This a workarround to implement Password Policy in Oracle. Note, i have commented all alter commands, if you have decided to go for it. Please remove comments. Look at this sql function. Create a sql script file. (XXX.SQL) anyname.sql as you like.
--SQL script Starts Here
CREATE OR REPLACE FUNCTION verify_function
(username varchar2,
password varchar2,
old_password varchar2)
RETURN boolean IS
n boolean;
m integer;
differ integer;
isdigit boolean;
ischar boolean;
ispunct boolean;
digitarray varchar2(20);
punctarray varchar2(25);
chararray varchar2(52);
BEGIN
digitarray:= '0123456789';
chararray:= 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
punctarray:='!"#$%&()``*+,-/:;<=>?_';
-- Check if the password is same as the username
IF NLS_LOWER(password) = NLS_LOWER(username) THEN
raise_application_error(-20001, 'Password same as or similar to user');
END IF;
-- Check for the minimum length of the password
IF length(password) < 8 THEN
raise_application_error(-20002, 'Password length less than 8');
END IF;
-- Check if the password is too simple. A dictionary of words may be
-- maintained and a check may be made so as not to allow the words
-- that are too simple for the password.
IF NLS_LOWER(password) IN ('welcome', 'database', 'account', 'user', 'password', 'oracle', 'computer', 'abcd', 'siemens', 'snemeis', '12345678') THEN
raise_application_error(-20002, 'Password too simple');
END IF;
-- Check if the password contains at least one letter, one digit and one
-- punctuation mark.
-- 1. Check for the digit
isdigit:=FALSE;
m := length(password);
FOR i IN 1..10 LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(digitarray,i,1) THEN
isdigit:=TRUE;
GOTO findchar;
END IF;
END LOOP;
END LOOP;
IF isdigit = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one digit, one character and one punctuation');
END IF;
-- 2. Check for the character
<<findchar>>
ischar:=FALSE;
FOR i IN 1..length(chararray) LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(chararray,i,1) THEN
ischar:=TRUE;
GOTO findpunct;
END IF;
END LOOP;
END LOOP;
IF ischar = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one \
digit, one character and one punctuation');
END IF;
-- 3. Check for the punctuation
<<findpunct>>
ispunct:=FALSE;
FOR i IN 1..length(punctarray) LOOP
FOR j IN 1..m LOOP
IF substr(password,j,1) = substr(punctarray,i,1) THEN
ispunct:=TRUE;
GOTO endsearch;
END IF;
END LOOP;
END LOOP;
IF ispunct = FALSE THEN
raise_application_error(-20003, 'Password should contain at least one \
digit, one character and one punctuation');
END IF;
<<endsearch>>
-- Check if the password differs from the previous password by at least
-- 3 letters
IF old_password IS NOT NULL THEN
differ := length(old_password) - length(password);
IF abs(differ) < 3 THEN
IF length(password) < length(old_password) THEN
m := length(password);
ELSE
m := length(old_password);
END IF;
differ := abs(differ);
FOR i IN 1..m LOOP
IF substr(password,i,1) != substr(old_password,i,1) THEN
differ := differ + 1;
END IF;
END LOOP;
IF differ < 3 THEN
raise_application_error(-20004, 'Password should differ by at \
least 3 characters');
END IF;
END IF;
END IF;
-- Everything is fine; return TRUE ;
RETURN(TRUE);
END;
/
-- You can revove comment mark (--) adapt the following if you want to apply this Verification & Password.
--ALTER PROFILE DEFAULT LIMIT
--PASSWORD_LIFE_TIME 90
--PASSWORD_GRACE_TIME 3
--PASSWORD_REUSE_TIME_LIMIT UNLIMITED
--PASSWORD_REUSE_MAX 5
--FAILED_LOGIN_ATTEMPTS 10
--PASSWORD_LOCK_TIME 1/1440 * 30
--PASSWORD_VERIFY_FUNCTION verify_function;
--SQL script ends
Thanks Regards
