如何从字符串中删除所有非字母的字符?

非字母数字呢?

这必须是一个自定义函数还是也有更通用的解决方案?


当前回答

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 |

其他回答

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'

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

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]', '')

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。

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 |