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

下面是使用iTVF删除非字母字符的另一种方法。首先，需要一个基于模式的字符串分配器。以下是Dwain Camp文章中的一段:

-- 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)

样本

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

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
   */

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

SQL Server 2017+的另一个可能的选项，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