本篇文章主要介绍了"Oracle11g R2创建PASSWORD_VERIFY_FUNCTION对应密码复杂度验证函数步骤"，主要涉及到PL/SQL,表空间方面的内容，对于其他数据库感兴趣的同学可以参考一下：
Oracle11g R2创建PASSWORD_VERIFY_FUNCTION对应密码复杂度验证函数步骤 运行测试环境：数据库服务器Oracle Linux 5....
-- 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;
