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?
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
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
Here's a solution that doesn't require creating a function or listing all instances of characters to replace. It uses a recursive WITH statement in combination with a PATINDEX to find unwanted chars. It will replace all unwanted chars in a column - up to 100 unique bad characters contained in any given string. (E.G. "ABC123DEF234" would contain 4 bad characters 1, 2, 3 and 4) The 100 limit is the maximum number of recursions allowed in a WITH statement, but this doesn't impose a limit on the number of rows to process, which is only limited by the memory available. If you don't want DISTINCT results, you can remove the two options from the code.
-- 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;
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
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'
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