The Impossible Password and default accounts. Oracle #LockDown #QuickTip #Infosec

Let’s revisit a customer who was about to go through an IG Audit. There is one finding that always seems to come up. Default accounts with default passwords. I don’t care if the accounts are expired and locked, if the user still has the default password set, then it’s a finding. Now you can go through all the default accounts and change the password if you want. I’m personally fond of having a script that does the work for me.

This is the script I use to go through and change all the default passwords. Now, once I’m done with this, I go back on set any necessary default accounts such as DBSNMP to a known password. Why, I find it easier to lock all the doors then methodically move through and unlock only the doors that are necessary.

First we are going to need a function that returns an impossible password. Well, in all honesty, it’s not impossible to crack, (that is because given enough horsepower any password can be eventually cracked.) This function will do one thing, return a 30 character randomly generated password of any printable character. There are a couple characters you can’t put in a password, so when we encounter those, we are going to replace them with an integer between 0 – 9.

Next we are going to go through all the default accounts and change the password.

<CODE>

DECLARE
    -- get the list of users with default passwords.
    CURSOR users_with_defpwd_cur IS
        SELECT username
        FROM sys.dba_users_with_defpwd;
    stmt     VARCHAR2(2000);    -- the base sql statement
    passwd   VARCHAR2(32);      -- the impossible_password.

    FUNCTION impossible_password RETURN VARCHAR2 AS
    -- will create a 30 character password wrapped in double quotes.
    passwd           VARCHAR2(32);        -- this is the password we are returning.
                                          -- we need 32 characters because we are
                                          -- wrapping the password in double quotes.
    p_invalid_char_3 VARCHAR2(1) := '"';  -- invalid password character 3 is '"'
    p_invalid_char_4 VARCHAR2(1) := ';';  -- invalid password character 4 is ';'
    BEGIN 
        passwd := SYS.dbms_random.STRING('p',30); -- get 30 printable characters. 
        -- find all the invalid characters and replace them with a random integer
        -- between 0 and 9.
        passwd := REPLACE(passwd, p_invalid_char_3, ceil(SYS.dbms_random.VALUE(-1,9)));
        passwd := REPLACE(passwd, p_invalid_char_4, ceil(SYS.dbms_random.VALUE(-1,9)));
        -- before we pass back the password, we need to put a double quote 
        -- on either side of it. This is because sometime we are going to 
        -- get a strange character that will cause oracle to cough up a hairball.
        passwd := '"' || passwd || '"';
        RETURN passwd;
    END;
-- main procedure.
BEGIN
    FOR users_with_defpwd_rec IN users_with_defpwd_cur LOOP
        passwd := impossible_password;
        stmt := 'alter user ' || users_with_defpwd_rec.username || ' identified by ' || passwd;
        EXECUTE IMMEDIATE stmt;
    END LOOP;
EXCEPTION WHEN OTHERS THEN
    sys.dbms_output.put_line(sqlerrm);
    sys.dbms_output.put_line(stmt);
END;
/
</CODE>