从4个方面实战oracle的密码操作(三)

2014-11-24 12:33:01 · 作者: · 浏览: 1
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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;
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-- 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; www.2cto.com
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;
/