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
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
其他回答
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。
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;
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 |
看过所有给出的解决方案后,我认为必须有一个纯SQL方法,它不需要函数或CTE / XML查询,并且不涉及难以维护的嵌套REPLACE语句。以下是我的解决方案:
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,便于为不同的字符集重新配置。
缺点是您必须为每个字符添加一行SQL,直到您的列的大小。为了让这个任务更容易,我只是使用了下面的Powershell脚本,这个例子如果是VARCHAR(64):
1..64 | % {
" + CASE WHEN a NOT LIKE '%' + SUBSTRING(x, {0}, 1) + '%' THEN '' ELSE SUBSTRING(x, {0}, 1) END" -f $_
} | clip.exe