问题
I have the following string: 'BOB*', how do I trim the * so it shows up as 'BOB'
I tried the RTRIM('BOB*','*') but does not work as says needs only 1 parameter.
回答1:
LEFT('BOB*', LEN('BOB*')-1)
should do it.
回答2:
Another pretty good way to implement Oracle's TRIM char FROM string in MS SQL Server is the following:
- First, you need to identify a char that will never be used in your string, for example
~ - You replace all spaces with that character
- You replace the character
*you want to trim with a space - You
LTrim+RTrimthe obtained string - You replace back all spaces with the trimmed character
* - You replace back all never-used characters with a space
For example:
REPLACE(REPLACE(LTrim(RTrim(REPLACE(REPLACE(string,' ','~'),'*',' '))),' ','*'),'~',' ')
回答3:
CREATE FUNCTION dbo.TrimCharacter
(
@Value NVARCHAR(4000),
@CharacterToTrim NVARCHAR(1)
)
RETURNS NVARCHAR(4000)
AS
BEGIN
SET @Value = LTRIM(RTRIM(@Value))
SET @Value = REVERSE(SUBSTRING(@Value, PATINDEX('%[^'+@CharacterToTrim+']%', @Value), LEN(@Value)))
SET @Value = REVERSE(SUBSTRING(@Value, PATINDEX('%[^'+@CharacterToTrim+']%', @Value), LEN(@Value)))
RETURN @Value
END
GO
--- Example
----- SELECT dbo.TrimCharacter('***BOB*********', '*')
----- returns 'BOB'
回答4:
If you want to remove all asterisks then it's obvious:
SELECT REPLACE('Hello*', '*', '')
However, If you have more than one asterisk at the end and multiple throughout, but are only interested in trimming the trailing ones, then I'd use this:
DECLARE @String VarChar(50) = '**H*i****'
SELECT LEFT(@String, LEN(REPLACE(@String, '*', ' '))) --Returns: **H*i
I updated this answer to include show how to remove leading characters:
SELECT RIGHT(@String, LEN(REPLACE(REVERSE(@String), '*', ' '))) --Returns: H*i****
LEN() has a "feature" (that looks a lot like a bug) where it does not count trailing spaces.
回答5:
If you wanted behavior similar to how RTRIM handles spaces i.e. that "B*O*B**" would turn into "B*O*B" without losing the embedded ones then something like -
REVERSE(SUBSTRING(REVERSE('B*O*B**'), PATINDEX('%[^*]%',REVERSE('B*O*B**')), LEN('B*O*B**') - PATINDEX('%[^*]%', REVERSE('B*O*B**')) + 1))
Should do it.
回答6:
If you only want to remove a single '*' character from the value when the value ends with a '*', a simple CASE expression will do that for you:
SELECT CASE WHEN RIGHT(foo,1) = '*' THEN LEFT(foo,LEN(foo)-1) ELSE foo END AS foo
FROM (SELECT 'BOB*' AS foo)
To remove all trailing '*' characters, then you'd need a more complex expression, making use of the REVERSE, PATINDEX, LEN and LEFT functions.
NOTE: Be careful with the REPLACE function, as that will replace all occurrences of the specified character within the string, not just the trailing ones.
回答7:
RRIM() LTRIM() only remove spaces try http://msdn.microsoft.com/en-us/library/ms186862.aspx
Basically just replace the * with empty space
REPLACE('TextWithCharacterToReplace','CharacterToReplace','CharacterToReplaceWith')
So you want
REPLACE ('BOB*','*','')
回答8:
How about.. (in this case to trim off trailing comma or period)
For a variable:
-- Trim commas and full stops from end of City
WHILE RIGHT(@CITY, 1) IN (',', '.'))
SET @CITY = LEFT(@CITY, LEN(@CITY)-1)
For table values:
-- Trim commas and full stops from end of City
WHILE EXISTS (SELECT 1 FROM [sap_out_address] WHERE RIGHT([CITY], 1) IN (',', '.'))
UPDATE [sap_out_address]
SET [CITY] = LEFT([CITY], LEN([CITY])-1)
WHERE RIGHT([CITY], 1) IN (',', '.')
回答9:
I really like Teejay's answer, and almost stopped there. It's clever, but I got the "almost too clever" feeling, as, somehow, your string at some point will actually have a ~ (or whatever) in it on purpose. So that's not defensive enough for me to put into production.
I like Chris' too, but the PATINDEX call seems like overkill.
Though it's probably a micro-optimization, here's one without PATINDEX:
CREATE FUNCTION dbo.TRIMMIT(@stringToTrim NVARCHAR(MAX), @charToTrim NCHAR(1))
RETURNS NVARCHAR(MAX)
AS
BEGIN
DECLARE @retVal NVARCHAR(MAX)
SET @retVal = @stringToTrim
WHILE 1 = charindex(@charToTrim, reverse(@retVal))
SET @retVal = SUBSTRING(@retVal,0,LEN(@retVal))
WHILE 1 = charindex(@charToTrim, @retVal)
SET @retVal = SUBSTRING(@retVal,2,LEN(@retVal))
RETURN @retVal
END
--select dbo.TRIMMIT('\\trim\asdfds\\\', '\')
--trim\asdfds
Returning a MAX nvarchar bugs me a little, but that's the most flexible way to do this..
回答10:
Trim with many cases
--id = 100 101 102 103 104 105 106 107 108 109 110 111
select right(id,2)+1 from ordertbl -- 1 2 3 4 5 6 7 8 9 10 11 -- last two positions are taken
select LEFT('BOB', LEN('BOB')-1) -- BO
select LEFT('BOB*',1) --B
select LEFT('BOB*',2) --BO
回答11:
I've used a similar approach to some of the above answers of using pattern matching and reversing the string to find the first non-trimmable character, then cutting that off. The difference is this version does less work than those above, so should be a little more efficient.
- This creates
RTRIMfunctionality for any specified character. - It includes an additional step
set @charToFind = case...to escape the chosen character. - There is currently an issue if
@charToReplaceis a right crotchet (]) as there appears to be no way to escape this.
.
declare @stringToSearch nvarchar(max) = '****this is****a ** demo*****'
, @charToFind nvarchar(5) = '*'
--escape @charToFind so it doesn't break our pattern matching
set @charToFind = case @charToFind
when ']' then '[]]' --*this does not work / can't find any info on escaping right crotchet*
when '^' then '\^'
--when '%' then '%' --doesn't require escaping in this context
--when '[' then '[' --doesn't require escaping in this context
--when '_' then '_' --doesn't require escaping in this context
else @charToFind
end
select @stringToSearch
, left
(
@stringToSearch
,1
+ len(@stringToSearch)
- patindex('%[^' + @charToFind + ']%',reverse(@stringToSearch))
)
回答12:
Try this:
Original
select replace('BOB*','*','')
Fixed to be an exact replacement
select replace('BOB*','BOB*','BOB')
回答13:
Solution for one char parameter:
rtrim('0000100','0') -> select left('0000100',len(rtrim(replace('0000100','0',' '))))
ltrim('0000100','0') -> select right('0000100',len(replace(ltrim(replace('0000100','0',' ')),' ','.')))
回答14:
SqlServer2017 has a new way to do it: https://docs.microsoft.com/en-us/sql/t-sql/functions/trim-transact-sql?view=sql-server-2017
SELECT TRIM('0' FROM '00001900'); -> 19
SELECT TRIM( '.,! ' FROM '# test .'); -> # test
SELECT TRIM('*' FROM 'BOB*'); --> BOB
Unfortunately, RTRIM does not support trimming a specific character.
回答15:
@teejay solution is great. But the code below can be more understandable:
declare @X nvarchar(max)='BOB *'
set @X=replace(@X,' ','^')
set @X=replace(@X,'*',' ')
set @X= ltrim(rtrim(@X))
set @X=replace(@X,'^',' ')
回答16:
An other approach ONLY if you want to remove leading and trailing characters is the use of TRIM function. By default removes white spaces but have te avility of remove other characters if you specify its.
SELECT TRIM('=' FROM '=SPECIALS=') AS Result;
Result
--------
SPECIALS
Unfortunately LTRIM and RTRIM does not work in the same way and only removes white spaces instead of specified characters like TRIM does if you specify its.
Reference and more examples: https://database.guide/how-to-remove-leading-and-trailing-characters-in-sql-server/
来源：https://stackoverflow.com/questions/7838676/sql-server-trim-character