TrimEnd Equivalent in SQL Server - sql

Question

I have column (Numbers) which has values as follows:
1,2,3
1,2,3,
1,2,3,,,
1,2,3,,,,,,
I want to Trim all the Commas at the end of string, So that result would be
1,2,3
1,2,3
1,2,3
1,2,3
I have tried below Query but by this we can remove only one last comma
DECLARE #String as VARCHAR(50)
SET #String='1,2,3,4,,,,,,,,,,,,,,,,'
SELECT CASE WHEN right(rtrim(#String),1) = ',' then substring(rtrim(#String),1,len(rtrim(#String))-1)
ELSE #String
END AS TruncString
How can I remove all the commas at the end of string?

Answer 1

You can do this using:
LEFT(Numbers, LEN(Numbers) - (PATINDEX('%[^,]%', REVERSE(Numbers)) - 1))
The premise of this is you first reverse the string using REVERSE:
REVERSE(Numbers) --> ,,,,,,3,2,1
You then find the position of the first character that is not a comma using PATINDEX and the pattern match [^,]:
PATINDEX('%[^,]%', REVERSE(Numbers)) --> ,,,,,,3,2,1 = 7
Then you can use the length of the string using LEN, to get the inverse position, i.e. if the position of the first character that is not a comma is 7 in the reversed string, and the length of the string is 10, then you need the first 4 characters of the string. You then use SUBSTRING to extract the relevant part
A full example would be
SELECT Numbers,
Reversed = REVERSE(Numbers),
Position = PATINDEX('%[^,]%', REVERSE(Numbers)),
TrimEnd = LEFT(Numbers, LEN(Numbers) - (PATINDEX('%[^,]%', REVERSE(Numbers)) - 1))
FROM (VALUES
('1,2,3'),
('1,2,3,'),
('1,2,3,,,'),
('1,2,3,,,,,,'),
('1,2,3,,,5,,,'),
(',,1,2,3,,,5,,')
) t (Numbers);
EDIT
In response to an edit, that had some errors in the syntax, the below has functions to trim the start, and trim both sides of commas:
SELECT Numbers,
Reversed = REVERSE(Numbers),
Position = PATINDEX('%[^,]%', REVERSE(Numbers)),
TrimEnd = LEFT(Numbers, LEN(Numbers) - (PATINDEX('%[^,]%', REVERSE(Numbers)) - 1)),
TrimStart = SUBSTRING(Numbers, PATINDEX('%[^,]%', Numbers), LEN(Numbers)),
TrimBothSide = SUBSTRING(Numbers,
PATINDEX('%[^,]%', Numbers),
LEN(Numbers) -
(PATINDEX('%[^,]%', REVERSE(Numbers)) - 1) -
(PATINDEX('%[^,]%', Numbers) - 1)
)
FROM (VALUES
('1,2,3'),
('1,2,3,'),
('1,2,3,,,'),
('1,2,3,,,,,,'),
('1,2,3,,,5,,,'),
(',,1,2,3,,,5,,')
) t (Numbers);

Answer 2

Because there are multiple occurrences you can't do it with a simple builtin function expression, but a simple user defined function can do the job.
create function dbo.MyTrim(#text varchar(max)) returns varchar(max)
as
-- function to remove all commas from the right end of the input.
begin
while (right(#text, 1) = ','
begin
set #text = left(#text, len(#text) - 1)
end
return #text
end
go

Answer 3

You can search for the first occurrence of ',,' and take everything before that:
select (case when numbers like '%,,'
then left(numbers, charindex(',,', numbers) - 1)
when numbers like '%,'
then left(numbers, len(numbers) - 1)
else numbers
end)
Note: it would seem that you are storing lists of things in a comma-delimited string. It is usually better to store these using a junction table.
EDIT:
Or, an alternative way of formulating this without the case:
select left(numbers + ',,', charindex(',,', numbers + ',,') - 1)

Answer 4

SQL Server 2017 has implemented an enhanced version of TRIM function.
You can use TRIM(',' FROM '1,2,3,,,') to get the string, '1,2,3'

Answer 5

Run below query and get expected results
declare #sql varchar(500)
set #sql ='1,2,3,,,,,,'
select left(#sql,case charindex(',,',#sql,0)
when 0 then len(#sql)-1
else charindex(',,',#sql,0)-1
end)

Answer 6

Create FUNCTION TrimStartEndAll
(
#string varchar(max),
#trimValue varchar(5),
#removeall int=0
)
RETURNS varchar(max)
AS
BEGIN
if #removeall=1
while CHARINDEX(#trimValue,#string) >0 and #removeall=1
begin
set #string = REPLACE(#string,#trimValue,'')
end
if #removeall = 0
begin
while CHARINDEX(#trimValue,#string) =1
begin
set #string = SUBSTRING(#string,len(#trimValue)+1, len(#string))
end
while substring(#string,len(#string)-len(#trimValue)+1, len(#trimValue)) = #trimValue
begin
set #string =substring(#string,0, (len(#string)-len(#trimValue)+1))
end
end
return #string
END
GO
output
select dbo.TrimStartEndAll( ',,1,2,3,,,5,,,,,,,,,',',,',1) => 1,2,3,5,
select dbo.TrimStartEndAll( ',,1,2,3,,,5,,,,,,,,,',',,',0) => 1,2,3,,,5,