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

CREATE FUNCTION remove_spc_char(@str VARCHAR(MAX))
  RETURNS VARCHAR(MAX) 
AS
BEGIN
  DECLARE @resp    VARCHAR(MAX) = '';
  DECLARE @str_val   VARCHAR(MAX) = UPPER(@str);
  DECLARE @i       INTEGER= 1;
  DECLARE @v_asc   INTEGER;
   WHILE @i <= (LEN(@str_val))
   BEGIN
     SET @v_asc = (ASCII(SUBSTRING(@str_val, @i, 1))) 
        BEGIN
        IF @v_asc in (192,193,194,195,196,65) 
            begin
                SET @v_asc = 65;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (200,201,202,203,233,69)
            begin
                SET @v_asc = 69;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (204,205,206,207,296,73)
            begin
                SET @v_asc = 73;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (210,211,212,213,214,79)
            begin
                SET @v_asc = 79;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (217,218,219,220,85)
            begin
                SET @v_asc = 85;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (199,231,67)
            begin
                SET @v_asc = 67;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (209,78)
            begin
                SET @v_asc = 78;
                SET @resp = concat(@resp, CHAR(@v_asc));
            end;
        IF @v_asc in (924,181,358,216,222,330,272,208,198,42,37,38,34,36,35,
64,33,39,41,40,43,61,95,45,62,60,63,47,176,183,124,166,174,359,248,254,
180,170,186,126,312,331,273,172,178,179,163,162,123,91,93,125,92,167,240,
223,230,171,187,169,185,168)
            begin
                SET @resp = concat(@resp, '');
            end;
        ELSE 
            begin
                if @v_asc not in (65,67,69,73,78,79,85)
                begin
                    SET @resp = concat(@resp, CHAR(@v_asc));
                end;
            end;
        END;
      SET @i = @i + 1
    END;
    RETURN @resp;
END;

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

这是另一个递归CTE解决方案，基于@Gerhard Weiss的回答。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

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

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]', '')