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  • 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;
    /

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  • 原文地址:https://www.cnblogs.com/weiyi1314/p/10209501.html
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