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
从性能角度来看,我会使用内联函数:
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
*/
其他回答
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
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
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 $_
} | clip.exe
SQL Server 2017+的另一个可能的选项,没有循环和/或递归,是使用TRANSLATE()和REPLACE()的基于字符串的方法。
t - sql声明:
DECLARE @pattern varchar(52) = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
SELECT
v.[Text],
REPLACE(
TRANSLATE(
v.[Text],
REPLACE(TRANSLATE(v.[Text], @pattern, REPLICATE('a', LEN(@pattern))), 'a', ''),
REPLICATE('0', LEN(REPLACE(TRANSLATE(v.[Text], @pattern, REPLICATE('a', LEN(@pattern))), 'a', '')))
),
'0',
''
) AS AlphabeticCharacters
FROM (VALUES
('abc1234def5678ghi90jkl#@$&'),
('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
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