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]', '')

虽然这篇文章有点老了，但我想说以下几点。 我有上述解决方案的问题是，它没有过滤出字符，如ç， ë， ï等。我调整了一个函数如下(我只使用80 varchar字符串来节省内存):

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

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 |

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]', '')

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

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