Question

有没有办法在SQL Server中生成带有已定义字符数的随机 base36 identifiers？

我搜索过并发现了很多将base 36转换为int的例子，反之亦然，但不是随机生成唯一ID。

Answer 1

这个解决方案有点冗长但有效，可以很容易地适应各种要求。这是一些示例输出：

aapx0k  k4fdbb  vzbl5x
8vr1bs  gbix1q  g5kctv
he6e50  m9j0m0  2vz53l
yw72hs  hgbo5h  3oen9v
6t4q75  337670  5sf3h4
yqr35s  xoh4hh  tc0wtf
w7trkj  lnnpdk  zk2ln1
1gt7qr  l6m72n  ja5kvm
kg6f9y  6t3b7a  ujfr0i
2jatgo  0yv8rv  wvbjfa

请注意，您需要创建一个视图以包含在UDF中不允许使用的RAND。所以这个解决方案需要两个db对象，一个视图和一个udf。

CREATE VIEW ViewRandInt AS (SELECT RAND() * 36 as RandInt)
GO

CREATE FUNCTION GetRandomBase36Id
(
    @charCount AS INT
) 
RETURNS VARCHAR(50) AS BEGIN

    DECLARE @characters CHAR(36),
            @result VARCHAR(MAX),
            @counter INT,
            @randNum INT

    SELECT @characters = '0123456789abcdefghijklmnopqrstuvwxyz',
           @result = '',
           @counter = 0;


    WHILE @counter < @charCount
    BEGIN
        SELECT @randNum = RandInt FROM ViewRandInt
        SET @result = @result + SUBSTRING(@characters, @randNum+1, 1)
        SET @counter = @counter + 1
    END

    RETURN @result;

END

-- Test:
SELECT dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)
, dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6), dbo.GetRandomBase36Id(6)

Answer 2

MySQL 版本的 Paul Sasik 的回答：

注意视图中的重要变化和：FLOOR(RandInt * 36) INTO v_randNum

CREATE VIEW `ViewRandInt` AS (select rand() AS `RandInt`);

DROP function IF EXISTS `GetRandomBase36Id`;
DELIMITER $$
CREATE FUNCTION `GetRandomBase36Id`(
    p_charCount INT
) RETURNS varchar(50) CHARSET latin1
BEGIN

    DECLARE v_characters CHAR(36);
    DECLARE v_result LONGTEXT;
    DECLARE v_counter INT;
    DECLARE v_randNum INT;

    SET v_characters = '0123456789abcdefghijklmnopqrstuvwxyz',
           v_result = '',
           v_counter = 0;
           
    WHILE v_counter < p_charCount
    DO
        SELECT FLOOR(RandInt * 36) INTO v_randNum FROM ViewRandInt;
        SET v_result = Concat(v_result , SUBSTRING(v_characters, v_randNum +1, 1));
        SET v_counter = v_counter + 1;
    END WHILE;

    RETURN v_result;

END$$
DELIMITER ;

-- Test
SELECT GetRandomBase36Id(6)
, GetRandomBase36Id(6)
, GetRandomBase36Id(6)
, GetRandomBase36Id(6)
, GetRandomBase36Id(6)
, GetRandomBase36Id(6);

如何生成随机Base36 ID

2 个答案: