如何在SQLServer2008中使用函数生成字母数字随机数

我需要生成6个字符长度的字母数字随机数，它应该包含数字、字母（小写和大写）检查下面的查询

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我需要在函数中实现。（在函数中是否可以使用NEWID（），RAND（）

输出：

我需要输出为：

试试这个：

 select cast((Abs(Checksum(NewId()))%10) as varchar(1)) + 
       char(ascii('a')+(Abs(Checksum(NewId()))%25)) +
       char(ascii('A')+(Abs(Checksum(NewId()))%25)) +
       left(newid(),5) Random_Number

而且

DECLARE@exclude varchar（50）
设置@exclude='0:@O[]`^\/'
声明@char
声明@lenchar
声明@output varchar（50）
设置@output=''
设置为@len=8
当@len>0开始时
选择@char=char（圆形（rand（）*74+48,0））
如果charindex（@char，@exclude）=0，则开始
设置@output=@output+@char
设置为@len=@len-1
结束
结束
选择@output

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可以使用。

在函数中，我们不能使用NEWID（）或RAND（）首先需要创建视图

功能

输出：

FORSELECT语句

输出：

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Dinesh答案的扩展

create view NewID  as select newid() as [newid] , RAND() as [rand]


CREATE FUNCTION [dbo].[GenerateRandomID]
(
@length as int
)
RETURNS   VARCHAR(32)
AS
BEGIN
declare @randIndex as int 
declare @randstring  as varchar(36)

select  @randIndex = CEILING( (30 - @length) * [rand]) , @randstring=   Replace(CONVERT (varchar(40) , [newid]) , '-','') from getNewID

-- Return the result of the function
RETURN SUBSTRING(@randstring,@randIndex, @length)

END

---

生成GUID的方式可以生成随机数。

下面是在SQL中生成4或8个字符长的随机字母数字字符串的方式。

select LEFT(CONVERT(VARCHAR(36),NEWID()),4)+RIGHT(CONVERT(VARCHAR(36),NEWID()),4)

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---

可能需要一个CLR函数。听起来从C#开始更容易。在函数中是否可以使用NEWID（）@mohan111I在函数中不能使用RAND（）或NEWID（）。

 select cast((Abs(Checksum(NewId()))%10) as varchar(1)) + 
       char(ascii('a')+(Abs(Checksum(NewId()))%25)) +
       char(ascii('A')+(Abs(Checksum(NewId()))%25)) +
       left(newid(),5) Random_Number

    DECLARE @exclude varchar(50) 
    SET @exclude = '0:;<=>?@O[]`^\/'
    DECLARE @char char
    DECLARE @len char
    DECLARE @output varchar(50)
    set @output = ''
    set @len = 8

    while @len > 0 begin
       select @char = char(round(rand() * 74 + 48, 0))
       if charindex(@char, @exclude) = 0 begin
           set @output = @output + @char
           set @len = @len - 1
       end
    end

   SELECT @output

CREATE VIEW NewID as select newid() as new_id


DECLARE @new_id VARCHAR(255)


SELECT @new_id = new_id FROM newid


SELECT @Password = CAST((ABS(CHECKSUM(@new_id))%10) AS VARCHAR(1)) + 
CHAR(ASCII('a')+(ABS(CHECKSUM(@new_id))%25)) +
CHAR(ASCII('A')+(ABS(CHECKSUM(@new_id))%25)) +
LEFT(@new_id,3)


SELECT @PASSWORD

9eEF44
5uUFA2
7hHFA7
.
.
.

DECLARE @new_id VARCHAR(200)

SELECT @new_id = NEWID()

SELECT CAST((ABS(CHECKSUM(@new_id))%10) AS VARCHAR(1)) + 
CHAR(ASCII('a')+(ABS(CHECKSUM(@new_id))%25)) +
CHAR(ASCII('A')+(ABS(CHECKSUM(@new_id))%25)) +
LEFT(@new_id,3)

0aAF3C
5pP3CE
2wW85E
.
.
.

create view NewID  as select newid() as [newid] , RAND() as [rand]


CREATE FUNCTION [dbo].[GenerateRandomID]
(
@length as int
)
RETURNS   VARCHAR(32)
AS
BEGIN
declare @randIndex as int 
declare @randstring  as varchar(36)

select  @randIndex = CEILING( (30 - @length) * [rand]) , @randstring=   Replace(CONVERT (varchar(40) , [newid]) , '-','') from getNewID

-- Return the result of the function
RETURN SUBSTRING(@randstring,@randIndex, @length)

END

select LEFT(CONVERT(VARCHAR(36),NEWID()),4)+RIGHT(CONVERT(VARCHAR(36),NEWID()),4)

DECLARE @chars NCHAR(36)

SET @chars = N’0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ’

DECLARE @result NCHAR(5)

SET @result = SUBSTRING(@chars, CAST((RAND() * LEN(@chars)) AS INT) + 1, 1)
+ SUBSTRING(@chars, CAST((RAND() * LEN(@chars)) AS INT) + 1, 1)
+ SUBSTRING(@chars, CAST((RAND() * LEN(@chars)) AS INT) + 1, 1)
+ SUBSTRING(@chars, CAST((RAND() * LEN(@chars)) AS INT) + 1, 1)
+ SUBSTRING(@chars, CAST((RAND() * LEN(@chars)) AS INT) + 1, 1)

SELECT @result