oracle中验证身份证是否合法的函数脚本

--创建函数 入参是身份证   返回1 合法 0不合法

CREATE OR REPLACE FUNCTION fn_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 0;
END IF;
IF p_idcard ='111111111111111' THEN
RETURN 0;
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 :=
'^[1-8][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}$';
ELSE
v_regstr :=
'^[1-8][0-9]{5}[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}$';
END IF;

IF REGEXP_LIKE (p_idcard, v_regstr) THEN
RETURN 1;
ELSE
RETURN 0;
END IF;

WHEN 18 THEN -- 18位
IF not REGEXP_LIKE (p_idcard, '^1[12345]|^2[123]|^3[1234567]|^4[123456]|^5[01234]|^6[12345]|^71|^8[12]') THEN --增加行政区划的条件
RETURN 0;
END IF;

IF INSTRB (v_areacode, SUBSTRB (p_idcard, 1, 2) || ',') = 0 THEN
RETURN 0;
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 :=
'^[1-8][0-9]{5}(19|20)[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]$';
ELSE
v_regstr :=
'^[1-8][0-9]{5}(19|20)[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}[0-9Xx]$';
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 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;
/

 

posted @ 2019-01-02 16:56  唯一520  阅读(1711)  评论(0编辑  收藏  举报