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+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
其他回答
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;
这是另一个递归CTE解决方案,基于@Gerhard Weiss的回答。您应该能够将整个代码块复制并粘贴到SSMS中,并在那里使用它。结果包括一些额外的列,以帮助我们理解发生了什么。我花了一段时间才理解了PATINDEX (RegEx)和递归CTE的全部原理。
DECLARE @DefineBadCharPattern varchar(30)
SET @DefineBadCharPattern = '%[^A-z]%' --Means anything NOT between A and z characters (according to ascii char value) is "bad"
SET @DefineBadCharPattern = '%[^a-z0-9]%' --Means anything NOT between a and z characters or numbers 0 through 9 (according to ascii char value) are "bad"
SET @DefineBadCharPattern = '%[^ -~]%' --Means anything NOT between space and ~ characters (all non-printable characters) is "bad"
--Change @ReplaceBadCharWith to '' to strip "bad" characters from string
--Change to some character if you want to 'see' what's being replaced. NOTE: It must be allowed accoring to @DefineBadCharPattern above
DECLARE @ReplaceBadCharWith varchar(1) = '#' --Change this to whatever you want to replace non-printable chars with
IF patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, @ReplaceBadCharWith) > 0
BEGIN
RAISERROR('@ReplaceBadCharWith value (%s) must be a character allowed by PATINDEX pattern of %s',16,1,@ReplaceBadCharWith, @DefineBadCharPattern)
RETURN
END
--A table of values to play with:
DECLARE @temp TABLE (OriginalString varchar(100))
INSERT @temp SELECT ' 1hello' + char(13) + char(10) + 'there' + char(30) + char(9) + char(13) + char(10)
INSERT @temp SELECT '2hello' + char(30) + 'there' + char(30)
INSERT @temp SELECT ' 3hello there'
INSERT @temp SELECT ' tab' + char(9) + ' character'
INSERT @temp SELECT 'good bye'
--Let the magic begin:
;WITH recurse AS (
select
OriginalString,
OriginalString as CleanString,
patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString) as [Position],
substring(OriginalString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString),1) as [InvalidCharacter],
ascii(substring(OriginalString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN, OriginalString),1)) as [ASCIICode]
from @temp
UNION ALL
select
OriginalString,
CONVERT(varchar(100),REPLACE(CleanString,InvalidCharacter,@ReplaceBadCharWith)),
patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) as [Position],
substring(CleanString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString),1),
ascii(substring(CleanString,patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString),1))
from recurse
where patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) > 0
)
SELECT * FROM recurse
--optionally comment out this last WHERE clause to see more of what the recursion is doing:
WHERE patindex(@DefineBadCharPattern COLLATE Latin1_General_BIN,CleanString) = 0
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
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
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