如何从SQL Server中的字符串中删除所有非字母字符？

时间：2009-06-17 15:16:31

标签： sql-server parsing user-defined-functions alphanumeric alphabetic

如何删除字符串中不是字母的所有字符？

非字母数字怎么样？

这是必须是自定义功能还是还有更通用的解决方案？

21 个答案:

答案 0 :(得分：334)

尝试此功能：

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

这样称呼：

Select dbo.RemoveNonAlphaCharacters('abc1234def5678ghi90jkl')

一旦理解了代码，就应该看到更改代码以删除其他字符也相对简单。你甚至可以使这种动态足以传递你的搜索模式。

希望它有所帮助。

答案 1 :(得分：150)

G Mastros'awesome answer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

仅限字母：

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

仅限数字：

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

仅限字母数字：

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')

答案 2 :(得分：5)

信不信由你，在我的系统中，这个丑陋的功能比G Mastros优雅的功能更好。

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

答案 3 :(得分：4)

在查看了所有给定的解决方案后，我认为必须有一个不需要函数或CTE / XML查询的纯SQL方法，并且不涉及难以维护嵌套的REPLACE语句。这是我的解决方案：

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，可以轻松地为不同的字符集重新配置。

缺点是你必须为每个字符添加一行SQL，直到列的大小。为了使这个任务更容易，我只使用下面的Powershell脚本，这个例子对于VARCHAR（64）：

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

答案 4 :(得分：4)

我知道SQL在字符串操作方面很糟糕，但我认为这不会很困难。这是一个从字符串中去掉所有数字的简单函数。有更好的方法可以做到这一点，但这是一个开始。

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

<强>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

<强>输出 

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

我对读者的挑战：你能提高效率吗？使用递归怎么样？

答案 5 :(得分：4)

如果您像我一样，并且无法仅为生产数据添加功能但仍希望执行此类过滤，那么这里是使用PIVOT表进行过滤的纯SQL解决方案碎片又回来了。

N.B。我将表格硬编码为40个字符，如果您有更长的字符串需要过滤，则必须添加更多字符。

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

答案 6 :(得分：3)

以下是使用iTVF删除非字母字符的另一种方法。首先，您需要一个基于模式的字符串拆分器。这是一个取自Dwain Camp的article：

-- 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

现在您有了基于模式的拆分器，您需要拆分与模式匹配的字符串：

[a-z]

然后将它们连接起来以获得所需的结果：

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)

SAMPLE

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 |

答案 7 :(得分：2)

这是一种非常笨重的方式来拍摄你不想要的所有角色。问题是你必须指定你不想要的字符。如果一个新角色进入你，它将通过，除非你将它添加到列表中。

好处是您不必创建特殊功能。我没有写权限，所以这使我可以从简单的查询运行。 

REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
p.Name
,'®','')
,'©','')
,'ö','o')
,'ë','e')
,'ä','a')
,'ü','u')
,'ú','u')
,'í','i')
,'ï','i')
,'™','')
,'é','e')
,'²','2')
,'è','e')
,'—','-')
,'–','-')
,'ó','o')
,'•',' ')
,'…','.')
,'ô','o')
,'â','a')
,'á','a')
,'ê','e')
,'è','e')
,'’',' ')
,'·',' ')
,'à','a')
,'å','a')
,'ã','a')
,'’',' ')
,'a€s','as')
,'ø','o')
,'ñ','n')
,'î','i')
,'ç','c')
,'Ç','C')
,'Ã','A')
,'”','"')
,'“','"')
,'Á','A')
,'¢','c')
,'Ã','A')
,'Å','A')
,'¶','S')
,'×','x')
,'†','')
,'š','')
,'¤','')
,'µ','')
,'õ','')
,'€','')
,'‘','')
,'Õ','')
,'ð','')
,'Ò','')
,'¨','')
,'º','')
,'°','')
,'ì','')
,'ƒ','')
,'ÿ','')
,'ß','')
,'«','')
,'»','')
,'Æ','')
,'¬','')
,'Ù','')
,'ý','')
,'û','')
,'|','')
    as Name

答案 8 :(得分：2)

这个解决方案受到艾伦先生解决方案的启发，需要一个Numbers整数表（如果你想进行性能良好的严格查询操作，你应该有这个表）。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表达式以仅保留特定字符。

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 ...

答案 9 :(得分：2)

这是一个不需要创建函数或列出要替换的所有字符实例的解决方案。它使用递归WITH语句与PATINDEX结合来查找不需要的字符。它将替换列中所有不需要的字符 - 任何给定字符串中包含多达100个唯一的错误字符。 （EG“ABC123DEF234”将包含4个错误字符1,2,3和4）100限制是WITH语句中允许的最大递归数，但这不会对要处理的行数施加限制，仅受可用内存的限制 如果您不想要DISTINCT结果，可以从代码中删除这两个选项。

-- 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;

答案 10 :(得分：1)

- 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

现在调用此函数，如

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

其结果如

1223

答案 11 :(得分：1)

我把它放在两个叫PatIndex的地方。

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

用于RemoveNonAlphaCharacters上方的自定义函数并将其重命名为RemoveNonAlphaNumericCharacters

答案 12 :(得分：1)

从性能角度来看，我使用内联函数：

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
   */

答案 13 :(得分：1)

这是另一个递归CTE解决方案，基于@Gerhard Weiss的回答here。您应该能够将整个代码块复制并粘贴到SSMS中并在那里进行播放。结果包括一些额外的列，以帮助我们了解正在发生的事情。我花了一段时间，直到我理解了所有与PATINDEX（RegEx）和递归CTE相关的事情。

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

答案 14 :(得分：1)

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 语句：

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

答案 15 :(得分：1)

对于 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'

答案 16 :(得分：0)

使用CTE生成的数字表来检查每个字符，然后使用FOR XML连接到保留值的字符串，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

答案 17 :(得分：0)

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

答案 18 :(得分：0)

这种方式对我来说没有用，因为我试图保留阿拉伯字母，我试图取代正则表达式，但它也没有用。我写了另一种方法来处理ASCII级别，因为它是我唯一的选择而且它有效。

 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

答案 19 :(得分：-1)

虽然帖子有点旧，但我想说以下内容。 我在上面的解决方案中遇到的问题是它没有过滤掉像ç，ë，ï等字符。我调整了一个函数如下（我只使用了一个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

答案 20 :(得分：-3)

我刚刚发现它内置于Oracle 10g中，如果这是您正在使用的内容。我不得不将所有特殊字符去掉，以进行电话号码比较。

regexp_replace(c.phone, '[^0-9]', '')
相关问题
最新问题