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
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;
其他回答
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
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
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
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
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;
