問題描述

有沒有辦法生成隨機base36 標識符SQL Server 中定義的字符數?

Is there a way of generating random base36 identifiers with a defined character count in SQL Server?

我搜索并發現了許多將 base 36 轉換為 int 以及將其轉換為 int 的示例，但不是用于隨機生成唯一 ID 的示例.

I have searched and found many examples of converting base 36 to int and vice versa, but not for randomly generating unique IDs.

推薦答案

這個解決方案有點冗長但有效并且可以輕松適應各種需求.以下是一些示例輸出:

This solution a bit verbose but works and can easily be adapted to a variety of requirements. Here's some sample output:

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 UDF 中不允許使用的 RAND.所以這個解決方案需要兩個 db 對象，一個視圖和一個 udf.

Note that you need to create a view to wrap the use of RAND which is not allowed inside a UDF. So this solution needs two db objects, a view and a 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)

這篇關于如何生成隨機 Base36 ID的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！