大虾们推荐个性能高一点的方法吧,有200多W条记录需要过滤,谢谢啦。
数据库是oracle 9的。
数据库是oracle 9的。
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小弟又是菜鸟,改不来这个。
18位的号码substr(xx,7,4)必须>1900 <2008
RETURN INT
IS
v_regstr VARCHAR2 (2000);
v_sum NUMBER;
v_mod NUMBER;
v_checkcode CHAR (11) := '10X98765432';
v_checkbit CHAR (1);
v_areacode VARCHAR2 (2000) :='11,12,13,14,15,21,22,23,31,32,33,34,35,36,37,41,42,43,44,45,46,50,51,52,53,54,61,62,63,64,65,71,81,82,91,';
BEGIN
CASE LENGTHB (p_idcard)
WHEN 15
THEN -- 15位
IF INSTRB (v_areacode, SUBSTR (p_idcard, 1, 2) || ',') = 0
THEN
RETURN 0;
END IF; IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 2)) + 1900, 400) = 0
OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 2)) + 1900, 100) <>
0
AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 2)) + 1900, 4) = 0
)
THEN -- 闰年
v_regstr :=
'^[1-9][0-9]{5}[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|[1-2][0-9]))[0-9]{3}$';
ELSE
v_regstr :=
'^[1-9][0-9]{5}[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}$';
END IF; IF REGEXP_LIKE (p_idcard, v_regstr)
THEN
RETURN 1;
ELSE
RETURN 0;
END IF;
WHEN 18
THEN -- 18位
IF INSTRB (v_areacode, SUBSTRB (p_idcard, 1, 2) || ',') = 0
THEN
RETURN 0;
END IF; IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 4)), 400) = 0
OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 4)), 100) <> 0
AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 6, 4)), 4) = 0
)
THEN -- 闰年
v_regstr :=
'^[1-9][0-9]{5}19[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|[1-2][0-9]))[0-9]{3}[0-9Xx]$';
ELSE
v_regstr :=
'^[1-9][0-9]{5}19[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}[0-9Xx]$';
END IF; IF REGEXP_LIKE (p_idcard, v_regstr)
THEN
v_sum :=
( TO_NUMBER (SUBSTRB (p_idcard, 1, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 11, 1))
)
* 7
+ ( TO_NUMBER (SUBSTRB (p_idcard, 2, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 12, 1))
)
* 9
+ ( TO_NUMBER (SUBSTRB (p_idcard, 3, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 13, 1))
)
* 10
+ ( TO_NUMBER (SUBSTRB (p_idcard, 4, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 14, 1))
)
* 5
+ ( TO_NUMBER (SUBSTRB (p_idcard, 5, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 15, 1))
)
* 8
+ ( TO_NUMBER (SUBSTRB (p_idcard, 6, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 16, 1))
)
* 4
+ ( TO_NUMBER (SUBSTRB (p_idcard, 7, 1))
+ TO_NUMBER (SUBSTRB (p_idcard, 17, 1))
)
* 2
+ TO_NUMBER (SUBSTRB (p_idcard, 8, 1)) * 1
+ TO_NUMBER (SUBSTRB (p_idcard, 9, 1)) * 6
+ TO_NUMBER (SUBSTRB (p_idcard, 10, 1)) * 3;
v_mod := MOD (v_sum, 11);
v_checkbit := SUBSTRB (v_checkcode, v_mod + 1, 1); IF v_checkbit = SUBSTRB (p_idcard, 18, 1)
THEN
RETURN 1;
ELSE
RETURN 0;
END IF;
ELSE
RETURN 0;
END IF;
ELSE
RETURN 0; -- 身份证号码位数不对
END CASE;
EXCEPTION
WHEN OTHERS
THEN
RETURN 0;
END fn_checkidcard;
/
也可以看看这个 很傻的...
http://topic.csdn.net/u/20090903/16/73e786aa-4907-4ca7-aac8-63414faeb42b.html?83960