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
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)='''.,!?/<>"[]{}|`~@#$%^&*()-+=/\:;'+char(13)+char(10)
-- interim replaced with
declare @replace_with char(1)=' '
-- call the translate function that will change unwanted characters to spaces
-- in this sample
declare @translated varchar(max)
select @translated=TRANSLATE(@text,@unwanted,REPLICATE(@replace_with,len(@unwanted)))
-- In this case, I want to preserve one space
select string_agg(trim(value),' ')
from STRING_SPLIT(@translated,' ')
where trim(value)<>''
-- Result
'Lorem ipsum dolor sit amet consectetur adipiscing elit Vivamus commodo justo metus sed facilisis ante congue eget Proin ac bibendum sem'
其他回答
乔治·马斯特罗斯精彩回答的参数化版本:
CREATE FUNCTION [dbo].[fn_StripCharacters]
(
@String NVARCHAR(MAX),
@MatchExpression VARCHAR(255)
)
RETURNS NVARCHAR(MAX)
AS
BEGIN
SET @MatchExpression = '%['+@MatchExpression+']%'
WHILE PatIndex(@MatchExpression, @String) > 0
SET @String = Stuff(@String, PatIndex(@MatchExpression, @String), 1, '')
RETURN @String
END
字母只有:
SELECT dbo.fn_StripCharacters('a1!s2@d3#f4$', '^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')
信不信由你,在我的系统中,这个丑陋的函数比G masters的优雅函数表现得更好。
CREATE FUNCTION dbo.RemoveSpecialChar (@s VARCHAR(256))
RETURNS VARCHAR(256)
WITH SCHEMABINDING
BEGIN
IF @s IS NULL
RETURN NULL
DECLARE @s2 VARCHAR(256) = '',
@l INT = LEN(@s),
@p INT = 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
SET @s2 = @s2 + CHAR(@c)
SET @p = @p + 1
END
IF LEN(@s2) = 0
RETURN NULL
RETURN @s2
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。
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
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]', '')