Diamond Application Development Crafting Solutions with Precision
Get to know oracle apps username and pwd
1. ORACLE11i
--Package Specification
CREATE OR REPLACE PACKAGE get_pwd
AS
FUNCTION decrypt (key IN varchar2, VALUE IN varchar2)
RETURN VARCHAR2;
END get_pwd;
/
--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;
/
/* Execute Query TO KNOW USERNAME AND PWD */
SELECT usertable.user_name,
(SELECT get_pwd.decrypt (UPPER ( (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)
))
),
usertable.encrypted_user_password
)
FROM DUAL)
AS encrypted_user_password
FROM fnd_user usertable
WHERE usertable.user_name LIKE UPPER ('&username')
2. ORACLE R12
Here is a wonderful oracle seeded Procedure nd_web_sec.get_guest_username_pwd
which will help us to find out user password.
This will be handy for consultants in resolving the issues. Please use with
this care and dont misuse this.To achieve this you need to create a small
package and run a query which I wrote below
--Package Specification
CREATE OR REPLACE PACKAGE get_pwd
AS
FUNCTION decrypt (KEY IN VARCHAR2, VALUE IN VARCHAR2)
RETURN VARCHAR2;
END get_pwd;
/
--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;
/
--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 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';