USPS ACS Keyline Check Digit

Question

我已经使用SQL实现了“MOD 10”校验位算法，根据其文档中的方法为美国邮政服务地址更改服务Keyline，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。当我将结果与测试应用程序的结果进行比较时，我得到的数字不同。我不明白发生了什么事？有没有人看到我的算法有什么问题？它一定是明显的东西...

该方法的文档，可以在这个文件的12-13页上找到： http://ribbs.usps.gov/acs/documents/tech_guides/KEYLINE.EXE

： http://www.usps.com/cpim/ftp/pubs/pub8a.pdf

示例应用程序可以在这里找到请注意：我根据论坛用户的帮助修复了下面的代码。这是为了让未来的读者能够完整地使用这些代码。

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 = '' 
    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 

Answer 1

set @check_digit = convert(varchar, (10 - (@total_value % 10)) % 10) 

Answer 2

为什么我们有一个额外的国防部：

转换（VARCHAR，10％ < < -

文件称，只有最后一个数字需要从扣除？ 10.我错过了什么吗？

Answer 3

我不确定你为什么搞乱当你使用基于集合的语言工作时，整个字符串表示。

我可能会像下面这样做。我跑了四次测试，他们都成功了。如果你真的想这么做，你可以很容易地扩展这个角色来处理角色，甚至可以让表格永久化。

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