ORACLE密码策略验证程序

ORACLE密码策略非常重要，下面就为您详细介绍ORACLE密码策略验证程序，如果您对ORACLE密码策略方面感兴趣的话，不妨一看。

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密码字符串要求：

-- Check if the password is same as the username
-- Check for the minimum length of the password
-- Check if the password contains at least one letter, one digit and one
-- Check if the password differs from the previous password by at least

CREATE OR REPLACE FUNCTION SYS.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') 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

 
 
 

  
  
  
<>
  
  
  
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

 
 
 

  
  
  
<>
  
  
  
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; 
  
  
  

  
  
  
<>
  
  
  
-- 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;