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
如果您像我一样,不能仅向生产数据添加函数,但仍然想执行这种过滤,那么这里有一个纯SQL解决方案,使用PIVOT表将过滤后的部分重新组合在一起。
注意:我硬编码表高达40个字符,如果你有更长的字符串要过滤,你将不得不添加更多。
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
其他回答
乔治·马斯特罗斯精彩回答的参数化版本:
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
字母只有:
SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^a-z')
数字只有:
SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^0-9')
字母数字只有:
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')
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
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 |
SQL Server 2017+的另一个可能的选项,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声明:
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])
或作为一个函数:
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
使用CTE生成的数字表来检查每个字符,然后FOR XML连接到一个保留值的字符串,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