查询 postgres 中表的授权

Question

如何查询 postgres 中授予某个对象的所有 GRANTS？

例如，我有表“mytable”：

GRANT SELECT, INSERT ON mytable TO user1
GRANT UPDATE ON mytable TO user2 

我需要一些可以给我的东西：

user1: SELECT, INSERT
user2: UPDATE

Answer 1

我已经找到了：

SELECT grantee, privilege_type 
FROM information_schema.role_table_grants 
WHERE table_name='mytable'

Answer 2

psql 中的 \z mytable 为您提供表中的所有授权，但您必须将其按单个用户拆分。

Answer 3

下面的查询将为您提供所有用户及其对架构中表的权限的列表。

select a.schemaname, a.tablename, b.usename,
  HAS_TABLE_PRIVILEGE(usename, quote_ident(schemaname) || '.' || quote_ident(tablename), 'select') as has_select,
  HAS_TABLE_PRIVILEGE(usename, quote_ident(schemaname) || '.' || quote_ident(tablename), 'insert') as has_insert,
  HAS_TABLE_PRIVILEGE(usename, quote_ident(schemaname) || '.' || quote_ident(tablename), 'update') as has_update,
  HAS_TABLE_PRIVILEGE(usename, quote_ident(schemaname) || '.' || quote_ident(tablename), 'delete') as has_delete, 
  HAS_TABLE_PRIVILEGE(usename, quote_ident(schemaname) || '.' || quote_ident(tablename), 'references') as has_references 
from pg_tables a, pg_user b 
where a.schemaname = 'your_schema_name' and a.tablename='your_table_name';

有关 has_table_privilages 的更多详细信息，请访问此处 。

Answer 4

如果您确实希望每个用户一行，您可以按受让人分组（string_agg 需要 PG9+）

SELECT grantee, string_agg(privilege_type, ', ') AS privileges
FROM information_schema.role_table_grants 
WHERE table_name='mytable'   
GROUP BY grantee;

这应该输出类似以下内容的内容：

 grantee |   privileges   
---------+----------------
 user1   | INSERT, SELECT
 user2   | UPDATE
(2 rows)

Answer 5

此查询将列出所有数据库和架构中的所有表（取消注释 WHERE 子句中的行以过滤特定数据库、架构或表），其权限如下所示以便很容易地查看是否授予了特定权限：

SELECT grantee
      ,table_catalog
      ,table_schema
      ,table_name
      ,string_agg(privilege_type, ', ' ORDER BY privilege_type) AS privileges
FROM information_schema.role_table_grants 
WHERE grantee != 'postgres' 
--  and table_catalog = 'somedatabase' /* uncomment line to filter database */
--  and table_schema  = 'someschema'   /* uncomment line to filter schema  */
--  and table_name    = 'sometable'    /* uncomment line to filter table  */
GROUP BY 1, 2, 3, 4;

示例输出：

grantee |table_catalog   |table_schema  |table_name     |privileges     |
--------|----------------|--------------|---------------|---------------|
PUBLIC  |adventure_works |pg_catalog    |pg_sequence    |SELECT         |
PUBLIC  |adventure_works |pg_catalog    |pg_sequences   |SELECT         |
PUBLIC  |adventure_works |pg_catalog    |pg_settings    |SELECT, UPDATE |
...

Answer 6

添加@shruti的答案

查询给定用户模式中所有表的授权

select a.tablename, 
       b.usename, 
       HAS_TABLE_PRIVILEGE(usename,tablename, 'select') as select,
       HAS_TABLE_PRIVILEGE(usename,tablename, 'insert') as insert, 
       HAS_TABLE_PRIVILEGE(usename,tablename, 'update') as update, 
       HAS_TABLE_PRIVILEGE(usename,tablename, 'delete') as delete, 
       HAS_TABLE_PRIVILEGE(usename,tablename, 'references') as references 
from pg_tables a, 
     pg_user b 
where schemaname='your_schema_name' 
      and b.usename='your_user_name' 
order by tablename;

Answer 7

这是一个为特定表生成授权查询的脚本。它忽略了所有者的特权。

SELECT 
    format (
      'GRANT %s ON TABLE %I.%I TO %I%s;',
      string_agg(tg.privilege_type, ', '),
      tg.table_schema,
      tg.table_name,
      tg.grantee,
      CASE
        WHEN tg.is_grantable = 'YES' 
        THEN ' WITH GRANT OPTION' 
        ELSE '' 
      END
    )
  FROM information_schema.role_table_grants tg
  JOIN pg_tables t ON t.schemaname = tg.table_schema AND t.tablename = tg.table_name
  WHERE
    tg.table_schema = 'myschema' AND
    tg.table_name='mytable' AND
    t.tableowner <> tg.grantee
  GROUP BY tg.table_schema, tg.table_name, tg.grantee, tg.is_grantable;

如果您想获取视图，请执行以下声明：

 SELECT 
    format (
      'GRANT %s ON TABLE %I.%I TO %I%s;',
      string_agg(tg.privilege_type, ', '),
      tg.table_schema,
      tg.table_name,
      tg.grantee,
      CASE
        WHEN tg.is_grantable = 'YES' 
        THEN ' WITH GRANT OPTION' 
        ELSE '' 
      END
    )
  FROM information_schema.role_table_grants tg
  JOIN pg_views v ON v.schemaname = tg.table_schema AND v.viewname = tg.table_name
  WHERE
    tg.table_schema = 'myschema' AND
    tg.table_name='myview' 
    AND tg.grantee <> 'myschema'
  GROUP BY tg.table_schema, tg.table_name, tg.grantee, tg.is_grantable