此，我对schema，table，views，columns，sequences，functions甚至还有database特权查询default特权。

此外，还有一个有用的扩展名pg_permission，在该扩展名中我可以获取所提供查询的逻辑，并出于我的目的对其进行了升级。

Answer 10

去年没有关于这个问题的更新。然而，据我所知，这个问题还有一个答案。

SELECT grantor, grantee, table_schema, table_name, privilege_type
FROM information_schema.table_privileges
WHERE grantee = 'userName';

这可以详细查看表权限。

我看到这个答案适用于不属于假定用户的数据库。对于部分访问权限，您可以使用它来验证对架构表的访问权限。