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
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'
这个解决方案受到Allen先生的解决方案的启发,需要一个整数的Numbers表(如果您想进行具有良好性能的严肃查询操作,您应该手头有这个表)。它不需要CTE。您可以更改NOT IN(…)表达式以排除特定字符,或将其更改为IN(…)只保留某些字符的OR LIKE表达式。
SELECT (
SELECT SUBSTRING([YourString], N, 1)
FROM dbo.Numbers
WHERE N > 0 AND N <= CONVERT(INT, LEN([YourString]))
AND SUBSTRING([YourString], N, 1) NOT IN ('(',')',',','.')
FOR XML PATH('')
) AS [YourStringTransformed]
FROM ...
CREATE FUNCTION remove_spc_char(@str VARCHAR(MAX))
RETURNS VARCHAR(MAX)
AS
BEGIN
DECLARE @resp VARCHAR(MAX) = '';
DECLARE @str_val VARCHAR(MAX) = UPPER(@str);
DECLARE @i INTEGER= 1;
DECLARE @v_asc INTEGER;
WHILE @i <= (LEN(@str_val))
BEGIN
SET @v_asc = (ASCII(SUBSTRING(@str_val, @i, 1)))
BEGIN
IF @v_asc in (192,193,194,195,196,65)
begin
SET @v_asc = 65;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (200,201,202,203,233,69)
begin
SET @v_asc = 69;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (204,205,206,207,296,73)
begin
SET @v_asc = 73;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (210,211,212,213,214,79)
begin
SET @v_asc = 79;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (217,218,219,220,85)
begin
SET @v_asc = 85;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (199,231,67)
begin
SET @v_asc = 67;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (209,78)
begin
SET @v_asc = 78;
SET @resp = concat(@resp, CHAR(@v_asc));
end;
IF @v_asc in (924,181,358,216,222,330,272,208,198,42,37,38,34,36,35,
64,33,39,41,40,43,61,95,45,62,60,63,47,176,183,124,166,174,359,248,254,
180,170,186,126,312,331,273,172,178,179,163,162,123,91,93,125,92,167,240,
223,230,171,187,169,185,168)
begin
SET @resp = concat(@resp, '');
end;
ELSE
begin
if @v_asc not in (65,67,69,73,78,79,85)
begin
SET @resp = concat(@resp, CHAR(@v_asc));
end;
end;
END;
SET @i = @i + 1
END;
RETURN @resp;
END;
首先创建一个函数
CREATE FUNCTION [dbo].[GetNumericonly]
(@strAlphaNumeric VARCHAR(256))
RETURNS VARCHAR(256)
AS
BEGIN
DECLARE @intAlpha INT
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric)
BEGIN
WHILE @intAlpha > 0
BEGIN
SET @strAlphaNumeric = STUFF(@strAlphaNumeric, @intAlpha, 1, '' )
SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric )
END
END
RETURN ISNULL(@strAlphaNumeric,0)
END
现在把这个函数叫做
select [dbo].[GetNumericonly]('Abhi12shek23jaiswal')
它的结果是
1223
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]', '')