For Apps DBA sometimes it is necessary to decrypt the EBS Application user’s password. Below package can be used to decrypt the FND_USER password. You will need password of “apps” database userid to be able to decrypt the application user’s password.
Step 1: Create Package Specfication
SQL> create or replace PACKAGE test_package12 AS FUNCTION decrypt (KEY IN VARCHAR2, VALUE IN VARCHAR2) RETURN VARCHAR2; END test_package12; /
Step 2: Create Package Body
SQL> create or replace PACKAGE BODY test_package12 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 test_package12; /
Step 3: Decrypt the Password
SQL> SELECT USERTABLE.USER_NAME ,(SELECT test_package12.decrypt (UPPER ('&apps_password') ,usertable.encrypted_user_password) FROM DUAL) AS encrypted_user_password FROM FND_USER USERTABLE WHERE usertable.user_name LIKE UPPER ('&USER_NAME_TO_DECRYPT');
Step 4: Drop Package
SQL> drop package test_package12;
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This doesn’t work in R12.2.. does it?
Works with R12.2 as we have tried in one of our R12.2 instance.
Enter value for apps_password: myappspassword
old 2: ,(SELECT test_package12.decrypt (UPPER (‘&apps_password’)
new 2: ,(SELECT test_package12.decrypt (UPPER (‘myappspassword’)
Enter value for user_name: SYSADMIN
old 6: WHERE usertable.user_name LIKE UPPER (‘&USER_NAME’)
new 6: WHERE usertable.user_name LIKE UPPER (‘SYSADMIN’)
USER_NAME
——————————————————————————–
ENCRYPTED_USER_PASSWORD
——————————————————————————–
SYSADMIN
mysysadminpassword
SQL> select release_name from apps.fnd_product_groups;
RELEASE_NAME
————————————————–
12.2.4
Thanks Brijesh Sir…..:-)
Oracle Apps, all users lock including EAMADMIN, only SYADMIN is able to access Apps. Please suggest any solution?
not working in 11.5