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2023-12-29
今天就跟大家聊聊有关怎么进行Oracle Data Redaction数据加密,可能很多人都不太了解,为了让大家更加了解,小编给大家总结了以下内容,希望大家根据这篇文章可以有所收获。
Oracle Data Redaction是Oracle安全加密类的高级功能,可用于对于敏感数据的加密处理,加密配置处理均在Oracle层面实现。这是一项和安全相关的技术类别,对于指定的用户可以限制某些表的某些列显示被加密改过的值。对于Redaction之前,可能需要自定义加密函数、创建特定的视图,或者在存储到数据库的时候就用加密算法进行加密。而Redaction可以直接对数据进行加密,不会影响到数据真实的存储,对应用透明,不需要改动。
对于权限,Redaction不能对sys和system用户进行数据的加密。因为他们都有EXP_FULL_DATABASE这个角色, 而这个角色又包含了EXEMPT REDACTION POLICY系统权限。同时,也不能直接赋予用户dba权限,dba自动包含EXP_FULL_DATABASE角色。测试过程中发现,对于拥有dba权限的用户来说,表的数据可以加密操作,但没有实际加密效果。
对于常用的加密类型说明如下:
1.Full redaction:对某字段数据全部加密,number类型的列将全部返回为0,character类型的列将全部返回为空格,日期类型返回为yyyy-mm-dd;
2.Partial redaction:对列中的一部分数据进行redact,比如,可以对身份证号或手机号的中间几位设置返回为*,剩下的几位保持不变,这种场景适用于固定长度;
3.Regular expressions:对于非固定长度的character类型数据进行部分加密;
4.Random redaction:随机加密,每次展现的加密结果是不一定一样;
本次加密测试环境介绍信息如下:
OS版本
Oracle版本
是否RAC
RHEL6.5
11.2.0.4.170418
是
本次只模拟几种常用加密场景,创建用户、表,并赋予相应权限;若无特殊说明时,调用DBMS_REDACT所使用的均为zhangxg用户。
SQL> create user zhangxg identified by zhangxg;
User created.
SQL> grant connect,resource to zhangxg;
Grant succeeded.
SQL> grant select on sys.redaction_policies TO zhangxg;
Grant succeeded.
SQL> grant select on sys.redaction_columns TO zhangxg;
Grant succeeded.
SQL> grant execute on dbms_redact TO zhangxg;
Grant succeeded.
SQL> create user nosee identified by zhangxg;
User created.
SQL> grant connect,resource to nosee;
Grant succeeded.
SQL> CREATE TABLE ZHANGXG.TAB1 (
2 "EMPLOYEE_ID" NUMBER(6,0),
3 "FIRST_NAME" VARCHAR2(20),
4 "LAST_NAME" VARCHAR2(25),
5 "SOCIAL_SECURITY" VARCHAR2(11),
6 "SALARY" NUMBER(4,0)
7 );
Table created.
SQL> insert into tab1 values (100,steven,king,247-85-9056,7000);
1 row created.
SQL> insert into tab1 values (101,neena,kochhar,334-08-6578,5000);
1 row created.
SQL> commit;
commit complete.
SQL> grant select on zhangxg.tab1 to nosee;
grant succeeded.
SQL> select * from tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-6578 5000
调用DBMS_REDACT包创建policy策略
SQL> BEGIN
2 DBMS_REDACT.ADD_POLICY (
3 object_schema => ZHANGXG,
4 object_name => TAB1,
5 policy_name => REDACT_1,
6 column_name => SOCIAL_SECURITY,
7 function_type => DBMS_REDACT.PARTIAL,
8 EXPRESSION =>1=1,
9 function_parameters => VVVFVVFVVVV,VVV-VV-VVVV,*,1,5
10 );
11 END;
12 /
PL/SQL procedure successfully completed.
SQL>
SQL> select * from tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- -------------------- -------------------- ------------------ ----------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
SQL> SQL> conn nosee/zhangxg
Connected.
SQL>
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
基于某列加密的基础上,再增加一个乱码显示的列,即多列加密显示
SQL> BEGIN
2 DBMS_REDACT.ALTER_POLICY(
3 object_schema => ZHANGXG,
4 object_name => TAB1,
5 policy_name => REDACT_1,
6 action => DBMS_REDACT.ADD_COLUMN,
7 column_name => LAST_NAME,
8 function_type => DBMS_REDACT.RANDOM);
9 END;
10 /
PL/SQL procedure successfully completed.
SQL> select * from tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven K5r. ***-**-9056 7000
101 Neena ymP@Ea ***-**-6578 5000
SQL> conn nosee/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven YZ$z ***-**-9056 7000
101 Neena HSTk}5l ***-**-6578 5000
对于权限的控制,我们可以直接选择用户去过滤,但用户过多时可以使用role来进行权限的控制。
nosee用户看到的结果也是加密的
SQL> show user
USER is "ZHANGXG"
SQL>
SQL> select * from tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven NPnj ***-**-9056 7000
101 Neena ~<`utz ***-**-6578 5000
SQL> conn nosee/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven @a8H ***-**-9056 7000
101 Neena `&&3P-- ***-**-6578 5000
更改策略,除zhangxg本身用户外,其他用户访问均为加密
SQL> BEGIN
2 DBMS_REDACT.ALTER_POLICY (
3 object_schema =>ZHANGXG,
4 object_name =>TAB1,
5 policy_name =>REDACT_1,
6 column_name =>SOCIAL_SECURITY,
7 action => DBMS_REDACT.MODIFY_EXPRESSION,
8 expression =>SYS_CONTEXT ( USERENV,SESSION_USER ) !=ZHANGXG
9 );
10 END;
11 /
PL/SQL procedure successfully completed.
SQL> select * from tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-6578 5000
SQL> conn nosee/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven 5}2g ***-**-9056 7000
101 Neena $b=Z%,W ***-**-6578 5000
创建角色和用户用于模拟
SQL> create user cansee identified by zhangxg;
User created.
SQL> grant connect,resource to cansee;
Grant succeeded.
SQL> grant select on zhangxg.tab1 to cansee;
Grant succeeded.
SQL> create role redac_role;
Role created.
SQL> grant redac_role to cansee;
Grant succeeded.
下面,我们尝试去掉一个列的加密效果,即去掉LAST_NAME字段的全加密策略
SQL> BEGIN
2 DBMS_REDACT.ALTER_POLICY(
3 object_schema => ZHANGXG,
4 object_name => TAB1,
5 policy_name => REDACT_1,
6 action => DBMS_REDACT.DROP_COLUMN,
7 column_name => LAST_NAME,
8 expression =>1=1);
9 END;
10 /
PL/SQL procedure successfully completed.
SQL>
SQL> conn nosee/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- -------------- ----------- ----------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
再次更改策略,只让zhangxg和cansee用户查看,其他用户为加密结果;这里我们新建了REDAC_ROLE角色去控制,拥有该角色的用户可正常查看,没有该角色的用户查看结果为加密;
SQL> BEGIN
2 DBMS_REDACT.ALTER_POLICY (
3 object_schema =>ZHANGXG,
4 object_name =>TAB1,
5 policy_name =>REDACT_1,
6 column_name =>SOCIAL_SECURITY,
7 action => DBMS_REDACT.MODIFY_EXPRESSION,
8 expression => SYS_CONTEXT(SYS_SESSION_ROLES,REDAC_ROLE) = FALSE
9 );
10 END;
11 /
PL/SQL procedure successfully completed.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
SQL>
SQL> conn / as sysdba
Connected.
SQL>
SQL> grant redac_role to zhangxg;
Grant succeeded.
SQL> conn zhangxg/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-6578 5000
SQL> conn cansee/zhangxg
Connected.
SQL>
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-6578 5000
SQL> conn nosee/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECU SALARY
----------- -------------------- ------------------------- ----------- ----------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
通过测试可以得出以下结论:
1.对加密的列不能同时使用distinct和order by,如果使用,必须加一层select,无论对于使用的用户是否是加密可见;
2.加密的结果优先级高于函数,比如distinct;
3.对于group by不影响结果的准确性,与未加密结果一致;
4.加密后的表无法进行CTAS(create table as select)操作;
对于加密后剩余字符不一样时,distinct结果是不影响的
SQL> show user
USER is "NOSEE"
SQL> col SOCIAL_SECURITY for a25
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ----------- ------------------ -------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-6578 5000
SQL> select distinct SOCIAL_SECURITY from zhangxg.tab1;
SOCIAL_SECURITY
-------------------------
***-**-9056
***-**-6578
SQL> conn zhangxg/zhangxg
Connected.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ------------ ---------------- ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-6578 5000
手动update,将SOCIAL_SECURITY后4位改成一致
SQL> update tab1 set SOCIAL_SECURITY=334-08-9056 where EMPLOYEE_ID=101;
1 row updated.
SQL> commit;
Commit complete.
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ----------- ------------------ ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-9056 5000
此时,对于加密不限制用户来说,distinct结果是2行
SQL> select distinct SOCIAL_SECURITY from zhangxg.tab1;
SOCIAL_SECURITY
-------------------------
334-08-9056
247-85-9056
对于加密用户来说,distinct结果是1行,说明是先进行的加密,后进行distinct
SQL> conn nosee/zhangxg
Connected.
SQL> select distinct SOCIAL_SECURITY from zhangxg.tab1;
SOCIAL_SECURITY
-------------------------
***-**-9056
当对加密列同时使用distinct和order by,报语法错误,该问题已确认为bug,而且没有补丁,Bug 19558306 ;
SQL> select distinct SOCIAL_SECURITY from zhangxg.tab1 order by SOCIAL_SECURITY;
select distinct SOCIAL_SECURITY from zhangxg.tab1 order by SOCIAL_SECURITY
*
ERROR at line 1:
ORA-01791: not a SELECTed expression
做为workaround,可以将SQL改写为子查询
SQL> select * from (select distinct SOCIAL_SECURITY from zhangxg.tab1 order by SOCIAL_SECURITY);
SOCIAL_SECURITY
-------------------------
***-**-9056
***-**-9056
SQL> conn zhangxg/zhangxg
Connected.
SQL> select distinct SOCIAL_SECURITY from zhangxg.tab1 order by SOCIAL_SECURITY;
select distinct SOCIAL_SECURITY from zhangxg.tab1 order by SOCIAL_SECURITY
*
ERROR at line 1:
ORA-01791: not a SELECTed expression
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- -------------- ----------- ------------------ ----------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-9056 5000
对于group by,加密后的效果是不影响分组判断
SQL> conn zhangxg/zhangxg
Connected.
SQL> select SOCIAL_SECURITY,count(*) from zhangxg.tab1 group by SOCIAL_SECURITY;
SOCIAL_SECURITY COUNT(*)
------------------------- ----------
334-08-9056 1
247-85-9056 1
SQL> conn nosee/zhangxg
Connected.
SQL>
SQL> select SOCIAL_SECURITY,count(*) from zhangxg.tab1 group by SOCIAL_SECURITY;
SOCIAL_SECURITY COUNT(*)
------------------------- ----------
***-**-9056 1
***-**-9056 1
对于加密后的表无法进行CTAS操作,其实逻辑上是对的,既然加密了,如果能ctas成功那么加密就是去了意义
SQL> show user
USER is "NOSEE"
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ---------- ----------------- ------
100 Steven King ***-**-9056 7000
101 Neena Kochhar ***-**-9056 5000
SQL> create table ctastab as select * from zhangxg.tab1;
create table ctastab as select * from zhangxg.tab1
*
ERROR at line 1:
ORA-28081: Insufficient privileges - the command references a redacted object.
SQL> conn / as sysdba
Connected.
SQL> grant exempt redaction policy to NOSEE;
Grant succeeded.
SQL> conn nosee/zhangxg
Connected.
SQL> create table ctastab as select * from zhangxg.tab1;
Table created.
SQL> select * from ctastab;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ----------- ----------------- -------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-9056 5000
同时,有了该权限后加密也是去了意义
SQL> select * from zhangxg.tab1;
EMPLOYEE_ID FIRST_NAME LAST_NAME SOCIAL_SECURITY SALARY
----------- ------------ ----------- ----------------- -------
100 Steven King 247-85-9056 7000
101 Neena Kochhar 334-08-9056 5000
1.对于已添加的策略,可通过下面视图进行查询
select * from redaction_policies;
selectobject_owner,object_name,column_name,function_type,function_parameters from redaction_columns;
2.删除加密策略
BEGIN
DBMS_REDACT.DROP_POLICY(
object_schema => ZHANGXG,
object_name => TAB1,
policy_name => REDACT_1);
END;
看完上述内容,你们对怎么进行Oracle Data Redaction数据加密有进一步的了解吗?如果还想了解更多知识或者相关内容,请关注行业资讯频道,感谢大家的支持。
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