查询postgres中表的授予

问题描述 投票:75回答:7

如何在postgres中查询授予对象的所有GRANTS?

例如,我有表“mytable”:

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

我需要能给我的东西:

user1: SELECT, INSERT
user2: UPDATE
sql postgresql grant
7个回答
85
投票

来自psql的\z mytable为您提供了表格中的所有授权,但您必须由个人用户将其拆分。


84
投票

我已经找到了:

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

23
投票

如果你真的想要每个用户一行,你可以按受让人分组(要求PG9 +用于string_agg)

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

这应该输出如下:

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

20
投票

请尝试以下查询。它将为您提供表中所有用户及其权限的列表。

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 a.tablename='your_table_name';

7
投票

此查询将列出所有数据库和模式中的所有表(取消注释WHERE子句中的行以过滤特定数据库,模式或表),并按顺序显示权限,以便于查看如果授予特定权限:

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;

样本输出:

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 |
...

1
投票

加上@ shruti的答案

查询给定用户的架构中所有表的授权

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;

0
投票

这是一个为特定表生成授权查询的脚本。它省略了所有者的特权。

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;
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