如何查询在postgres中授予对象的所有格兰仕?
例如,我有表“ mytable”:
GRANT SELECT, INSERT ON mytable TO user1
GRANT UPDATE ON mytable TO user2
我需要带给我的东西:
user1: SELECT, INSERT
user2: UPDATE
Answers:
\z mytable
来自psql的命令为您提供了表中的所有授予,但随后您必须按单个用户将其拆分。
\z
,都用于psql。psql是PostgreSQL的命令行界面。
如果您确实希望每位用户一行,则可以按受赠者分组(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:
grantee | privileges
---------+----------------
user1 | INSERT, SELECT
user2 | UPDATE
(2 rows)
GRANT
pg_dump之类的确切输出吗?
下面的查询将为您提供所有用户及其对表中表的权限的列表。
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';
有关更多详细信息,请has_table_privilages
参见此处。
has_table_privilege(usename, contact(schemaname, '.', tablename), ...)
避免歧义。
该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的数据库,模式或表),并按顺序显示特权,以便轻松查看是否是否授予特定特权:
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 |
...
添加到@shruti的答案
查询架构中给定用户的所有表的授权
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;
FROM pg_tables AS a CROSS JOIN pg_user AS b
而不是用逗号进行SQL 92 编写的方式from pg_tables a, pg_user b
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;