使用SQL生成唯一的随机数

DECLARE @Random1 INT,@Random2 INT,@Random3 INT,@Random4 INT,@Random5 INT,@Random6 INT,@Upper INT,@Lower INT
---- This will create a random number between 1 and 49
SET @Lower = 1 ---- The lowest random number
SET @Upper = 49; ---- The highest random number


with nums as (
    select @lower as n
    union all
    select nums.n+1
    from nums
    where nums.n < @Upper
   ),randnums as 
   (select nums.n,ROW_NUMBER() over (order by newid()) as seqnum
    from nums
   )
select @Random1 = MAX(case when rn.seqnum = 1 then rn.n end),@Random2 = MAX(case when rn.seqnum = 2 then rn.n end),@Random3 = MAX(case when rn.seqnum = 3 then rn.n end),@Random4 = MAX(case when rn.seqnum = 4 then rn.n end),@Random5 = MAX(case when rn.seqnum = 5 then rn.n end),@Random6 = MAX(case when rn.seqnum = 6 then rn.n end)
from randnums rn;

select @Random1,@Random2,@Random3,@Random4,@Random5,@Random6

我的问题是这个数字生成是多么随机？还有另一种方法可以做到这一点,更“随机”.

我在用：

Microsoft SQL Server 2008 (SP3) - 10.0.5512.0 (X64)   Aug 22 2012 19:25:47   Copyright (c) 1988-2008 Microsoft Corporation  Developer Edition (64-bit) on Windows NT 6.1 <X64> (Build 7601: Service Pack 1)

大多数解决方案的问题是你最终会得到这样的值：14,29,8,14,27,27我不能有重复的数字！

解决方法

DECLARE @Upper INT;
DECLARE @Lower INT;
SET @Lower = 1     -- The lowest random number
SET @Upper = 49    -- The highest random number


SELECT @Lower + CONVERT(INT,(@Upper-@Lower+1)*RAND())

为了获得随机数而不重复,我认为这将完成这项工作

;with CTE 
as
 (
    SELECT  randomNumber,COUNT(1) countOfRandomNumber
    FROM (
    SELECT ABS(CAST(NEWID() AS binary(6)) %49) + 1 randomNumber
    FROM sysobjects) sample
    GROUP BY randomNumber
 )
SELECT TOP 5 randomNumber
FROM CTE
ORDER BY newid()

设置最高限制,您可以用最高限制数替换49

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!