这种方式没有为我工作，因为我试图保持阿拉伯字母，我试图取代正则表达式，但它也不起作用。我写了另一个方法工作在ASCII级别，因为这是我唯一的选择，它工作。

 Create function [dbo].[RemoveNonAlphaCharacters] (@s varchar(4000)) returns varchar(4000)
   with schemabinding
begin
   if @s is null
      return null
   declare @s2 varchar(4000)
   set @s2 = ''
   declare @l int
   set @l = len(@s)
   declare @p int
   set @p = 1
   while @p <= @l begin
      declare @c int
      set @c = ascii(substring(@s, @p, 1))
      if @c between 48 and 57 or @c between 65 and 90 or @c between 97 and 122 or @c between 165 and 253 or @c between 32 and 33
         set @s2 = @s2 + char(@c)
      set @p = @p + 1
      end
   if len(@s2) = 0
      return null
   return @s2
   end

GO

使用CTE生成的数字表来检查每个字符，然后FOR XML连接到一个保留值的字符串，您可以…

CREATE FUNCTION [dbo].[PatRemove](
    @pattern varchar(50),
    @expression varchar(8000) 
    )
RETURNS varchar(8000)
AS
BEGIN
    WITH 
        d(d) AS (SELECT d FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) digits(d)),
        nums(n) AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM d d1, d d2, d d3, d d4),
        chars(c) AS (SELECT SUBSTRING(@expression, n, 1) FROM nums WHERE n <= LEN(@expression))
    SELECT 
        @expression = (SELECT c AS [text()] FROM chars WHERE c NOT LIKE @pattern FOR XML PATH(''));

    RETURN @expression;
END

Here's a solution that doesn't require creating a function or listing all instances of characters to replace. It uses a recursive WITH statement in combination with a PATINDEX to find unwanted chars. It will replace all unwanted chars in a column - up to 100 unique bad characters contained in any given string. (E.G. "ABC123DEF234" would contain 4 bad characters 1, 2, 3 and 4) The 100 limit is the maximum number of recursions allowed in a WITH statement, but this doesn't impose a limit on the number of rows to process, which is only limited by the memory available. If you don't want DISTINCT results, you can remove the two options from the code.

-- Create some test data:
SELECT * INTO #testData 
FROM (VALUES ('ABC DEF,K.l(p)'),('123H,J,234'),('ABCD EFG')) as t(TXT)

-- Actual query:
-- Remove non-alpha chars: '%[^A-Z]%'
-- Remove non-alphanumeric chars: '%[^A-Z0-9]%'
DECLARE @BadCharacterPattern VARCHAR(250) = '%[^A-Z]%';

WITH recurMain as (
    SELECT DISTINCT CAST(TXT AS VARCHAR(250)) AS TXT, PATINDEX(@BadCharacterPattern, TXT) AS BadCharIndex
    FROM #testData
    UNION ALL
    SELECT CAST(TXT AS VARCHAR(250)) AS TXT, PATINDEX(@BadCharacterPattern, TXT) AS BadCharIndex
    FROM (
        SELECT 
            CASE WHEN BadCharIndex > 0 
                THEN REPLACE(TXT, SUBSTRING(TXT, BadCharIndex, 1), '')
                ELSE TXT 
            END AS TXT
        FROM recurMain
        WHERE BadCharIndex > 0
    ) badCharFinder
)
SELECT DISTINCT TXT
FROM recurMain
WHERE BadCharIndex = 0;

我把它放在调用PatIndex的两个地方。

PatIndex('%[^A-Za-z0-9]%', @Temp)

为上面的自定义函数RemoveNonAlphaCharacters并重命名为RemoveNonAlphaNumericCharacters

SQL Server >= 2017…

declare @text varchar(max)

-- create some sample text
select
@text=
'
Lorem @ipsum  *&dolor-= sit?! amet, {consectetur } adipiscing\ elit. Vivamus commodo justo metus, sed facilisis ante 
congue eget. Proin ac bibendum sem/.
'

-- the characters to be removed
declare @unwanted varchar(max)='''.,!?/<>"[]{}|`~@#$%^&*()-+=/\:;'+char(13)+char(10)

-- interim replaced with
declare @replace_with char(1)=' '

-- call the translate function that will change unwanted characters to spaces
-- in this sample
declare @translated varchar(max)
select @translated=TRANSLATE(@text,@unwanted,REPLICATE(@replace_with,len(@unwanted)))

-- In this case, I want to preserve one space
select  string_agg(trim(value),' ')
from    STRING_SPLIT(@translated,' ')
where   trim(value)<>''

-- Result
'Lorem ipsum dolor sit amet consectetur adipiscing elit Vivamus commodo justo metus sed facilisis ante congue eget Proin ac bibendum sem'

乔治·马斯特罗斯精彩回答的参数化版本:

CREATE FUNCTION [dbo].[fn_StripCharacters]
(
    @String NVARCHAR(MAX), 
    @MatchExpression VARCHAR(255)
)
RETURNS NVARCHAR(MAX)
AS
BEGIN
    SET @MatchExpression =  '%['+@MatchExpression+']%'
    
    WHILE PatIndex(@MatchExpression, @String) > 0
        SET @String = Stuff(@String, PatIndex(@MatchExpression, @String), 1, '')
    
    RETURN @String
    
END

字母只有:

SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^a-z')

数字只有:

SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^0-9')

字母数字只有:

SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^a-z0-9')

非字母数字：

SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', 'a-z0-9')