如果您想使用T-SQL生成伪随机字母数字字符串，您会怎么做？你会如何排除美元符号，破折号和斜线等字符？
答案 0 :(得分：168)
使用guid
SELECT @randomString = CONVERT(varchar(255), NEWID())
很短......
答案 1 :(得分：45)
与第一个例子类似，但具有更大的灵活性：
-- min_length = 8, max_length = 12
SET @Length = RAND() * 5 + 8
-- SET @Length = RAND() * (max_length - min_length + 1) + min_length
-- define allowable character explicitly - easy to read this way an easy to
-- omit easily confused chars like l (ell) and 1 (one) or 0 (zero) and O (oh)
SET @CharPool =
'abcdefghijkmnopqrstuvwxyzABCDEFGHIJKLMNPQRSTUVWXYZ23456789.,-_!$@#%^&*'
SET @PoolLength = Len(@CharPool)
SET @LoopCount = 0
SET @RandomString = ''
WHILE (@LoopCount < @Length) BEGIN
SELECT @RandomString = @RandomString +
SUBSTRING(@Charpool, CONVERT(int, RAND() * @PoolLength), 1)
SELECT @LoopCount = @LoopCount + 1
END
我忘了提及使其更灵活的其他功能之一。通过重复@CharPool中的字符块，您可以增加某些字符的权重，以便更有可能选择它们。
答案 2 :(得分：32)
当生成随机数据时，特别是用于测试，使数据随机，但可重现是非常有用的。秘诀是对随机函数使用显式种子，这样当使用相同的种子再次运行测试时，它再次产生完全相同的字符串。以下是以可重现的方式生成对象名称的函数的简化示例：
alter procedure usp_generateIdentifier
@minLen int = 1
, @maxLen int = 256
, @seed int output
, @string varchar(8000) output
as
begin
set nocount on;
declare @length int;
declare @alpha varchar(8000)
, @digit varchar(8000)
, @specials varchar(8000)
, @first varchar(8000)
declare @step bigint = rand(@seed) * 2147483647;
select @alpha = 'qwertyuiopasdfghjklzxcvbnm'
, @digit = '1234567890'
, @specials = '_@# '
select @first = @alpha + '_@';
set @seed = (rand((@seed+@step)%2147483647)*2147483647);
select @length = @minLen + rand(@seed) * (@maxLen-@minLen)
, @seed = (rand((@seed+@step)%2147483647)*2147483647);
declare @dice int;
select @dice = rand(@seed) * len(@first),
@seed = (rand((@seed+@step)%2147483647)*2147483647);
select @string = substring(@first, @dice, 1);
while 0 < @length
begin
select @dice = rand(@seed) * 100
, @seed = (rand((@seed+@step)%2147483647)*2147483647);
if (@dice < 10) -- 10% special chars
begin
select @dice = rand(@seed) * len(@specials)+1
, @seed = (rand((@seed+@step)%2147483647)*2147483647);
select @string = @string + substring(@specials, @dice, 1);
end
else if (@dice < 10+10) -- 10% digits
begin
select @dice = rand(@seed) * len(@digit)+1
, @seed = (rand((@seed+@step)%2147483647)*2147483647);
select @string = @string + substring(@digit, @dice, 1);
end
else -- rest 80% alpha
begin
declare @preseed int = @seed;
select @dice = rand(@seed) * len(@alpha)+1
, @seed = (rand((@seed+@step)%2147483647)*2147483647);
select @string = @string + substring(@alpha, @dice, 1);
end
select @length = @length - 1;
end
end
go
运行测试时，调用者会生成一个与测试运行相关联的随机种子（将其保存在结果表中），然后传递种子，类似于：
declare @seed int;
declare @string varchar(256);
select @seed = 1234; -- saved start seed
exec usp_generateIdentifier
@seed = @seed output
, @string = @string output;
print @string;
exec usp_generateIdentifier
@seed = @seed output
, @string = @string output;
print @string;
exec usp_generateIdentifier
@seed = @seed output
, @string = @string output;
print @string;
更新2016-02-17：请参阅下面的评论，原始程序在推进随机种子的方式上存在问题。我更新了代码，并修复了提到的一对一问题。
答案 3 :(得分：30)
使用以下代码返回一个短字符串：
SELECT SUBSTRING(CONVERT(varchar(40), NEWID()),0,9)
答案 4 :(得分：16)
如果您运行的是SQL Server 2008或更高版本，则可以使用新的加密函数crypt_gen_random（），然后使用base64编码将其作为字符串。这最多可以使用8000个字符。
declare @BinaryData varbinary(max)
, @CharacterData varchar(max)
, @Length int = 2048
set @BinaryData=crypt_gen_random (@Length)
set @CharacterData=cast('' as xml).value('xs:base64Binary(sql:variable("@BinaryData"))', 'varchar(max)')
print @CharacterData
答案 5 :(得分：11)
我不是T-SQL的专家，但是我已经使用它的最简单方式是这样的：
select char((rand()*25 + 65))+char((rand()*25 + 65))
这会产生两个字符（A-Z，在ascii 65-90中）。
答案 6 :(得分：10)
select left(NEWID(),5)
这将返回guid字符串的最左边5个字符
Example run
------------
11C89
9DB02
答案 7 :(得分：6)
这是一个随机字母数字生成器
print left(replace(newid(),'-',''),@length) //--@length is the length of random Num.
答案 8 :(得分：4)
这对我有用：我需要为ID生成三个随机的字母数字字符，但它可以在任何长度达到15左右的情况下工作。
declare @DesiredLength as int = 3;
select substring(replace(newID(),'-',''),cast(RAND()*(31-@DesiredLength) as int),@DesiredLength);
答案 9 :(得分：3)
对于一封随机字母，您可以使用：
select substring('ABCDEFGHIJKLMNOPQRSTUVWXYZ',
(abs(checksum(newid())) % 26)+1, 1)
使用newid()与rand()之间的一个重要区别是，如果您返回多行，则newid()会针对每一行单独计算，而rand()则会计算一次整个查询。
答案 10 :(得分：2)
我意识到这是一个很老的问题，有很多很好的答案。然而，当我发现这一点时，我还发现了一篇由Saeid Hasani撰写的最近一篇关于TechNet的文章
T-SQL: How to Generate Random Passwords
虽然解决方案侧重于密码，但它适用于一般情况。赛义德通过各种考虑来达成解决方案。这非常有启发性。
包含文章所有代码块的脚本可以通过TechNet Gallery单独获得，但我肯定会从文章开始。
答案 11 :(得分：1)
对于SQL Server 2016及更高版本，这是一个非常简单且相对有效的表达式，用于生成给定字节长度的加密随机字符串：
--Generates 36 bytes (48 characters) of base64 encoded random data
select r from OpenJson((select Crypt_Gen_Random(36) r for json path))
with (r varchar(max))
请注意，字节长度与编码大小不同；使用this文章中的以下内容进行转换：
Bytes = 3 * (LengthInCharacters / 4) - Padding
答案 12 :(得分：1)
我使用我开发的这个程序只是简单地规定了你想要在输入变量中显示的字符，你也可以定义它的长度。 希望这种格式很好，我是堆栈溢出的新手。
IF EXISTS (SELECT * FROM sys.objects WHERE type = 'P' AND object_id = OBJECT_ID(N'GenerateARandomString'))
DROP PROCEDURE GenerateARandomString
GO
CREATE PROCEDURE GenerateARandomString
(
@DESIREDLENGTH INTEGER = 100,
@NUMBERS VARCHAR(50)
= '0123456789',
@ALPHABET VARCHAR(100)
='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz',
@SPECIALS VARCHAR(50)
= '_=+-$£%^&*()"!@~#:',
@RANDOMSTRING VARCHAR(8000) OUT
)
AS
BEGIN
-- Author David Riley
-- Version 1.0
-- You could alter to one big string .e.e numebrs , alpha special etc
-- added for more felxibility in case I want to extend i.e put logic in for 3 numbers, 2 pecials 3 numbers etc
-- for now just randomly pick one of them
DECLARE @SWAP VARCHAR(8000); -- Will be used as a tempoary buffer
DECLARE @SELECTOR INTEGER = 0;
DECLARE @CURRENTLENGHT INTEGER = 0;
WHILE @CURRENTLENGHT < @DESIREDLENGTH
BEGIN
-- Do we want a number, special character or Alphabet Randonly decide?
SET @SELECTOR = CAST(ABS(CHECKSUM(NEWID())) % 3 AS INTEGER); -- Always three 1 number , 2 alphaBET , 3 special;
IF @SELECTOR = 0
BEGIN
SET @SELECTOR = 3
END;
-- SET SWAP VARIABLE AS DESIRED
SELECT @SWAP = CASE WHEN @SELECTOR = 1 THEN @NUMBERS WHEN @SELECTOR = 2 THEN @ALPHABET ELSE @SPECIALS END;
-- MAKE THE SELECTION
SET @SELECTOR = CAST(ABS(CHECKSUM(NEWID())) % LEN(@SWAP) AS INTEGER);
IF @SELECTOR = 0
BEGIN
SET @SELECTOR = LEN(@SWAP)
END;
SET @RANDOMSTRING = ISNULL(@RANDOMSTRING,'') + SUBSTRING(@SWAP,@SELECTOR,1);
SET @CURRENTLENGHT = LEN(@RANDOMSTRING);
END;
END;
GO
DECLARE @RANDOMSTRING VARCHAR(8000)
EXEC GenerateARandomString @RANDOMSTRING = @RANDOMSTRING OUT
SELECT @RANDOMSTRING
答案 13 :(得分：0)
基于本文的各种有用回答，我结合了一些我喜欢的选项。
DECLARE @UserId BIGINT = 12345 -- a uniqueId in my system
SELECT LOWER(REPLACE(NEWID(),'-','')) + CONVERT(VARCHAR, @UserId)
答案 14 :(得分：0)
以下是在SQL中生成4或8个字符的长随机字母数字字符串的方法
select LEFT(CONVERT(VARCHAR(36),NEWID()),4)+RIGHT(CONVERT(VARCHAR(36),NEWID()),4)
SELECT RIGHT(REPLACE(CONVERT(VARCHAR(36),NEWID()),'-',''),8)
答案 15 :(得分：0)
我首先遇到this blog post，然后想出了我正在使用的当前项目的以下存储过程（抱歉奇怪的格式化）：
CREATE PROCEDURE [dbo].[SpGenerateRandomString]
@sLength tinyint = 10,
@randomString varchar(50) OUTPUT
AS
BEGIN
SET NOCOUNT ON
DECLARE @counter tinyint
DECLARE @nextChar char(1)
SET @counter = 1
SET @randomString = ”
WHILE @counter <= @sLength
BEGIN
SELECT @nextChar = CHAR(48 + CONVERT(INT, (122-48+1)*RAND()))
IF ASCII(@nextChar) not in (58,59,60,61,62,63,64,91,92,93,94,95,96)
BEGIN
SELECT @randomString = @randomString + @nextChar
SET @counter = @counter + 1
END
END
END
答案 16 :(得分：0)
这将产生一个长度为96个字符的字符串，其长度来自Base64范围（上，下，数字，+和/）。添加3个“ NEWID（）”将使长度增加32，而没有Base64填充（=）。
SELECT
CAST(
CONVERT(NVARCHAR(MAX),
CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
,2)
AS XML).value('xs:base64Binary(xs:hexBinary(.))', 'VARCHAR(MAX)') AS StringValue
如果要将其应用于集合，请确保引入该集合中的内容，以便重新计算NEWID（），否则每次将获得相同的值：
SELECT
U.UserName
, LEFT(PseudoRandom.StringValue, LEN(U.Pwd)) AS FauxPwd
FROM Users U
CROSS APPLY (
SELECT
CAST(
CONVERT(NVARCHAR(MAX),
CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), NEWID())
+CONVERT(VARBINARY(8), U.UserID) -- Causes a recomute of all NEWID() calls
,2)
AS XML).value('xs:base64Binary(xs:hexBinary(.))', 'VARCHAR(MAX)') AS StringValue
) PseudoRandom
答案 17 :(得分：0)
非常简单。使用它并享受它。
CREATE VIEW [dbo].[vwGetNewId]
AS
SELECT NEWID() AS Id
Creat FUNCTION [dbo].[fnGenerateRandomString](@length INT = 8)
RETURNS NVARCHAR(MAX)
AS
BEGIN
DECLARE @result CHAR(2000);
DECLARE @String VARCHAR(2000);
SET @String = 'abcdefghijklmnopqrstuvwxyz' + --lower letters
'ABCDEFGHIJKLMNOPQRSTUVWXYZ' + --upper letters
'1234567890'; --number characters
SELECT @result =
(
SELECT TOP (@length)
SUBSTRING(@String, 1 + number, 1) AS [text()]
FROM master..spt_values
WHERE number < DATALENGTH(@String)
AND type = 'P'
ORDER BY
(
SELECT TOP 1 Id FROM dbo.vwGetNewId
) --instead of using newid()
FOR XML PATH('')
);
RETURN @result;
END;
答案 18 :(得分：0)
这是我今天想出的一个（因为我不喜欢任何现有的答案）。
这个生成随机字符串的临时表，基于newid()，但也支持自定义字符集（因此不仅仅是0-9＆amp; AF），自定义长度（最多255，限制是硬编码的，但可以更改），以及自定义数量的随机记录。
这是源代码（希望评论有帮助）：
/**
* First, we're going to define the random parameters for this
* snippet. Changing these variables will alter the entire
* outcome of this script. Try not to break everything.
*
* @var {int} count The number of random values to generate.
* @var {int} length The length of each random value.
* @var {char(62)} charset The characters that may appear within a random value.
*/
-- Define the parameters
declare @count int = 10
declare @length int = 60
declare @charset char(62) = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'
/**
* We're going to define our random table to be twice the maximum
* length (255 * 2 = 510). It's twice because we will be using
* the newid() method, which produces hex guids. More later.
*/
-- Create the random table
declare @random table (
value nvarchar(510)
)
/**
* We'll use two characters from newid() to make one character in
* the random value. Each newid() provides us 32 hex characters,
* so we'll have to make multiple calls depending on length.
*/
-- Determine how many "newid()" calls we'll need per random value
declare @iterations int = ceiling(@length * 2 / 32.0)
/**
* Before we start making multiple calls to "newid", we need to
* start with an initial value. Since we know that we need at
* least one call, we will go ahead and satisfy the count.
*/
-- Iterate up to the count
declare @i int = 0 while @i < @count begin set @i = @i + 1
-- Insert a new set of 32 hex characters for each record, limiting to @length * 2
insert into @random
select substring(replace(newid(), '-', ''), 1, @length * 2)
end
-- Now fill the remaining the remaining length using a series of update clauses
set @i = 0 while @i < @iterations begin set @i = @i + 1
-- Append to the original value, limit @length * 2
update @random
set value = substring(value + replace(newid(), '-', ''), 1, @length * 2)
end
/**
* Now that we have our base random values, we can convert them
* into the final random values. We'll do this by taking two
* hex characters, and mapping then to one charset value.
*/
-- Convert the base random values to charset random values
set @i = 0 while @i < @length begin set @i = @i + 1
/**
* Explaining what's actually going on here is a bit complex. I'll
* do my best to break it down step by step. Hopefully you'll be
* able to follow along. If not, then wise up and come back.
*/
-- Perform the update
update @random
set value =
/**
* Everything we're doing here is in a loop. The @i variable marks
* what character of the final result we're assigning. We will
* start off by taking everything we've already done first.
*/
-- Take the part of the string up to the current index
substring(value, 1, @i - 1) +
/**
* Now we're going to convert the two hex values after the index,
* and convert them to a single charset value. We can do this
* with a bit of math and conversions, so function away!
*/
-- Replace the current two hex values with one charset value
substring(@charset, convert(int, convert(varbinary(1), substring(value, @i, 2), 2)) * (len(@charset) - 1) / 255 + 1, 1) +
-- (1) -------------------------------------------------------^^^^^^^^^^^^^^^^^^^^^^^-----------------------------------------
-- (2) ---------------------------------^^^^^^^^^^^^^^^^^^^^^^11111111111111111111111^^^^-------------------------------------
-- (3) --------------------^^^^^^^^^^^^^2222222222222222222222222222222222222222222222222^------------------------------------
-- (4) --------------------333333333333333333333333333333333333333333333333333333333333333---^^^^^^^^^^^^^^^^^^^^^^^^^--------
-- (5) --------------------333333333333333333333333333333333333333333333333333333333333333^^^4444444444444444444444444--------
-- (6) --------------------5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555^^^^----
-- (7) ^^^^^^^^^^^^^^^^^^^^66666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666^^^^
/**
* (1) - Determine the two hex characters that we'll be converting (ex: 0F, AB, 3C, etc.)
* (2) - Convert those two hex characters to a a proper hexadecimal (ex: 0x0F, 0xAB, 0x3C, etc.)
* (3) - Convert the hexadecimals to integers (ex: 15, 171, 60)
* (4) - Determine the conversion ratio between the length of @charset and the range of hexadecimals (255)
* (5) - Multiply the integer from (3) with the conversion ratio from (4) to get a value between 0 and (len(@charset) - 1)
* (6) - Add 1 to the offset from (5) to get a value between 1 and len(@charset), since strings start at 1 in SQL
* (7) - Use the offset from (6) and grab a single character from @subset
*/
/**
* All that is left is to add in everything we have left to do.
* We will eventually process the entire string, but we will
* take things one step at a time. Round and round we go!
*/
-- Append everything we have left to do
substring(value, 2 + @i, len(value))
end
-- Select the results
select value
from @random
它不是一个存储过程，但它很难将它变成一个存储过程。它也不是非常慢（我花了大约0.3秒来生成长度为60的1000个结果，这比我个人所需要的还多），这是我最初关注的所有字符串之一突然我正在做。
这里的主要内容是我没有尝试创建自己的随机数生成器，而且我的字符集不受限制。我只是使用SQL具有的随机生成器（我知道那里有rand()，但这对于表结果来说并不好）。希望这种方法在这里结合两种答案，从过于简单（即只是newid()）和过于复杂（即自定义随机数算法）。
它也很短（减去评论），并且易于理解（至少对我而言），这在我的书中总是有优势。
但是，这种方法不能播种，因此每次都会真正随机，并且您无法以任何可靠的方式复制同一组数据。 OP没有列出这个要求，但我知道有些人会寻找那种东西。
我知道我在这里参加派对很晚，但希望有人会觉得这很有用。
答案 19 :(得分：0)
CREATE OR ALTER PROC USP_GENERATE_RANDOM_CHARACTER (@NO_OF_CHARS INT, @RANDOM_CHAR VARCHAR(40) OUTPUT) 作为 开始
SELECT @RANDOM_CHAR = SUBSTRING (REPLACE(CONVERT(VARCHAR(40), NEWID()), '-',''), 1, @NO_OF_CHARS)
结束 /* 用法： 声明@OUT VARCHAR(40) EXEC USP_GENERATE_RANDOM_CHARACTER 13,@RANDOM_CHAR = @OUT OUTPUT 选择@OUT */
答案 20 :(得分：0)
所以我喜欢上面的很多答案，但我一直在寻找一些本质上更随意的东西。我还想要一种明确调出排除字符的方法。下面是我的解决方案，使用一个调用CRYPT_GEN_RANDOM来获取加密随机数的视图。在我的例子中，我只选择了一个8字节的随机数。请注意，您可以增加此大小，并根据需要使用函数的种子参数。以下是文档的链接：https://docs.microsoft.com/en-us/sql/t-sql/functions/crypt-gen-random-transact-sql
CREATE VIEW [dbo].[VW_CRYPT_GEN_RANDOM_8]
AS
SELECT CRYPT_GEN_RANDOM(8) as [value];
创建视图的原因是因为无法直接从函数调用CRYPT_GEN_RANDOM。
从那里，我创建了一个标量函数，它接受一个长度和一个字符串参数，该参数可以包含逗号分隔的排除字符串。
CREATE FUNCTION [dbo].[fn_GenerateRandomString]
(
@length INT,
@excludedCharacters VARCHAR(200) --Comma delimited string of excluded characters
)
RETURNS VARCHAR(Max)
BEGIN
DECLARE @returnValue VARCHAR(Max) = ''
, @asciiValue INT
, @currentCharacter CHAR;
--Optional concept, you can add default excluded characters
SET @excludedCharacters = CONCAT(@excludedCharacters,',^,*,(,),-,_,=,+,[,{,],},\,|,;,:,'',",<,.,>,/,`,~');
--Table of excluded characters
DECLARE @excludedCharactersTable table([asciiValue] INT);
--Insert comma
INSERT INTO @excludedCharactersTable SELECT 44;
--Stores the ascii value of the excluded characters in the table
INSERT INTO @excludedCharactersTable
SELECT ASCII(TRIM(value))
FROM STRING_SPLIT(@excludedCharacters, ',')
WHERE LEN(TRIM(value)) = 1;
--Keep looping until the return string is filled
WHILE(LEN(@returnValue) < @length)
BEGIN
--Get a truly random integer values from 33-126
SET @asciiValue = (SELECT TOP 1 (ABS(CONVERT(INT, [value])) % 94) + 33 FROM [dbo].[VW_CRYPT_GEN_RANDOM_8]);
--If the random integer value is not in the excluded characters table then append to the return string
IF(NOT EXISTS(SELECT *
FROM @excludedCharactersTable
WHERE [asciiValue] = @asciiValue))
BEGIN
SET @returnValue = @returnValue + CHAR(@asciiValue);
END
END
RETURN(@returnValue);
END
以下是如何调用该函数的示例。
SELECT [dbo].[fn_GenerateRandomString](8,'!,@,#,$,%,&,?');
〜干杯
答案 21 :(得分：0)
在 SQL Server 2012 + 中，我们可以连接某些（G）UID的二进制文件，然后对结果进行base64转换。
SELECT
textLen.textLen
, left((
select CAST(newid() as varbinary(max)) + CAST(newid() as varbinary(max))
where textLen.textLen is not null /*force evaluation for each outer query row*/
FOR XML PATH(''), BINARY BASE64
),textLen.textLen) as randomText
FROM ( values (2),(4),(48) ) as textLen(textLen) --define lengths here
;
如果您需要更长的字符串（或者您在结果中看到=个字符），则需要在子选择中添加更多+ CAST(newid() as varbinary(max))。
答案 22 :(得分：0)
有时我们需要很多随机的东西：爱，善良，度假等。 多年来我收集了一些随机发生器，这些都是来自Pinal Dave和我曾经发现的一个堆栈溢出的答案。参考下面。
--Adapted from Pinal Dave; http://blog.sqlauthority.com/2007/04/29/sql-server-random-number-generator-script-sql-query/
SELECT
ABS( CAST( NEWID() AS BINARY( 6)) %1000) + 1 AS RandomInt
, CAST( (ABS( CAST( NEWID() AS BINARY( 6)) %1000) + 1)/7.0123 AS NUMERIC( 15,4)) AS RandomNumeric
, DATEADD( DAY, -1*(ABS( CAST( NEWID() AS BINARY( 6)) %1000) + 1), GETDATE()) AS RandomDate
--This line from http://stackoverflow.com/questions/15038311/sql-password-generator-8-characters-upper-and-lower-and-include-a-number
, 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) AS RandomChar
, ABS(CHECKSUM(NEWID()))%50000+1 AS RandomID
答案 23 :(得分：0)
有很多好的答案，但到目前为止，它们都不允许使用可自定义的字符池，并且可以作为列的默认值。我希望能够做到这样的事情：
alter table MY_TABLE add MY_COLUMN char(20) not null
default dbo.GenerateToken(crypt_gen_random(20))
所以我想出了这个。如果你修改它，请注意硬编码的数字32。
-- Converts a varbinary of length N into a varchar of length N.
-- Recommend passing in the result of CRYPT_GEN_RANDOM(N).
create function GenerateToken(@randomBytes varbinary(max))
returns varchar(max) as begin
-- Limit to 32 chars to get an even distribution (because 32 divides 256) with easy math.
declare @allowedChars char(32);
set @allowedChars = 'abcdefghijklmnopqrstuvwxyz012345';
declare @oneByte tinyint;
declare @oneChar char(1);
declare @index int;
declare @token varchar(max);
set @index = 0;
set @token = '';
while @index < datalength(@randomBytes)
begin
-- Get next byte, use it to index into @allowedChars, and append to @token.
-- Note: substring is 1-based.
set @index = @index + 1;
select @oneByte = convert(tinyint, substring(@randomBytes, @index, 1));
select @oneChar = substring(@allowedChars, 1 + (@oneByte % 32), 1); -- 32 is the number of @allowedChars
select @token = @token + @oneChar;
end
return @token;
end
答案 24 :(得分：0)
这使用rand和其他答案之一的种子，但没有必要在每次调用时提供种子。在第一次通话时提供它就足够了。
这是我修改后的代码。
IF EXISTS (SELECT * FROM sys.objects WHERE type = 'P' AND object_id = OBJECT_ID(N'usp_generateIdentifier'))
DROP PROCEDURE usp_generateIdentifier
GO
create procedure usp_generateIdentifier
@minLen int = 1
, @maxLen int = 256
, @seed int output
, @string varchar(8000) output
as
begin
set nocount on;
declare @length int;
declare @alpha varchar(8000)
, @digit varchar(8000)
, @specials varchar(8000)
, @first varchar(8000)
select @alpha = 'qwertyuiopasdfghjklzxcvbnm'
, @digit = '1234567890'
, @specials = '_@#$&'
select @first = @alpha + '_@';
-- Establish our rand seed and store a new seed for next time
set @seed = (rand(@seed)*2147483647);
select @length = @minLen + rand() * (@maxLen-@minLen);
--print @length
declare @dice int;
select @dice = rand() * len(@first);
select @string = substring(@first, @dice, 1);
while 0 < @length
begin
select @dice = rand() * 100;
if (@dice < 10) -- 10% special chars
begin
select @dice = rand() * len(@specials)+1;
select @string = @string + substring(@specials, @dice, 1);
end
else if (@dice < 10+10) -- 10% digits
begin
select @dice = rand() * len(@digit)+1;
select @string = @string + substring(@digit, @dice, 1);
end
else -- rest 80% alpha
begin
select @dice = rand() * len(@alpha)+1;
select @string = @string + substring(@alpha, @dice, 1);
end
select @length = @length - 1;
end
end
go
答案 25 :(得分：0)
我以为我会分享，或者回馈社区...... 它是基于ASCII的，解决方案并不完美，但效果很好。 请享用， Goran B。
/*
-- predictable masking of ascii chars within a given decimal range
-- purpose:
-- i needed an alternative to hashing alg. or uniqueidentifier functions
-- because i wanted to be able to revert to original char set if possible ("if", the operative word)
-- notes: wrap below in a scalar function if desired (i.e. recommended)
-- by goran biljetina (2014-02-25)
*/
declare
@length int
,@position int
,@maskedString varchar(500)
,@inpString varchar(500)
,@offsetAsciiUp1 smallint
,@offsetAsciiDown1 smallint
,@ipOffset smallint
,@asciiHiBound smallint
,@asciiLoBound smallint
set @ipOffset=null
set @offsetAsciiUp1=1
set @offsetAsciiDown1=-1
set @asciiHiBound=126 --> up to and NOT including
set @asciiLoBound=31 --> up from and NOT including
SET @inpString = '{"config":"some string value", "boolAttr": true}'
SET @length = LEN(@inpString)
SET @position = 1
SET @maskedString = ''
--> MASK:
---------
WHILE (@position < @length+1) BEGIN
SELECT @maskedString = @maskedString +
ISNULL(
CASE
WHEN ASCII(SUBSTRING(@inpString,@position,1))>@asciiLoBound AND ASCII(SUBSTRING(@inpString,@position,1))<@asciiHiBound
THEN
CHAR(ASCII(SUBSTRING(@inpString,@position,1))+
(case when @ipOffset is null then
case when ASCII(SUBSTRING(@inpString,@position,1))%2=0 then @offsetAsciiUp1 else @offsetAsciiDown1 end
else @ipOffset end))
WHEN ASCII(SUBSTRING(@inpString,@position,1))<=@asciiLoBound
THEN '('+CONVERT(varchar,ASCII(SUBSTRING(@Inpstring,@position,1))+1000)+')' --> wrap for decode
WHEN ASCII(SUBSTRING(@inpString,@position,1))>=@asciiHiBound
THEN '('+CONVERT(varchar,ASCII(SUBSTRING(@inpString,@position,1))+1000)+')' --> wrap for decode
END
,'')
SELECT @position = @position + 1
END
select @MaskedString
SET @inpString = @maskedString
SET @length = LEN(@inpString)
SET @position = 1
SET @maskedString = ''
--> UNMASK (Limited to within ascii lo-hi bound):
-------------------------------------------------
WHILE (@position < @length+1) BEGIN
SELECT @maskedString = @maskedString +
ISNULL(
CASE
WHEN ASCII(SUBSTRING(@inpString,@position,1))>@asciiLoBound AND ASCII(SUBSTRING(@inpString,@position,1))<@asciiHiBound
THEN
CHAR(ASCII(SUBSTRING(@inpString,@position,1))+
(case when @ipOffset is null then
case when ASCII(SUBSTRING(@inpString,@position,1))%2=1 then @offsetAsciiDown1 else @offsetAsciiUp1 end
else @ipOffset*(-1) end))
ELSE ''
END
,'')
SELECT @position = @position + 1
END
select @maskedString
答案 26 :(得分：0)
基于New Id的东西。
with list as
(
select 1 as id,newid() as val
union all
select id + 1,NEWID()
from list
where id + 1 < 10
)
select ID,val from list
option (maxrecursion 0)
答案 27 :(得分：0)
我是在SQL 2000中通过创建一个包含我想要使用的字符的表来创建的，创建一个视图，通过newid（）从该表中选择字符，然后从该视图中选择前1个字符。
CREATE VIEW dbo.vwCodeCharRandom
AS
SELECT TOP 100 PERCENT
CodeChar
FROM dbo.tblCharacter
ORDER BY
NEWID()
...
SELECT TOP 1 CodeChar FROM dbo.vwCodeCharRandom
然后你可以简单地从视图中提取字符并根据需要连接它们。
编辑：灵感来自斯蒂芬的回应......select top 1 RandomChar from tblRandomCharacters order by newid()
不需要视图（实际上我不知道为什么我这样做 - 代码来自几年前）。您仍然可以指定要在表格中使用的字符。