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Simple Steps to Find Password for an Application User in Oracle Applications

Anup - Sunday, April 28, 2013

To achieve this you need to create a small package and run a query which I wrote below:

Step 1:  Create Package Specification

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

Ramkumar said...

SQL> /
(fnd_web_sec.get_guest_username_pwd,
*
ERROR at line 4:
ORA-00904: "FND_WEB_SEC"."GET_GUEST_USERNAME_PWD": invalid identifier

Anonymous said...



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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