Oracle 身份证号码校验

来源：互联网发布：数据透视经典显示编辑：程序博客网时间：2024/05/17 05:03

今天来了身份证校验的需求，想在Oracle中校验后将错误数据查询出来。于是我在网上查到了身份证校验的Oracle函数。但是发现一些正确的身份证号也未校验通过！后来经过千辛万苦的努力终于发现了，网上给的函数中判断是否是闰年的 substrb 函数应该从第7为开始截取，而不是第6位！！还有要注意最后一位校验位的大小写转换后再做比较！坑爹啊！坑死我了！

CREATE OR REPLACE FUNCTION f_checkidcard (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 1;
END IF;

IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 2)) + 1900, 400) = 0 OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 2)) + 1900, 100) <> 0 AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 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 0;
ELSE
  RETURN 1;
END IF;

WHEN 18 THEN -- 18位
IF INSTRB (v_areacode, SUBSTRB (p_idcard, 1, 2) || ',') = 0 THEN
  RETURN 1;
END IF;

IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 4)), 400) = 0 OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 4)), 100) <> 0 AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 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 = upper(SUBSTRB (p_idcard, 18, 1))THEN
   RETURN 0;
  ELSE
   RETURN 1;
  END IF;
ELSE
  RETURN 1;
END IF;
ELSE
  RETURN 1; -- 身份证号码位数不对
END CASE;
EXCEPTION
  WHEN OTHERS THEN
   RETURN 1;
END f_checkidcard;