依据PASSWORD_VERIFY_FUNCTION可以设置oracle用户的密码复杂度，比如密码长度>=10，必须包含字母/数字等

首先需要创建一个密码验证的function，然后设置profile的PASSWORD_VERIFY_FUNCTION即可

SQL> select TEXT from dba_source where NAME='VERIFY_JUSTIN_USER';

TEXT

------------------------------------------------------------------------------------------------------------------------------------

FUNCTION verify_JUSTIN_user (  username VARCHAR2,

password VARCHAR2,

old_password varchar2 )

RETURN boolean

IS

passwordMinLength   INTEGER;

passwordLength      INTEGER;

differ              INTEGER;

differMinLength     INTEGER;

isDigit             BOOLEAN;

isChar              BOOLEAN;

isPunct             BOOLEAN;

digitArray          VARCHAR2(20);

punctArray          VARCHAR2(25);

charArray           VARCHAR2(52);

BEGIN

digitArray         := '0123456789';

charArray          := 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';

punctArray         := '!"#$%&()``*+,-/:;<=>?_';

passwordMinLength  := 10;

differMinLength    := 2;--HAD-1

passwordLength     := LENGTH(password);

isDigit            := FALSE;

isChar             := FALSE;

isPunct            := FALSE;

-- +------------------------------------------------+

-- | 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 that password is more than [x] characters |

-- | in length.                                      |

-- +-------------------------------------------------+

IF (LENGTH(password) < passwordMinLength) THEN

raise_application_error( -20002, 'Password must be greater than '

||

passwordMinLength

||

' characters.' );

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(-20003, '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                            |

-- +-----------------------------------------------------+

FOR i IN 1..10 LOOP

FOR j IN 1..passwordLength 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(-20004, 'Password should contain at least '

||

' one digit,'

||

' one character and'

||

' one punctuation');

END IF;

-- +-----------------------------------------------------+

-- | (2.) Check for the character                        |

-- +-----------------------------------------------------+

<>

FOR i IN 1..LENGTH(charArray) LOOP

FOR j IN 1..passwordLength 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(-20004, 'Password should contain at least '

||

' one digit,'

||

' one character and'

||

' one punctuation');

END IF;

-- +-----------------------------------------------------+

-- | (3.) Check for the punctuation                      |

-- +-----------------------------------------------------+

<>

FOR i IN 1..LENGTH(punctArray) LOOP

FOR j IN 1..passwordLength 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(-20004, 'Password should contain at least '

||

' one digit,'

||

' one character and'

||

' one punctuation');

END IF;

<>

-- +-----------------------------------------------------+

-- | Check that the new password is not null.            |

-- +-----------------------------------------------------+

IF old_password = '' THEN

raise_application_error(-20005, 'Old password is null');

END IF;

-- +-----------------------------------------------------+

-- | Check if the password differs from the previous     |

-- | password by at least [x] letters.                   |

-- +-----------------------------------------------------+

differ := ABS(LENGTH(old_password) - LENGTH(password));

IF differ < differMinLength THEN

IF LENGTH(password) < LENGTH(old_password) THEN

passwordLength := LENGTH(password);

ELSE

passwordLength := LENGTH(old_password);

END IF;

FOR i IN 1..passwordLength LOOP

IF SUBSTR(password,i,1) != SUBSTR(old_password,i,1) THEN

differ := differ + 1;

END IF;

END LOOP;

IF differ < differMinLength THEN

raise_application_error(-20006, 'Password should differ by at least '

||

differMinLength

||

' characters.');

END IF;

END IF;

-- +-----------------------------------------------------+

-- | Well, looks like we passed all of the requirements. |

-- | Simple return 'true'.                               |

-- +-----------------------------------------------------+

RETURN(true);

END;

188 rows selected.

SQL> select * from dba_profiles;

PROFILE                        RESOURCE_NAME                    RESOURCE LIMIT

------------------------------ -------------------------------- -------- ----------------------------------------

JUSTIN_PROFILE                 PASSWORD_VERIFY_FUNCTION         PASSWORD VERIFY_JUSTIN_USER

此后创建user可以指定该profile，密码若验证无法通过泽用户创建会失败

SQL> create user sagda identified by"asd245(" profile JUSTIN_PROFILE;

create user sagda identified by"asd245(" profile JUSTIN_PROFILE

*

ERROR at line 1:

ORA-28003: password verification for the specified password failed

ORA-20002: Password must be greater than 10 characters.

SQL> create user sagda identified by"asd245234155" profile JUSTIN_PROFILE;

create user sagda identified by"asd245234155" profile JUSTIN_PROFILE

*

ERROR at line 1:

ORA-28003: password verification for the specified password failed

ORA-20004: Password should contain at least  one digit, one character and one punctuation