Oracle11g R2创建PASSWORD_VERIFY_FUNCTION对应密码复杂度验证函数步骤
Oracle11g R2创建PASSWORD_VERIFY_FUNCTION对应密码复杂度验证函数步骤
运行测试环境:数据库服务器Oracle Linux 5.8 + Oracle 11g R2数据库
相关工具:PL/SQL软件(连接Linux下的Oracle数据库),SecureCRT软件(远程连接Linux服务器)
详细步骤:
1、连接上Linux数据库服务器,切换到Oracle数据库用户桌面,打开终端,进入到环境变量$ORACLE_HOME目录
Last login: Fri Dec 11 13:26:18 2015 from 192.168.1.100
[root@Linux主机名 ~]# su - oracle
[oracle@Linux主机名 dbhome_1]$ cd $ORACLE_HOME/rdbms/admin
[oracle@Linux主机名 admin]$
2、查看Oracle11g数据库提供的默认密码复杂度函数脚本(Oracle安装目录下的/rdbms/admin/utlpwdmg.sql文件)
[oracle@Linux主机名 admin]$ cat $ORACLE_HOME/rdbms/admin/utlpwdmg.sql
脚本详细内容详见文章末尾
3、登录Oracle数据库并执行Oracle11g数据库提供的默认密码复杂度函数脚本
[oracle@Linux主机名 admin]$ sqlplus /nolog
SQL*Plus: Release 11.2.0.1.0 Production on Fri Dec 11 13:33:58 2015
Copyright (c) 1982, 2009, Oracle. All rights reserved.
SQL> conn /as sysdba
Connected.
SQL> @?/rdbms/admin/utlpwdmg.sql
Function created.
Profile altered.
Function created.
SQL>
4、在PL/SQL中创建用户的资源文件,执行下面语句
CREATE PROFILE 资源文件名 LIMIT
SESSIONS_PER_USER UNLIMITED
CPU_PER_SESSION UNLIMITED
CPU_PER_CALL UNLIMITED
CONNECT_TIME UNLIMITED
IDLE_TIME 600 --10小时连续不活动的话系统自动断开连接
LOGICAL_READS_PER_SESSION UNLIMITED
LOGICAL_READS_PER_CALL UNLIMITED
COMPOSITE_LIMIT UNLIMITED
PRIVATE_SGA UNLIMITED
FAILED_LOGIN_ATTEMPTS 10 --指定锁定用户的登录失败次数为10次,超过10次则系统被自动锁定
PASSWORD_LIFE_TIME 180 --指定用户同一密码锁允许使用的天数为180天
PASSWORD_REUSE_TIME UNLIMITED
PASSWORD_REUSE_MAX UNLIMITED
PASSWORD_LOCK_TIME 1 --指定用户被锁定天数为1天
PASSWORD_GRACE_TIME 10 --数据库发出警告到登录失效前的宽限天数
PASSWORD_VERIFY_FUNCTION verify_function_11G
5、测试更新用户密码
--创建用户并使用自定义的配置文件
create user 用户名 identified by 密码 default tablespace 默认表空间名 temporary tablespace 临时表空间名 profile 资源文件名;
--用户授权
grant connect,resource,exp_full_database,imp_full_database to 用户名;
--更新用户密码为简单的字符串
alter user 用户名 identified by 123456;
--更新用户密码为复杂的字符串
alter user 用户名 identified by Csdn_20151211;
6、结论:发现简单密码无法更新,复杂的密码更新成功。
附:$ORACLE_HOME/rdbms/admin/utlpwdmg.sql脚本源文件内容
Rem
Rem $Header: utlpwdmg.sql 02-aug-2006.08:18:05 asurpur Exp $
Rem
Rem utlpwdmg.sql
Rem
Rem Copyright (c) 2006, Oracle. All rights reserved.
Rem
Rem NAME
Rem utlpwdmg.sql - script for Default Password Resource Limits
Rem
Rem DESCRIPTION
Rem This is a script for enabling the password management features
Rem by setting the default password resource limits.
Rem
Rem NOTES
Rem This file contains a function for minimum checking of password
Rem complexity. This is more of a sample function that the customer
Rem can use to develop the function for actual complexity checks that the
Rem customer wants to make on the new password.
Rem
Rem MODIFIED (MM/DD/YY)
Rem asurpur 05/30/06 - fix - 5246666 beef up password complexity check
Rem nireland 08/31/00 - Improve check for username=password. #1390553
Rem nireland 06/28/00 - Fix null old password test. #1341892
Rem asurpur 04/17/97 - Fix for bug479763
Rem asurpur 12/12/96 - Changing the name of password_verify_function
Rem asurpur 05/30/96 - New script for default password management
Rem asurpur 05/30/96 - Created
Rem
-- This script sets the default password resource parameters
-- This script needs to be run to enable the password features.
-- However the default resource parameters can be changed based
-- on the need.
-- A default password complexity function is also provided.
-- This function makes the minimum complexity checks like
-- the minimum length of the password, password not same as the
-- username, etc. The user may enhance this function according to
-- the need.
-- This function must be created in SYS schema.
-- connect sys/<password> as sysdba before running the script
CREATE OR REPLACE FUNCTION verify_function_11G
(username varchar2,
password varchar2,
old_password varchar2)
RETURN boolean IS
n boolean;
m integer;
differ integer;
isdigit boolean;
ischar boolean;
ispunct boolean;
db_name varchar2(40);
digitarray varchar2(20);
punctarray varchar2(25);
chararray varchar2(52);
i_char varchar2(10);
simple_password varchar2(10);
reverse_user varchar2(32);
BEGIN
digitarray:= '0123456789';
chararray:= 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
-- Check for the minimum length of the password
IF length(password) < 8 THEN
raise_application_error(-20001, 'Password length less than 8');
END IF;
-- Check if the password is same as the username or username(1-100)
IF NLS_LOWER(password) = NLS_LOWER(username) THEN
raise_application_error(-20002, 'Password same as or similar to user');
END IF;
FOR i IN 1..100 LOOP
i_char := to_char(i);
if NLS_LOWER(username)|| i_char = NLS_LOWER(password) THEN
raise_application_error(-20005, 'Password same as or similar to user name ');
END IF;
END LOOP;
-- Check if the password is same as the username reversed
FOR i in REVERSE 1..length(username) LOOP
reverse_user := reverse_user || substr(username, i, 1);
END LOOP;
IF NLS_LOWER(password) = NLS_LOWER(reverse_user) THEN
raise_application_error(-20003, 'Password same as username reversed');
END IF;
-- Check if the password is the same as server name and or servername(1-100)
select name into db_name from sys.v$database;
if NLS_LOWER(db_name) = NLS_LOWER(password) THEN
raise_application_error(-20004, 'Password same as or similar to server name');
END IF;
FOR i IN 1..100 LOOP
i_char := to_char(i);
if NLS_LOWER(db_name)|| i_char = NLS_LOWER(password) THEN
raise_application_error(-20005, 'Password same as or similar to server name ');
END IF;
END LOOP;
-- Check if the password is too simple. A dictionary of words may be
-- maintained and a check may be made so as not to allow the words
-- that are too simple for the password.
IF NLS_LOWER(password) IN ('welcome1', 'database1', 'account1', 'user1234', 'password1', 'oracle123', 'computer1', 'abcdefg1', 'change_on_install') THEN
raise_application_error(-20006, 'Password too simple');
END IF;
-- Check if the password is the same as oracle (1-100)
simple_password := 'oracle';
FOR i IN 1..100 LOOP
i_char := to_char(i);
if simple_password || i_char = NLS_LOWER(password) THEN
raise_application_error(-20007, 'Password too simple ');
END IF;
END LOOP;
-- Check if the password contains at least one letter, one digit
-- 1. Check for the digit
isdigit:=FALSE;
m := length(password);
FOR i IN 1..10 LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(digitarray,i,1) THEN
isdigit:=TRUE;
GOTO findchar;
END IF;
END LOOP;
END LOOP;
IF isdigit = FALSE THEN
raise_application_error(-20008, 'Password must contain at least one digit, one character');
END IF;
-- 2. Check for the character
<<findchar>>
ischar:=FALSE;
FOR i IN 1..length(chararray) LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(chararray,i,1) THEN
ischar:=TRUE;
GOTO endsearch;
END IF;
END LOOP;
END LOOP;
IF ischar = FALSE THEN
raise_application_error(-20009, 'Password must contain at least one \
digit, and one character');
END IF;
<<endsearch>>
-- Check if the password differs from the previous password by at least
-- 3 letters
IF old_password IS NOT NULL THEN
differ := length(old_password) - length(password);
differ := abs(differ);
IF differ < 3 THEN
IF length(password) < length(old_password) THEN
m := length(password);
ELSE
m := length(old_password);
END IF;
FOR i IN 1..m LOOP
IF substr(password,i,1) != substr(old_password,i,1) THEN
differ := differ + 1;
END IF;
END LOOP;
IF differ < 3 THEN
raise_application_error(-20011, 'Password should differ from the \
old password by at least 3 characters');
END IF;
END IF;
END IF;
-- Everything is fine; return TRUE ;
RETURN(TRUE);
END;
/
-- This script alters the default parameters for Password Management
-- This means that all the users on the system have Password Management
-- enabled and set to the following values unless another profile is
-- created with parameter values set to different value or UNLIMITED
-- is created and assigned to the user.
ALTER PROFILE DEFAULT LIMIT
PASSWORD_LIFE_TIME 180
PASSWORD_GRACE_TIME 7
PASSWORD_REUSE_TIME UNLIMITED
PASSWORD_REUSE_MAX UNLIMITED
FAILED_LOGIN_ATTEMPTS 10
PASSWORD_LOCK_TIME 1
PASSWORD_VERIFY_FUNCTION verify_function_11G;
-- Below is the older version of the script
-- This script sets the default password resource parameters
-- This script needs to be run to enable the password features.
-- However the default resource parameters can be changed based
-- on the need.
-- A default password complexity function is also provided.
-- This function makes the minimum complexity checks like
-- the minimum length of the password, password not same as the
-- username, etc. The user may enhance this function according to
-- the need.
-- This function must be created in SYS schema.
-- connect sys/<password> as sysdba before running the script
CREATE OR REPLACE FUNCTION verify_function
(username varchar2,
password varchar2,
old_password varchar2)
RETURN boolean IS
n boolean;
m integer;
differ integer;
isdigit boolean;
ischar boolean;
ispunct boolean;
digitarray varchar2(20);
punctarray varchar2(25);
chararray varchar2(52);
BEGIN
digitarray:= '0123456789';
chararray:= 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
punctarray:='!"#$%&()``*+,-/:;<=>?_';
-- Check if the password is same as the username
IF NLS_LOWER(password) = NLS_LOWER(username) THEN
raise_application_error(-20001, 'Password same as or similar to user');
END IF;
-- Check for the minimum length of the password
IF length(password) < 4 THEN
raise_application_error(-20002, 'Password length less than 4');
END IF;
-- Check if the password is too simple. A dictionary of words may be
-- maintained and a check may be made so as not to allow the words
-- that are too simple for the password.
IF NLS_LOWER(password) IN ('welcome', 'database', 'account', 'user', 'password', 'oracle', 'computer', 'abcd') THEN
raise_application_error(-20002, 'Password too simple');
END IF;
-- Check if the password contains at least one letter, one digit and one
-- punctuation mark.
-- 1. Check for the digit
isdigit:=FALSE;
m := length(password);
FOR i IN 1..10 LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(digitarray,i,1) THEN
isdigit:=TRUE;
GOTO findchar;
END IF;
END LOOP;
END LOOP;
IF isdigit = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one digit, one character and one punctuation');
END IF;
-- 2. Check for the character
<<findchar>>
ischar:=FALSE;
FOR i IN 1..length(chararray) LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(chararray,i,1) THEN
ischar:=TRUE;
GOTO findpunct;
END IF;
END LOOP;
END LOOP;
IF ischar = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one \
digit, one character and one punctuation');
END IF;
-- 3. Check for the punctuation
<<findpunct>>
ispunct:=FALSE;
FOR i IN 1..length(punctarray) LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(punctarray,i,1) THEN
ispunct:=TRUE;
GOTO endsearch;
END IF;
END LOOP;
END LOOP;
IF ispunct = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one \
digit, one character and one punctuation');
END IF;
<<endsearch>>
-- Check if the password differs from the previous password by at least
-- 3 letters
IF old_password IS NOT NULL THEN
differ := length(old_password) - length(password);
IF abs(differ) < 3 THEN
IF length(password) < length(old_password) THEN
m := length(password);
ELSE
m := length(old_password);
END IF;
differ := abs(differ);
FOR i IN 1..m LOOP
IF substr(password,i,1) != substr(old_password,i,1) THEN
differ := differ + 1;
END IF;
END LOOP;
IF differ < 3 THEN
raise_application_error(-20004, 'Password should differ by at \
least 3 characters');
END IF;
END IF;
END IF;
-- Everything is fine; return TRUE ;
RETURN(TRUE);
END;
/
-- This script alters the default parameters for Password Management
-- This means that all the users on the system have Password Management
-- enabled and set to the following values unless another profile is
-- created with parameter values set to different value or UNLIMITED
-- is created and assigned to the user.
-- Enable this if you want older version of the Password Profile parameters
-- ALTER PROFILE DEFAULT LIMIT
-- PASSWORD_LIFE_TIME 60
-- PASSWORD_GRACE_TIME 10
-- PASSWORD_REUSE_TIME 1800
-- PASSWORD_REUSE_MAX UNLIMITED
-- FAILED_LOGIN_ATTEMPTS 3
-- PASSWORD_LOCK_TIME 1/1440
-- PASSWORD_VERIFY_FUNCTION verify_function;
Oracle11g R2创建PASSWORD_VERIFY_FUNCTION对应密码复杂度验证函数步骤的更多相关文章
- Oracle11g R2创建PASSWORD_VERIFY_FUNCTION相应password复杂度验证函数步骤
Oracle11g R2创建PASSWORD_VERIFY_FUNCTION相应密码复杂度验证函数步骤 运行測试环境:数据库服务器Oracle Linux 5.8 + Oracle 11g R2数据库 ...
- oracle 11g/12c 密码复杂度验证设置
############################################################################### ###### 11g ###### ## ...
- Win7 32位安装Oracle11g R2 图解示例
Win7 32位操作系统安装Oracle11g R2 图解示例.废话不说了,直接上图. 1.下载的两个oracle 11gR2压缩包解压到单独的文件夹中. 2.找到解压的database文件夹中的Se ...
- ORACLE11g R2【RAC+ASM→单实例FS】
ORACLE11g R2[RAC+ASM→单实例FS] 11g R2 RAC+ASMà单实例FS的DG,建议禁用OMF. 本演示案例所用环境: primary standby OS Hostnam ...
- ORACLE11g R2【单实例 FS→单实例FS】
ORACLE11g R2[单实例 FS→单实例FS] 本演示案例所用环境: primary standby OS Hostname pry std OS Version RHEL6.5 RHEL6 ...
- ORACLE11g R2【RAC+ASM→RAC+ASM】
ORACLE11g R2[RAC+ASM→RAC+ASM] 本演示案例所用环境:RAC+ASM+OMF primary standby OS Hostname node1,node2 dgnode ...
- centos7远程安装oracle11g R2详细教程-解决一切问题
相关链接与资源: sqldevelper(各种操作系统的oracle客户端) http://www.oracle.com/technetwork/cn/developer-tools/sql-deve ...
- Oracle 11g R2创建数据库之手工建库方式
在之前的博文当中梳理了关于DBCA静默方式创建数据库的过程,本文就手工通过SQL*PLUS客户端采用CREATE DATABASE语句创建数据库.这种建库方式就是完全使用手工SQL语句创建数据库,通常 ...
- Oracle11g手动创建数据库方法
Oracle11g手动创建数据库方法 参考网页http://www.th7.cn/db/Oracle/201311/36926.shtml 安装路径 我的安装路径是:E:\app\admin\prod ...
随机推荐
- SSO-单点统一登录系统的设计与实现
本文主要基于web类应用展开讨论,提供的是一种通用机制和方法,所以不论何种技术栈都可进行相应的具体实现. 实现目标 可以在相同或跨域环境下完成各应用的统一登录/注销 方案原理 本质上是采用了web应用 ...
- [python]使用django快速生成自己的博客小站,含详细部署方法
前言 人生苦短,我用python 这是之前经常听到的一句笑谈.因为新公司很多业务是用的python语言,所以这几天也一直在学习python的一些东西. 作为一个之前一直java后端的开发人员,对比ja ...
- ActiveMQ笔记:一个高稳定,可扩展的的部署方案
本文介绍一个笔者在实际工作中的实施的基于ActiveMQ的一个高稳定,可扩展的异步消息系统. ActiveMQ是一个成熟的基于Java语言的开源消息系统,在实际应用中被大量使用.ActiveMQ在系统 ...
- [HNOI 2018]道路
Description 题库链接 给出一棵含有 \(n\) 个叶子节点的二叉树,对于每个非叶子节点的节点,其与左儿子相连的边为公路,其与右儿子相连的边为铁路.对于每个节点,选择一条与其儿子相连的铁路或 ...
- bzoj1272 Gate Of Babylon
[问题描述] [输入格式] [输出格式] [样例输入] 2 1 10 13 3 [样例输出] 12 [样例说明] [数据范围] 先容斥,考虑枚举哪些条件强制不满足,即直接选出b[i]+1件宝具 假设强 ...
- 组合数问题(zyys版)
[问题描述]定义"组合数"S(n,m)代表将 n 个不同的元素拆分成 m 个非空集合的方案数.举个栗子,将{1,2,3}拆分成 2 个集合有({1},{2,3}),({2},{1, ...
- hihocoder #1159 : 扑克牌
描述 一副不含王的扑克牌由52张牌组成,由红桃.黑桃.梅花.方块4组牌组成,每组13张不同的面值.现在给定52张牌中的若干张,请计算将它们排成一列,相邻的牌面值不同的方案数. 牌的表示方法为XY,其中 ...
- SpringMVC 处理映射
一.Spring MVC控制器名称处理映射 以下示例展示如何利用Spring MVC 框架使用控制器名称处理程序映射. ControllerClassNameHandlerMapping类是基于约定的 ...
- C语言程序设计第一次作业 冯亚杰
(一)实验总结 求圆面积和周长 (1)题目 输入圆的半径,计算圆的周长和面积. (2)流程图 (3)测试数据及运行结果 测试数据1:r=7 运行结果: 测试数据2:r=5 运行结果: (4)实验分析 ...
- Vue生命周期-手动挂载理解
改前端遇到个bug,console能够输出值,但是前端不能显示. 我简直一脸懵逼,vue的问题?网络的问题?浏览器的缓存问题? 公司网络,所以直接排除网络问题. 浏览器缓存,试了下确实一定概率可以显示 ...