如何在oracle中显示用户的所有特权? [英] How to show all privileges from a user in oracle?

问题描述

有人可以告诉我如何在sql-console中显示特定用户的所有特权/规则吗?

解决方案

您可以尝试以下视图.

SELECT * FROM USER_SYS_PRIVS; 
SELECT * FROM USER_TAB_PRIVS;
SELECT * FROM USER_ROLE_PRIVS;

DBA和其他超级用户可以使用这些视图的DBA_版本找到授予其他用户的特权.在文档中进行了介绍. >

这些视图仅显示直接 授予用户的特权.查找 all 特权，包括通过角色间接授予的特权，需要更复杂的递归SQL语句:

select * from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER' order by 1,2,3;
select * from dba_sys_privs  where grantee = '&USER' or grantee in (select granted_role from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER') order by 1,2,3;
select * from dba_tab_privs  where grantee = '&USER' or grantee in (select granted_role from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER') order by 1,2,3,4;

Can someone please tell me how to show all privileges/rules from a specific user in the sql-console?

解决方案

You can try these below views.

SELECT * FROM USER_SYS_PRIVS; 
SELECT * FROM USER_TAB_PRIVS;
SELECT * FROM USER_ROLE_PRIVS;

DBAs and other power users can find the privileges granted to other users with the DBA_ versions of these same views. They are covered in the documentation .

Those views only show the privileges granted directly to the user. Finding all the privileges, including those granted indirectly through roles, requires more complicated recursive SQL statements:

select * from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER' order by 1,2,3;
select * from dba_sys_privs  where grantee = '&USER' or grantee in (select granted_role from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER') order by 1,2,3;
select * from dba_tab_privs  where grantee = '&USER' or grantee in (select granted_role from dba_role_privs connect by prior granted_role = grantee start with grantee = '&USER') order by 1,2,3,4;

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