如何生成隨機Base36 ID

Question

有沒有辦法在SQL Server中生成隨機base36 identifiers與定義的字符數？

我已經搜索並找到許多將基數36轉換爲int的例子，反之亦然，但不是用於隨機生成唯一ID。

Answer 1

該解決方案有點冗長，但工作原理很容易適應各種需求。下面是一些示例輸出：

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 

請注意，您需要創建一個視圖來包裝這是不允許一個UDF內使用RAND的。所以這個解決方案需要兩個db對象，一個視圖和一個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 = 'abcdefghijklmnopqrstuvwxyz', 
      @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)