通过expdp/impdp获取oracle普通用户的密码

很多时候，可能需要获取普通用户的密码。但是在oracle 11g，通过视图也无法获取。
最近在做如下导入导出

突然联想到这点，于是就入了如下测试。实现了获取用户密码(准确的说应该是oracle内部记录形式的密码)的目标，以下是测试过程。

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[oracle@PMOTSTdb0 admin]$ sqlplus / as sysdba

SQL> @utlsampl
Disconnected from Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
With the Partitioning, OLAP, Data Mining and Real Application Testing options
[oracle@PMOTSTdb0 admin]$ sqlplus scott/tiger

SQL*Plus: Release 11.2.0.3.0 Production on Wed Jul 20 14:02:10 2016

Copyright (c) 1982, 2011, Oracle.  All rights reserved.

SQL> alter user scott identified by abc12345;
User altered.

expdp system SCHEMAS=scott directory=DATA_PUMP_DIR DUMPFILE=scott.dmp logfile=scott.log &

impdp system SCHEMAS=scott directory=DATA_PUMP_DIR DUMPFILE=scott.dmp logfile=scott.log sqlfile=scott.sql

cd /home/db/oracle/admin/sitdb1/dpdump

cat scott.sql | grep 'IDENTIFIED BY VALUES'
 CREATE USER "SCOTT" IDENTIFIED BY VALUES 'S:D9D841B328D1386A6CD7BB7E3362F635B0FA80262DC4C4A576B23C882F90;238AE5553644174C'
[oracle@PMOTSTdb0 dpdump]$ 

执行创建同名用户，并赋权：
SQL> grant connect,resource to scott identified by values 'S:D9D841B328D1386A6CD7BB7E3362F635B0FA80262DC4C4A576B23C882F90;238AE5553644174C';

Grant succeeded.

SQL> conn scott/abc12345;
Connected.

执行创建异名用户，并赋权：

SQL> conn / as sysdba
Connected.
SQL> grant connect,resource to scott2 identified by values 'S:D9D841B328D1386A6CD7BB7E3362F635B0FA80262DC4C4A576B23C882F90;238AE5553644174C';

Grant succeeded.

SQL> conn scott2/abc12345;
Connected.
SQL> 

到此，就获取并验证了scott用户的密码： abc12345 对应的values='S:D9D841B328D1386A6CD7BB7E3362F635B0FA80262DC4C4A576B23C882F90;238AE5553644174C'
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