我要评论在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 2 from dba_tab_privs where table_name='test'; owner table_name grantor grantee privilege gra type ------------ ------------ ------------ ------------ ------------ --- ------------------------ t1 test t1 t2 select no table 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 2 from 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')) and type='view'
);
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