Question

我想为SQL服务器表创建一个递增的字母数字序列，它使用数字0-9，小写字母a-z和大写字母A-Z。所以系列会像0,1,2,3 ...... 9，a，b，c，d ...，x，y，z，A，B，C，D ...... X，Y，Z然后超过它将成为11,12,13 ...... 1a，1b，1c ...... 1X，1Y，1Z，然后21,22,23 ...... 2A，2B等等。

目标是增加一个字符串短的字符串ID，基本上使用基本的62字母数字系统递增。

我的第一个倾向是创建一个包含许多列的表，每列都会在系统中存储一个位置。所以像0,0,0,0,0,1这样的列;每次需要一个新值时，它会循环遍历列，根据需要递增，最终达到0,0,0,3，D，z等等。

有没有人对如何做到更聪明或更简洁？同样，目标是创建一个递增序列，即＆＃34; short＆＃34;就代表所需的地点或字符数而言。

Answer 1

感谢@JohnCappelletti和Daniel Ballinger的回答：http://www.fishofprey.com/2011/08/convert-between-base-10-and-base-62-in.html

要记录：

在T-SQL中从Base 10转换为Base 62

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
ALTER FUNCTION [dbo].[Base62Encode](@a_number_to_convert [decimal](36, 0))
RETURNS [char](12) WITH EXECUTE AS CALLER
AS 
BEGIN

DECLARE @v_modulo INTEGER;  
DECLARE @v_temp_int decimal(38) = @a_number_to_convert;  
DECLARE @v_temp_val VARCHAR(256) = '';  
DECLARE @v_temp_char VARCHAR(1);    

--DECLARE @c_base62_digits VARCHAR(62) = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz';
DECLARE @c_base62_digits VARCHAR(62) = '0123456789aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ'; 

 IF ( @a_number_to_convert = 0 )
 BEGIN
     SET @v_temp_val = '0';  
 END

 WHILE ( @v_temp_int <> 0 )
 BEGIN
     SET @v_modulo = @v_temp_int % 62;  
     SET @v_temp_char = substring( @c_base62_digits, @v_modulo + 1, 1 );  
     SET @v_temp_val = @v_temp_char + @v_temp_val;   
     SET @v_temp_int = floor(@v_temp_int / 62);  

 END

 RETURN @v_temp_val;  

END

在T-SQL中从Base 62转换为Base 10

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
ALTER FUNCTION [dbo].[Base62Decode](@a_value_to_convert [char](12))
RETURNS [decimal](36, 0) WITH EXECUTE AS CALLER
AS 
BEGIN

DECLARE @v_iterator int;  
DECLARE @v_length int;  
DECLARE @v_temp_char VARCHAR(1);  
DECLARE @v_temp_int bigint;  
DECLARE @v_return_value decimal(38) = 0;  
DECLARE @v_multiplier decimal(38) = 1;  
DECLARE @v_temp_convert_val VARCHAR(256) = @a_value_to_convert;  

--DECLARE @c_base62_digits VARCHAR(62) = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz';
DECLARE @c_base62_digits VARCHAR(62) = '0123456789aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ'; 


SET @v_length = len( @v_temp_convert_val );  
SET @v_iterator = @v_length; 

WHILE ( @v_iterator > 0 )   
BEGIN
    -- The character being converted
    SET @v_temp_char = substring( @v_temp_convert_val, @v_iterator, 1 );  
    -- The index of the character being converted
    SET @v_temp_int = charindex( @v_temp_char collate  SQL_Latin1_General_CP1_CS_AS, @c_base62_digits collate  SQL_Latin1_General_CP1_CS_AS ) - 1;  

    SET @v_return_value = @v_return_value + ( @v_temp_int * @v_multiplier );  
    SET @v_multiplier = @v_multiplier * 62;  
    SET @v_iterator = @v_iterator - 1;  

END

RETURN @v_return_value; 

END

如何在SQL Server中创建递增基础62字母数字序列

1 个答案: