如何从SQL Server中的字符串中剥离所有非字母字符?

我知道SQL不擅长字符串操作，但我没想到它会这么难。下面是一个简单的函数，用于从字符串中剥离所有数字。当然还有更好的办法，但这只是个开始。

CREATE FUNCTION dbo.AlphaOnly (
@String varchar(100)
)
RETURNS varchar(100)
AS BEGIN
RETURN (
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
REPLACE(
@String,
'9', ''),
'8', ''),
'7', ''),
'6', ''),
'5', ''),
'4', ''),
'3', ''),
'2', ''),
'1', ''),
'0', '')
)
END
GO


-- ==================
DECLARE @t TABLE (
ColID       int,
ColString   varchar(50)
)


INSERT INTO @t VALUES (1, 'abc1234567890')


SELECT ColID, ColString, dbo.AlphaOnly(ColString)
FROM @t

输出

ColID ColString
----- ------------- ---
1 abc1234567890 abc

第2轮-数据驱动黑名单

-- ============================================
-- Create a table of blacklist characters
-- ============================================
IF EXISTS (SELECT * FROM sys.tables WHERE [object_id] = OBJECT_ID('dbo.CharacterBlacklist'))
DROP TABLE dbo.CharacterBlacklist
GO
CREATE TABLE dbo.CharacterBlacklist (
CharID              int         IDENTITY,
DisallowedCharacter nchar(1)    NOT NULL
)
GO
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'0')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'1')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'2')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'3')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'4')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'5')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'6')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'7')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'8')
INSERT INTO dbo.CharacterBlacklist (DisallowedCharacter) VALUES (N'9')
GO


-- ====================================
IF EXISTS (SELECT * FROM sys.objects WHERE [object_id] = OBJECT_ID('dbo.StripBlacklistCharacters'))
DROP FUNCTION dbo.StripBlacklistCharacters
GO
CREATE FUNCTION dbo.StripBlacklistCharacters (
@String nvarchar(100)
)
RETURNS varchar(100)
AS BEGIN
DECLARE @blacklistCt  int
DECLARE @ct           int
DECLARE @c            nchar(1)


SELECT @blacklistCt = COUNT(*) FROM dbo.CharacterBlacklist


SET @ct = 0
WHILE @ct < @blacklistCt BEGIN
SET @ct = @ct + 1


SELECT @String = REPLACE(@String, DisallowedCharacter, N'')
FROM dbo.CharacterBlacklist
WHERE CharID = @ct
END


RETURN (@String)
END
GO


-- ====================================
DECLARE @s  nvarchar(24)
SET @s = N'abc1234def5678ghi90jkl'


SELECT
@s                  AS OriginalString,
dbo.StripBlacklistCharacters(@s)   AS ResultString

输出

OriginalString           ResultString
------------------------ ------------
abc1234def5678ghi90jkl   abcdefghijkl

我对读者的挑战是:你能让这个过程更有效率吗?那么使用递归呢?

试试这个函数:

Create Function [dbo].[RemoveNonAlphaCharacters](@Temp VarChar(1000))
Returns VarChar(1000)
AS
Begin


Declare @KeepValues as varchar(50)
Set @KeepValues = '%[^a-z]%'
While PatIndex(@KeepValues, @Temp) > 0
Set @Temp = Stuff(@Temp, PatIndex(@KeepValues, @Temp), 1, '')


Return @Temp
End

这样叫它:

Select dbo.RemoveNonAlphaCharacters('abc1234def5678ghi90jkl')

一旦您理解了代码，您就会发现更改它以删除其他字符也相对简单。您甚至可以使此动态到足以传入您的搜索模式。

乔治获取 ` 很棒的回答的参数化版本:

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')

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

create FUNCTION dbo.udf_Cleanchars (@InputString varchar(80))
RETURNS varchar(80)
AS


BEGIN
declare @return varchar(80) , @length int , @counter int , @cur_char char(1)
SET @return = ''
SET @length = 0
SET @counter = 1
SET @length = LEN(@InputString)
IF @length > 0
BEGIN WHILE @counter <= @length


BEGIN SET @cur_char = SUBSTRING(@InputString, @counter, 1) IF ((ascii(@cur_char) in (32,44,46)) or (ascii(@cur_char) between 48 and 57) or (ascii(@cur_char) between 65 and 90) or (ascii(@cur_char) between 97 and 122))
BEGIN SET @return = @return + @cur_char END
SET @counter = @counter + 1
END END


RETURN @return END

如果您像我一样，不能仅向生产数据添加函数，但仍然想执行这种过滤，那么这里有一个纯SQL解决方案，使用PIVOT表将过滤后的部分重新组合在一起。

注意:我硬编码表高达40个字符，如果你有更长的字符串要过滤，你将不得不添加更多。

SET CONCAT_NULL_YIELDS_NULL OFF;


with
ToBeScrubbed
as (
select 1 as id, '*SOME 222@ !@* #* BOGUS !@*&! DATA' as ColumnToScrub
),


Scrubbed as (
select
P.Number as ValueOrder,
isnull ( substring ( t.ColumnToScrub , number , 1 ) , '' ) as ScrubbedValue,
t.id
from
ToBeScrubbed t
left join master..spt_values P
on P.number between 1 and len(t.ColumnToScrub)
and type ='P'
where
PatIndex('%[^a-z]%', substring(t.ColumnToScrub,P.number,1) ) = 0
)


SELECT
id,
[1]+ [2]+ [3]+ [4]+ [5]+ [6]+ [7]+ [8] +[9] +[10]
+  [11]+ [12]+ [13]+ [14]+ [15]+ [16]+ [17]+ [18] +[19] +[20]
+  [21]+ [22]+ [23]+ [24]+ [25]+ [26]+ [27]+ [28] +[29] +[30]
+  [31]+ [32]+ [33]+ [34]+ [35]+ [36]+ [37]+ [38] +[39] +[40] as ScrubbedData
FROM (
select
*
from
Scrubbed
)
src
PIVOT (
MAX(ScrubbedValue) FOR ValueOrder IN (
[1], [2], [3], [4], [5], [6], [7], [8], [9], [10],
[11], [12], [13], [14], [15], [16], [17], [18], [19], [20],
[21], [22], [23], [24], [25], [26], [27], [28], [29], [30],
[31], [32], [33], [34], [35], [36], [37], [38], [39], [40]
)
) pvt

我刚在Oracle 10g中找到了这个，如果你用的就是它的话。为了进行电话号码比较，我必须去掉所有的特殊字符。

regexp_replace(c.phone, '[^0-9]', '')

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

这个解决方案受到Allen先生的解决方案的启发，需要一个Numbers整数表(如果你想进行严肃的查询操作并获得良好的性能，你应该手头有这个表)。它不需要CTE。您可以更改NOT IN (...)表达式以排除特定字符，或将其更改为IN (...)或LIKE表达式以仅保留某些字符。

SELECT (
SELECT  SUBSTRING([YourString], N, 1)
FROM    dbo.Numbers
WHERE   N > 0 AND N <= CONVERT(INT, LEN([YourString]))
AND SUBSTRING([YourString], N, 1) NOT IN ('(',')',',','.')
FOR XML PATH('')
) AS [YourStringTransformed]
FROM ...

看过所有给出的解决方案后，我认为必须有一个纯SQL方法，它不需要函数或CTE / XML查询，并且不涉及难以维护的嵌套REPLACE语句。以下是我的解决方案:

SELECT
x
,CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 1, 1) + '%' THEN '' ELSE SUBSTRING(x, 1, 1) END
+ CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 2, 1) + '%' THEN '' ELSE SUBSTRING(x, 2, 1) END
+ CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 3, 1) + '%' THEN '' ELSE SUBSTRING(x, 3, 1) END
+ CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 4, 1) + '%' THEN '' ELSE SUBSTRING(x, 4, 1) END
+ CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 5, 1) + '%' THEN '' ELSE SUBSTRING(x, 5, 1) END
+ CASE WHEN a NOT LIKE '%' + SUBSTRING(x, 6, 1) + '%' THEN '' ELSE SUBSTRING(x, 6, 1) END
-- Keep adding rows until you reach the column size
AS stripped_column
FROM (SELECT
column_to_strip AS x
,'ABCDEFGHIJKLMNOPQRSTUVWXYZ' AS a
FROM my_table) a

这样做的好处是，有效字符包含在子查询中的一个字符串中，便于为不同的字符集重新配置。

缺点是您必须为每个字符添加一行SQL，直到您的列的大小。为了让这个任务更容易，我只是使用了下面的Powershell脚本，这个例子如果是VARCHAR(64):

1..64 | % {
"    + CASE WHEN a NOT LIKE '%' + SUBSTRING(x, {0}, 1) + '%' THEN '' ELSE SUBSTRING(x, {0}, 1) END" -f $_
} | clip.exe

-- 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;

下面是另一种使用iTVF删除非字母字符的方法。首先，需要一个基于模式的字符串分配器。下面这段话摘自Dwain Camp的<强> < / >强条:

-- PatternSplitCM will split a string based on a pattern of the form
-- supported by LIKE and PATINDEX
--
-- Created by: Chris Morris 12-Oct-2012
CREATE FUNCTION [dbo].[PatternSplitCM]
(
@List                VARCHAR(8000) = NULL
,@Pattern            VARCHAR(50)
) RETURNS TABLE WITH SCHEMABINDING
AS


RETURN
WITH numbers AS (
SELECT TOP(ISNULL(DATALENGTH(@List), 0))
n = ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
FROM
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) d (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) e (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) f (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) g (n)
)


SELECT
ItemNumber = ROW_NUMBER() OVER(ORDER BY MIN(n)),
Item = SUBSTRING(@List,MIN(n),1+MAX(n)-MIN(n)),
[Matched]
FROM (
SELECT n, y.[Matched], Grouper = n - ROW_NUMBER() OVER(ORDER BY y.[Matched],n)
FROM numbers
CROSS APPLY (
SELECT [Matched] = CASE WHEN SUBSTRING(@List,n,1) LIKE @Pattern THEN 1 ELSE 0 END
) y
) d
GROUP BY [Matched], Grouper

现在你有了一个基于模式的拆分器，你需要拆分匹配模式的字符串:

[a-z]

然后将它们连接起来以得到想要的结果:

SELECT *
FROM tbl t
CROSS APPLY(
SELECT Item + ''
FROM dbo.PatternSplitCM(t.str, '[a-z]')
WHERE Matched = 1
ORDER BY ItemNumber
FOR XML PATH('')
) x (a)

< a href = " http://sqlfiddle.com/ !6/10c31/1/0 rel =“nofollow”> < >强示例< /强> < / >

结果:

| Id |              str |              a |
|----|------------------|----------------|
|  1 |    test“te d'abc |     testtedabc |
|  2 |            anr¤a |           anra |
|  3 |  gs-re-C“te d'ab |     gsreCtedab |
|  4 |         M‚fe, DF |          MfeDF |
|  5 |           R™temd |          Rtemd |
|  6 |          ™jad”ji |          jadji |
|  7 |      Cje y ret¢n |       Cjeyretn |
|  8 |        J™kl™balu |        Jklbalu |
|  9 |       le“ne-iokd |       leneiokd |
| 10 |   liode-Pyr‚n‚ie |    liodePyrnie |
| 11 |         V„s G”ta |          VsGta |
| 12 |        Sƒo Paulo |        SoPaulo |
| 13 |  vAstra gAtaland | vAstragAtaland |
| 14 |  ¥uble / Bio-Bio |     ubleBioBio |
| 15 | U“pl™n/ds VAsb-y |    UplndsVAsby |

首先创建一个函数

CREATE FUNCTION [dbo].[GetNumericonly]
(@strAlphaNumeric VARCHAR(256))
RETURNS VARCHAR(256)
AS
BEGIN
DECLARE @intAlpha INT
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric)
BEGIN
WHILE @intAlpha > 0
BEGIN
SET @strAlphaNumeric = STUFF(@strAlphaNumeric, @intAlpha, 1, '' )
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric )
END
END
RETURN ISNULL(@strAlphaNumeric,0)
END

现在把这个函数叫做

select [dbo].[GetNumericonly]('Abhi12shek23jaiswal')

它的结果是

1223

DECLARE @vchVAlue NVARCHAR(255) = 'SWP, Lettering Position 1: 4 Ω, 2: 8 Ω, 3: 16 Ω, 4:  , 5:  , 6:  , Voltage Selector, Solder, 6, Step switch, : w/o fuseholder '




WHILE PATINDEX('%?%' , CAST(@vchVAlue AS VARCHAR(255))) > 0
BEGIN
SELECT @vchVAlue = STUFF(@vchVAlue,PATINDEX('%?%' , CAST(@vchVAlue AS VARCHAR(255))),1,' ')
END


SELECT @vchVAlue

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

去

信不信由你，在我的系统中，这个丑陋的函数比G masters的优雅函数表现得更好。

CREATE FUNCTION dbo.RemoveSpecialChar (@s VARCHAR(256))
RETURNS VARCHAR(256)
WITH SCHEMABINDING
BEGIN
IF @s IS NULL
RETURN NULL
DECLARE @s2 VARCHAR(256) = '',
@l INT = LEN(@s),
@p INT = 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
SET @s2 = @s2 + CHAR(@c)
SET @p = @p + 1
END


IF LEN(@s2) = 0
RETURN NULL


RETURN @s2

从性能角度来看，我会使用内联函数:

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[udf_RemoveNumericCharsFromString]
(
@List NVARCHAR(4000)
)
RETURNS TABLE
AS RETURN


WITH GetNums AS (
SELECT TOP(ISNULL(DATALENGTH(@List), 0))
n = ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
FROM
(VALUES (0),(0),(0),(0)) d (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) e (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) f (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) g (n)
)


SELECT StrOut = ''+
(SELECT Chr
FROM GetNums
CROSS APPLY (SELECT SUBSTRING(@List , n,1)) X(Chr)
WHERE Chr LIKE '%[^0-9]%'
ORDER BY N
FOR XML PATH (''),TYPE).value('.','NVARCHAR(MAX)')




/*How to Use
SELECT StrOut FROM dbo.udf_RemoveNumericCharsFromString ('vv45--9gut')
Result: vv--gut
*/

这是另一个递归CTE解决方案，基于@Gerhard Weiss的答案在这里。您应该能够将整个代码块复制并粘贴到SSMS中，并在那里使用它。结果包括一些额外的列，以帮助我们理解发生了什么。我花了一段时间才理解了PATINDEX (RegEx)和递归CTE的全部原理。

DECLARE @DefineBadCharPattern varchar(30)
SET @DefineBadCharPattern = '%[^A-z]%'  --Means anything NOT between A and z characters (according to ascii char value) is "bad"
SET @DefineBadCharPattern = '%[^a-z0-9]%'  --Means anything NOT between a and z characters or numbers 0 through 9 (according to ascii char value) are "bad"
SET @DefineBadCharPattern = '%[^ -~]%'  --Means anything NOT between space and ~ characters (all non-printable characters) is "bad"
--Change @ReplaceBadCharWith to '' to strip "bad" characters from string
--Change to some character if you want to 'see' what's being replaced. NOTE: It must be allowed accoring to @DefineBadCharPattern above
DECLARE @ReplaceBadCharWith varchar(1) = '#'  --Change this to whatever you want to replace non-printable chars with
IF patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, @ReplaceBadCharWith) > 0
BEGIN
RAISERROR('@ReplaceBadCharWith value (%s) must be a character allowed by PATINDEX pattern of %s',16,1,@ReplaceBadCharWith, @DefineBadCharPattern)
RETURN
END
--A table of values to play with:
DECLARE @temp TABLE (OriginalString varchar(100))
INSERT @temp SELECT ' 1hello' + char(13) + char(10) + 'there' + char(30) + char(9) + char(13) + char(10)
INSERT @temp SELECT '2hello' + char(30) + 'there' + char(30)
INSERT @temp SELECT ' 3hello there'
INSERT @temp SELECT ' tab' + char(9) + ' character'
INSERT @temp SELECT 'good bye'


--Let the magic begin:
;WITH recurse AS (
select
OriginalString,
OriginalString as CleanString,
patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString) as [Position],
substring(OriginalString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString),1) as [InvalidCharacter],
ascii(substring(OriginalString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString),1)) as [ASCIICode]
from @temp
UNION ALL
select
OriginalString,
CONVERT(varchar(100),REPLACE(CleanString,InvalidCharacter,@ReplaceBadCharWith)),
patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) as [Position],
substring(CleanString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString),1),
ascii(substring(CleanString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString),1))
from recurse
where patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) > 0
)
SELECT * FROM recurse
--optionally comment out this last WHERE clause to see more of what the recursion is doing:
WHERE patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) = 0

SQL Server 2017+的另一个可能选项，没有循环和/或递归，是使用TRANSLATE()和REPLACE()的基于字符串的方法。

t - sql声明:

DECLARE @pattern varchar(52) = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'


SELECT
v.[Text],
REPLACE(
TRANSLATE(
v.[Text],
REPLACE(TRANSLATE(v.[Text], @pattern, REPLICATE('a', LEN(@pattern))), 'a', ''),
REPLICATE('0', LEN(REPLACE(TRANSLATE(v.[Text], @pattern, REPLICATE('a', LEN(@pattern))), 'a', '')))
),
'0',
''
) AS AlphabeticCharacters
FROM (VALUES
('abc1234def5678ghi90jkl#@$&'),
('1234567890'),
('JAHDBESBN%*#*@*($E*sd55bn')
) v ([Text])

或作为一个函数:

CREATE FUNCTION dbo.RemoveNonAlphabeticCharacters (@Text varchar(1000))
RETURNS varchar(1000)
AS BEGIN


DECLARE @pattern varchar(52) = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
SET @text = REPLACE(
TRANSLATE(
@Text,
REPLACE(TRANSLATE(@Text, @pattern, REPLICATE('a', LEN(@pattern))), 'a', ''),
REPLICATE('0', LEN(REPLACE(TRANSLATE(@Text, @pattern, REPLICATE('a', LEN(@pattern))), 'a', '')))
),
'0',
''
)
   

RETURN @Text
END

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'