oracle自定义函数验证身份证是否合法

--身份证号校验位算法：
1：把前17位号码从第高位到低位与下列17个数字分别相乘求和（N）：
『2,4,8,5,10,9,7,3,6,1,2,4,8,5,10,9,7』
比如身份证号码为：C1C2C3……C16C17
则N＝C17×2＋C16×4＋……＋C1×7；
2：将N除以11取余数R，根据余数计算校验位T：
1）如果R＝0，则T＝1；如果R＝1，则T＝0；如果R＝2，则T＝X；
2）如果R＝3，则T＝9；如果R＝4，则T＝8；依此类推……；如果R＝10，则T＝2；
-- 0代表身份证不合法，1代表身份证合法

create or replace function CHECK_IDCARD(p_idcard varchar2) return number 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                                                            -- 验证15位身份证
IF INSTRB (v_areacode, SUBSTR (p_idcard, 1, 2) || ',') = 0
THEN
RETURN 0;
END IF;

IF    MOD (TO_NUMBER (SUBSTR (p_idcard, 7, 2)) + 1900, 400) = 0
OR (    MOD (TO_NUMBER (SUBSTR (p_idcard, 7, 2)) + 1900, 100) <>
0
AND MOD (TO_NUMBER (SUBSTR (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 1;
ELSE
RETURN 0;
END IF;
WHEN 18
THEN                                                             --验证 18位身份证
IF INSTRB (v_areacode, SUBSTR (p_idcard, 1, 2) || ',') = 0
THEN
RETURN 0;
END IF;

IF    MOD (TO_NUMBER (SUBSTR (p_idcard, 7, 4)), 400) = 0
OR (    MOD (TO_NUMBER (SUBSTR (p_idcard, 7, 4)), 100) <> 0
AND MOD (TO_NUMBER (SUBSTR (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 = 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 CHECK_IDCARD;