0

我已经根据他们文档中的方法为美国邮政服务地址更改服务 Keyline 使用 SQL 实现了“MOD 10”校验位算法，但似乎我得到了错误的数字！我们的输入字符串中只有数字，使计算更容易一些。当我将我的结果与他们的测试应用程序的结果进行比较时，我得到了不同的数字。我不明白这是怎么回事？有人看到我的算法有什么问题吗？这一定是显而易见的……

该方法的文档可以在本文档的第 12-13 页找到： http ://www.usps.com/cpim/ftp/pubs/pub8a.pdf

示例应用程序可在以下位置找到： http ://ribbs.usps.gov/acs/documents/tech_guides/KEYLINE.EXE

请注意：我根据论坛用户的帮助修复了以下代码。这样未来的读者将能够完整地使用代码。

ALTER function [dbo].[udf_create_acs] (@MasterCustomerId varchar(26))
returns varchar(30)
as
begin
    --this implements the "mod 10" check digit calculation
    --for the US Postal Service ACS function, from "Publication 8A"
    --found at "http://www.usps.com/cpim/ftp/pubs/pub8a.pdf"
    declare @result varchar(30)
    declare @current_char int
    declare @char_positions_odd varchar(10)
    declare @char_positions_even varchar(10)
    declare @total_value int
    declare @check_digit varchar(1)

    --These strings represent the pre-calculated values of each character
    --Example: '7' in an odd position in the input becomes 14, which is 1+4=5
    -- so the '7' is in position 5 in the string - zero-indexed
    set @char_positions_odd = '0516273849'
    set @char_positions_even = '0123456789'
    set @total_value = 0
    set @current_char = 1

    --stepping through the string one character at a time
    while (@current_char <= len(@MasterCustomerId)) begin
        --this is the calculation for the character's weighted value
        if (@current_char % 2 = 0) begin
            --it is an even position, so just add the digit's value
            set @total_value = @total_value + convert(int, substring(@MasterCustomerId, @current_char, 1))
        end else begin
            --it is an odd position, so add the pre-calculated value for the digit
            set @total_value = @total_value + (charindex(substring(@MasterCustomerId, @current_char, 1), @char_positions_odd) - 1)
        end

        set @current_char = @current_char + 1
    end

    --find the check digit (character) using the formula in the USPS document
    set @check_digit = convert(varchar,(10 - (@total_value % 10)) % 10)

    set @result = '#' + @MasterCustomerId + '   ' + @check_digit + '#'

    return @result
end
4

3 回答 3

1

当您使用基于集合的语言时，我不确定您为什么要搞乱整个字符串表示。

我可能会像下面那样做。我进行了四次测试，它们都成功了。您也可以轻松扩展它以处理字符，如果您真的想这样做，您甚至可以使表格永久化。

CREATE FUNCTION dbo.Get_Mod10
(
    @original_string    VARCHAR(26)
)
RETURNS VARCHAR(30)
AS
BEGIN
    DECLARE
        @value_mapping TABLE (original_char CHAR(1) NOT NULL, odd_value TINYINT NOT NULL, even_value TINYINT NOT NULL)

    INSERT INTO @value_mapping
    (
        original_char,
        odd_value,
        even_value
    )
    SELECT '0', 0, 0 UNION
    SELECT '1', 2, 1 UNION
    SELECT '2', 4, 2 UNION
    SELECT '3', 6, 3 UNION
    SELECT '4', 8, 4 UNION
    SELECT '5', 1, 5 UNION
    SELECT '6', 3, 6 UNION
    SELECT '7', 5, 7 UNION
    SELECT '8', 7, 8 UNION
    SELECT '9', 9, 9

    DECLARE
        @i              INT,
        @clean_string   VARCHAR(26),
        @len_string     TINYINT,
        @sum            SMALLINT

    SET @clean_string = REPLACE(@original_string, ' ', '')
    SET @len_string = LEN(@clean_string)
    SET @i = 1
    SET @sum = 0

    WHILE (@i <= @len_string)
    BEGIN
        SELECT
            @sum = @sum + CASE WHEN @i % 2 = 0 THEN even_value ELSE odd_value END
        FROM
            @value_mapping
        WHERE
            original_char = SUBSTRING(@clean_string, @i, 1)

        SET @i = @i + 1
    END

    RETURN (10 - (@sum % 10)) % 10
END
GO
于 2009-02-19T17:48:00.300 回答
0

为什么我们有一个额外的模组：

转换（varchar，10％ <<--？

文件说只需要从 10 中减去最后一位。我错过了什么吗？

于 2009-02-19T17:26:02.697 回答
0 
set @check_digit = convert(varchar, (10 - (@total_value % 10)) % 10)
于 2009-02-19T17:33:49.713 回答