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标题: Oracle如何找出一个表的间接授权信息? [打印本页]

作者: mahan 时间: 2025-7-5 12:53
标题: Oracle如何找出一个表的间接授权信息?
在Oracle数据库中, 如果需要找出一张表授权给了哪一个用户,这个是比较简单的,如果有一些视图引用了这张表,然后这张视图授权给了其它用户的话, 那么这也属于这张表的授权信息,如果也要找出这类授权信息,那么如何找出来这些信息呢?

下面简单看一个例子, 在数据库中存在三个用户T1, T2, T3, 假设T1用户将表T1.TEST的查询权限授予了用户T2.

create user t1 identified by t123456;
create user t2 identified by t234561;
create user t3 identified by t345612;

alter user t1 quota unlimited on users;
alter user t2 quota unlimited on users;
alter user t3 quota unlimited on users;

grant connect, resource to t1;
grant connect, resource to t2;
grant connect, resource to t3;

grant create view to t2;
grant create view to t3;

具体授权操作如下所示:

SQL> show user;
USER is "T1"
SQL> create table test(id number(10), name varchar2(30));

Table created.
SQL> insert into test
2 select 1, 'k1' from dual union all
3 select 2, 'k2' from dual;

2 rows created.

SQL> commit;

Commit complete.

SQL>
SQL> grant select on test to t2;

Grant succeeded

那么此时查看关于表TEST的授权信息如下所示:

SET LINESIZE 820;
COL GRANTEE FOR A12
COL OWNER FOR A12
COL TABLE_NAME FOR A12
COL GRANTOR FOR A12
COL PRIVILEGE FOR A12
SELECT OWNER, TABLE_NAME, GRANTOR , GRANTEE, PRIVILEGE, GRANTABLE, TYPE
FROM DBA_TAB_PRIVS WHERE TABLE_NAME='TEST';

SQL> show user;
USER is "SYS"
SQL> SET LINESIZE 820;
SQL> COL GRANTEE FOR A12
SQL> COL OWNER FOR A12
SQL> COL TABLE_NAME FOR A12
SQL> COL GRANTOR FOR A12
SQL> COL PRIVILEGE FOR A12
SQL> SELECT OWNER, TABLE_NAME, GRANTOR , GRANTEE, PRIVILEGE, GRANTABLE, TYPE
2FROM DBA_TAB_PRIVS WHERE TABLE_NAME='TEST';

OWNER       TABLE_NAME GRANTOR    GRANTEE    PRIVILEGE GRA TYPE
------------ ------------ ------------ ------------ ------------ --- ------------------------
T1          TEST       T1          T2          SELECT    NOTABLE

SQL>

如果用户T1将表TEST的查询权限授予了用户T2,并且使用了选项GRANT OPTION的话

SQL> show user;
USER is "T1"
SQL> grant select on test to t2 with grant option;

Grant succeeded.

SQL>

那么此时,如果在T2用户下面创建一个视图,引用表TEST, 然后将视图T2.V_TEST的查询权限授权给了用户T3.

SQL> show user;
USER is "T2"
SQL> create or replace view v_test
2 as
3 select name from t1.test;

View created.
SQL> grant select on t2.v_test to t3;

Grant succeeded.

SQL>

此时用户T3就相当间接拥有了表TEST的查询权限. 如下所示:

SQL> show user;
USER is "T3"
SQL> select * from t2.v_test;

NAME
------------------------------
k1
k2

SQL>

但是,我们用上面的SQL来查询一下表TEST授予了哪些用户.如下所示, 这个查询结果不能体现表TEST间接授权给了用户T3

SQL> show user;
USER is "SYS"
SQL> SET LINESIZE 820;
SQL> COL GRANTEE FOR A12
SQL> COL OWNER FOR A12
SQL> COL TABLE_NAME FOR A12
SQL> COL GRANTOR FOR A12
SQL> COL PRIVILEGE FOR A12
SQL> SELECT OWNER, TABLE_NAME, GRANTOR , GRANTEE, PRIVILEGE, GRANTABLE, TYPE
2FROM DBA_TAB_PRIVS WHERE TABLE_NAME='TEST';

OWNER       TABLE_NAME GRANTOR    GRANTEE    PRIVILEGE GRA TYPE
------------ ------------ ------------ ------------ ------------ --- ------------------------
T1          TEST       T1          T2          SELECT    YES TABLE

SQL>

那么问题来了,如何查询这种情况下的授权呢? 其实我们可以用下面SQL实现这个需求.如下所示:

SET LINESIZE 820
COL OWNER FOR A10
COL TABLE_NAME FOR A16;
COL GRANTOR FOR A16
COL GRANTEE FOR A16
COL PRIVILEGE FOR A8;
SELECT OWNER, TABLE_NAME, GRANTOR , GRANTEE, PRIVILEGE, GRANTABLE, TYPE
FROM DBA_TAB_PRIVS
WHERE TABLE_NAME=UPPER(TRIM('&tb_name'))
UNION ALL
SELECT  OWNER, TABLE_NAME, GRANTOR , GRANTEE, PRIVILEGE, GRANTABLE, TYPE
FROM DBA_TAB_PRIVS
WHERE TABLE_NAME IN(
SELECT NAME FROM dba_dependencies WHERE
REFERENCED_NAME=UPPER(TRIM('&tb_name')) ANDTYPE='VIEW'
);

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