TrimEnd Equivalent in SQL Server

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)

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

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);