需求：统计 t 表中有效的证件号码，但是查了下表数据，很多这种乱七八糟的数据，请问这种怎么过滤掉呢？有效证件号码位数有15位也有18位。
版本：oracle 11.2.0.4
现在这样写的，但还是有些无效号码被统计处理
select zjhm from t
where (length(t.zjhm)=15 or length(t.zjhm)=18)
可以用此函数：
CREATE OR REPLACE FUNCTION F_CHECK_ID_NO (p_idcard IN VARCHAR2)
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 ;
评论
有用 0
用正则：
var reg =/(^[1-9]\d{5}(18|19|([23]\d))\d{2}((0[1-9])|(10|11|12))(([0-2][1-9])|10|20|30|31)\d{3}[0-9Xx]$)|(^[1-9]\d{5}\d{2}((0[1-9])|(10|11|12))(([0-2][1-9])|10|20|30|31)\d{3}$)/;
看18位的身份证正则：
[1-9]\d{5} 前六位地区，非0打头
(18|19|([23]\d))\d{2} 出身年份，覆盖范围为 1800-3999 年
((0[1-9])|(10|11|12)) 月份，01-12月
(([0-2][1-9])|10|20|30|31) 日期，01-31天
\d{3}[0-9Xx]： 顺序码三位 + 一位校验码
15位的身份证：
[1-9]\d{5} 前六位地区，非0打头
\d{2} 出生年份后两位00-99
((0[1-9])|(10|11|12)) 月份，01-12月
(([0-2][1-9])|10|20|30|31) 日期，01-31天
\d{3} 顺序码三位，没有校验码
评论 有用 1帮人帮到底
select 1 from dual where regexp_like('430000190001010101', '(^[1-9]\d{5}(18|19|([23]\d))\d{2}((0[1-9])|(10|11|12))(([0-2][1-9])|10|20|30|31)\d{3}[0-9Xx]$)|(^[1-9]\d{5}\d{2}((0[1-9])|(10|11|12))(([0-2][1-9])|10|20|30|31)\d{3}$)');
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