Oracle 用户user锁定lock如何知道是什么原因导致的

文章由LinuxBoy分享于2019-03-31 04:03:00热评（483）

Oracle 用户user锁定lock如何知道是什么原因导致的

问题
Oracle数据库用户被锁了，如何查看什么原因导致被锁的呢

结论
1,dba_users是由底层表user$,profile$,profname$相关表构成，当然还有ts$
2,dba_users与账户锁定或过期相关的列有：created,account_status,lock_date,expiry_date,profile
3,具体如何分析到底是dba_profiles中的哪个参数为用户锁定的原因，要结合上述这些列的数据，以及dba_profiles相关记录的含义
进行一一排除

4,查找profile的含义方法为：在oracle sql rererence找create profile即可
或者从administrator guide找resource plan,也可以导航到create profile即可

5,profile说白了，就是控制资源如何使用的，具体细节大家可参考官方手册

测试
--oracle version
SQL> select * from v$version where rownum=1;

BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production

--用户字典表
SQL> desc dba_users;
Name Null? Type
----------------------------------------- -------- ----------------------------
USERNAME NOT NULL VARCHAR2(30)
USER_ID NOT NULL NUMBER
PASSWORD VARCHAR2(30)
ACCOUNT_STATUS NOT NULL VARCHAR2(32)
LOCK_DATE DATE --账户锁定时间
EXPIRY_DATE DATE
DEFAULT_TABLESPACE NOT NULL VARCHAR2(30)
TEMPORARY_TABLESPACE NOT NULL VARCHAR2(30)
CREATED NOT NULL DATE
PROFILE NOT NULL VARCHAR2(30)
INITIAL_RSRC_CONSUMER_GROUP VARCHAR2(30)
EXTERNAL_NAME VARCHAR2(4000)
PASSWORD_VERSIONS VARCHAR2(8)
EDITIONS_ENABLED VARCHAR2(1)
AUTHENTICATION_TYPE VARCHAR2(8)

SQL> create user test_user identified by system account unlock;

User created.

可见默认创建用户使用default profile
SQL> select username,password,account_status,lock_date,profile from dba_users where lower(username)='test_user';

USERNAME PASSWORD ACCOUNT_ST LOCK_DATE PROFILE
-------------------- ------------------------------ ---------- ------------ ------------------------------------------------------------
TEST_USER OPEN DEFAULT

看看default profile
关于profile的含义，自己可以查查官方手册即知
SQL> select profile,resource_name,resource_type,limit from dba_profiles where profile='DEFAULT';

PROFILE RESOURCE_NAME RESOURCE_TYPE LIMIT
------------------------------ -------------------------------------------------- ---------------- --------------------------------------------------------------------------------
DEFAULT COMPOSITE_LIMIT KERNEL UNLIMITED
DEFAULT SESSIONS_PER_USER KERNEL UNLIMITED
DEFAULT CPU_PER_SESSION KERNEL UNLIMITED
DEFAULT CPU_PER_CALL KERNEL UNLIMITED
DEFAULT LOGICAL_READS_PER_SESSION KERNEL UNLIMITED
DEFAULT LOGICAL_READS_PER_CALL KERNEL UNLIMITED
DEFAULT IDLE_TIME KERNEL UNLIMITED
DEFAULT CONNECT_TIME KERNEL UNLIMITED
DEFAULT PRIVATE_SGA KERNEL UNLIMITED
DEFAULT FAILED_LOGIN_ATTEMPTS PASSWORD 10 --注意下与password相关的记录
DEFAULT PASSWORD_LIFE_TIME PASSWORD 180

16 rows selected.

测试下，连续多次用不正确密码登陆，直到10次，账户锁了
[oracle@seconary ~]$ sqlplus test_user/p1

SQL*Plus: Release 11.2.0.1.0 Production on Mon Nov 16 03:54:36 2015

ERROR:
ORA-01017: invalid username/password; logon denied

--中间类似内容略
Enter user-name:
ERROR:
ORA-01017: invalid username/password; logon denied

SP2-0157: unable to CONNECT to ORACLE after 3 attempts, exiting SQL*Plus
[oracle@seconary ~]$ sqlplus test_user/p1

SQL*Plus: Release 11.2.0.1.0 Production on Mon Nov 16 03:54:48 2015

ERROR:
ORA-28000: the account is locked

account_status及lock_date有值了，表明账户涣定了，如果你现在想知道到底为何被锁了，只能查找底层的相关基表了
SQL> select username,password,account_status,lock_date,profile from dba_users where lower(username)='test_user';

USERNAME PASSWORD ACCOUNT_STATUS LOCK_DATE PROFILE
-------------------- -------------------- ------------------------------ ------------------- ------------------------------------------------------------
TEST_USER LOCKED(TIMED) 2015-11-16 03:54:43 DEFAULT

从底层定义的基表看，只有user$及profile$是我想分析所关联的基表
SQL> set long 99999999
SQL> set pagesize 300
SQL> select view_name,text from dba_views where view_name='DBA_USERS';

VIEW_NAME TEXT
------------------------------------------------------------ --------------------------------------------------------------------------------
DBA_USERS select u.name, u.user#,
decode(u.password, 'GLOBAL', u.password,
'EXTERNAL', u.password,
NULL),
m.status,
decode(u.astatus, 4, u.ltime,
5, u.ltime,
6, u.ltime,
8, u.ltime,
9, u.ltime,
10, u.ltime, to_date(NULL)),
decode(u.astatus,
1, u.exptime,
2, u.exptime,
5, u.exptime,
6, u.exptime,
9, u.exptime,
10, u.exptime,
decode(u.ptime, '', to_date(NULL),
decode(pr.limit#, 2147483647, to_date(NULL),
decode(pr.limit#, 0,
decode(dp.limit#, 2147483647, to_date(NULL), u.ptime +
dp.limit#/86400),
u.ptime + pr.limit#/86400)))),
dts.name, tts.name, u.ctime, p.name,
nvl(cgm.consumer_group, 'DEFAULT_CONSUMER_GROUP'),
u.ext_username,
decode(length(u.password),16,'10G ',NULL)||NVL2(u.spare4, '11G ' ,NULL),
decode(bitand(u.spare1, 16),
16, 'Y',
'N'),
decode(u.password, 'GLOBAL', 'GLOBAL',
'EXTERNAL', 'EXTERNAL',
'PASSWORD')
from sys.user$ u left outer join sys.resource_group_mapping$ cgm
on (cgm.attribute = 'ORACLE_USER' and cgm.status = 'ACTIVE' and
cgm.value = u.name),
sys.ts$ dts, sys.ts$ tts, sys.profname$ p,
sys.user_astatus_map m, sys.profile$ pr, sys.profile$ dp
where u.datats# = dts.ts#
and u.resource$ = p.profile#
and u.tempts# = tts.ts#
and u.astatus = m.status#
and u.type# = 1
and u.resource$ = pr.profile#
and dp.profile# = 0
and dp.type#=1
and dp.resource#=1
and pr.type# = 1
and pr.resource# = 1
先看下user$,只有列resource$是我们关注的列
create table user$ /* user table */
( user# number not null, /* user identifier number */
name varchar2("M_IDEN") not null, /* name of user */
/* 0 = role, 1 = user, 2 = adjunct schema, 3 = schema synonym */
type# number not null,
password varchar2("M_IDEN"), /* encrypted password */
datats# number not null, /* default tablespace for permanent objects */
tempts# number not null, /* default tablespace for temporary tables */
ctime date not null, /* user account creation time */
ptime date, /* password change time */
exptime date, /* actual password expiration time */
ltime date, /* time when account is locked */
resource$ number not null, /* resource profile# */
audit$ varchar2("S_OPFL"), /* user audit options */
defrole number not null, /* default role indicator: */
/* 0 = no roles, 1 = all roles granted, 2 = roles in defrole$ */
defgrp# number, /* default undo group */
defgrp_seq# number, /* global sequence number for the grp *
spare varchar2("M_IDEN"), /* reserved for future */
astatus number default 0 not null, /* status of the account */
/* 0x00 = 0 = Open */
/* 0x01 = 1 = Locked */
/* 0x02 = 2 = Expired */
/* 0x03 = 3 = Locked and Expired */
/* 0x10 = 16 = Password matches a default value */
lcount number default 0 not null, /* count of failed login attempts */
defschclass varchar2("M_IDEN"), /* initial consumer group */
ext_username varchar2("M_VCSZ"), /* external username */
/* also as base schema name for adjunct schemas */
spare1 number, /* used for schema level supp. logging: see ktscts.h */
spare2 number, /* used to store edition id for adjunct schemas */
spare3 number,
spare4 varchar2(1000),
spare5 varchar2(1000),
spare6 date
)
cluster c_user#(user#)

而据上可知，user$的resource$是与表profile$的profile#关联，所以还是要分析profile$表

VIEW_NAME TEXT
------------------------------------------------------------ --------------------------------------------------------------------------------
DBA_PROFILES select
n.name, m.name,--对应dba_profiles的resource_name
decode(u.type#, 0, 'KERNEL', 1, 'PASSWORD', 'INVALID'),
decode(u.limit#,
0, 'DEFAULT',
2147483647, decode(u.resource#,
4, decode(u.type#,
1, 'NULL', 'UNLIMITED'),
'UNLIMITED'),
decode(u.resource#,
4, decode(u.type#, 1, o.name, u.limit#),--对应dba_profiles的resource_type
decode(u.type#,
0, u.limit#,
decode(u.resource#,
1, trunc(u.limit#/86400, 4),
2, trunc(u.limit#/86400, 4),
5, trunc(u.limit#/86400, 4),
6, trunc(u.limit#/86400, 4),
u.limit#)))) --对应dba_profiles的limit
from sys.profile$ u, sys.profname$ n, sys.resource_map m, sys.obj$ o
where u.resource# = m.resource#
and u.type#=m.type#
and o.obj# (+) = u.limit#
and n.profile# = u.profile#

SQL> desc dba_profiles;
Name Null? Type
----------------------------------------- -------- ----------------------------
PROFILE NOT NULL VARCHAR2(30)
RESOURCE_NAME NOT NULL VARCHAR2(32)
RESOURCE_TYPE VARCHAR2(8)
LIMIT VARCHAR2(40)

综上分析下，你只要关注resource_type='password',并且limit为unlimited及null不用你管了（原因不用我了），所以只有我标注几条记录
SQL> select profile,resource_name,resource_type,limit from dba_profiles where profile='DEFAULT';

PROFILE RESOURCE_NAME RESOURCE_TYPE LIMIT
------------------------------ -------------------------------------------------- ---------------- --------------------------------------------------------------------------------
DEFAULT COMPOSITE_LIMIT KERNEL UNLIMITED
DEFAULT SESSIONS_PER_USER KERNEL UNLIMITED
DEFAULT CPU_PER_SESSION KERNEL UNLIMITED
DEFAULT CPU_PER_CALL KERNEL UNLIMITED
DEFAULT LOGICAL_READS_PER_SESSION KERNEL UNLIMITED
DEFAULT LOGICAL_READS_PER_CALL KERNEL UNLIMITED
DEFAULT IDLE_TIME KERNEL UNLIMITED
DEFAULT CONNECT_TIME KERNEL UNLIMITED
DEFAULT PRIVATE_SGA KERNEL UNLIMITED
DEFAULT FAILED_LOGIN_ATTEMPTS PASSWORD 10 ---关注，经查，这是登陆的次数
DEFAULT PASSWORD_LIFE_TIME PASSWORD 180 --关注 --这是账户有效期的天数，即用同一个账户密码登陆

PROFILE RESOURCE_NAME RESOURCE_TYPE LIMIT
------------------------------ -------------------------------------------------- ---------------- --------------------------------------------------------------------------------
DEFAULT PASSWORD_REUSE_TIME PASSWORD UNLIMITED
DEFAULT PASSWORD_REUSE_MAX PASSWORD UNLIMITED
DEFAULT PASSWORD_VERIFY_FUNCTION PASSWORD NULL
DEFAULT PASSWORD_LOCK_TIME PASSWORD 1 ---关注，表明连续多次失败登陆后，账户锁定的天数，默认为1天
DEFAULT PASSWORD_GRACE_TIME PASSWORD 7 --关注，指定多少天，即如果你在指定天数之内不改密码就会提示密码失败无法登陆

关注上述参数的含义，请见：

Oracle? Database SQL Language Reference
11g Release 2 (11.2)
Part Number E26088-03

之
create profile

可见expiry_date即账期何时会过期，此列一直有数据，只要用户创建
SQL> select username,password,account_status,expiry_date,lock_date,profile from dba_users;

USERNAME PASSWORD ACCOUNT_STATUS EXPIRY_DATE LOCK_DATE PROFILE
-------------------- -------------------- ------------------------------ ------------------- ------------------- ------------------------------
SCOTT OPEN 2016-03-06 06:58:08 DEFAULT
TEST1 OPEN 2016-05-03 11:59:30 DEFAULT
TEST_USER LOCKED(TIMED) 2016-05-14 03:45:06 2015-11-16 03:54:43 DEFAULT

但是与password相关在dba_profiles有几条记录，我们知道是哪条记录是导致用户锁定的原因呢

所以结合测试用户创建的时间dba_users.created就可以排除password_life_time,password_lock_time也可以排除，因为它是账户被锁定的表数，也就是说已发生在账户锁定之后的动作了
这样就只有failed_login_attempts和password_grace_time
我们经查官方手册发现,password_grace_time一般和password_life_time搭配使用，即后者要小于前者，即如果在后者指定的天数不修改密码，用户就会过期不能登陆了，所以
只能是参数failed_login_attempts
SQL> select username,password,account_status,expiry_date,lock_date,profile from dba_users where lower(username)='test_user';

USERNAME PASSWORD ACCOUNT_STATUS EXPIRY_DATE LOCK_DATE PROFILE
-------------------- -------------------- ------------------------------ ------------------- ------------------- ------------------------------
TEST_USER LOCKED(TIMED) 2016-05-14 03:45:06 2015-11-16 03:54:43 DEFAULT

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