Remove the last character in a string in T-SQL?

问题:

How do I remove the last character in a string in T-SQL?

For example:

'TEST STRING'

to return:

'TEST STRIN'

回答1:

e.g.

DECLARE @String VARCHAR(100)
SET @String = 'TEST STRING'

-- Chop off the end character
SET @String = 
     CASE @String WHEN null THEN null 
     ELSE (
         CASE LEN(@String) WHEN 0 THEN @String 
            ELSE LEFT(@String, LEN(@String) - 1) 
         END 
     ) END


SELECT @String

回答2:

If for some reason your column logic is complex (case when ... then ... else ... end), then the above solutions causes you to have to repeat the same logic in the len() function. Duplicating the same logic becomes a mess. If this is the case then this is a solution worth noting. This example gets rid of the last unwanted comma. I finally found a use for the REVERSE function.

select reverse(stuff(reverse('a,b,c,d,'), 1, 1, ''))

回答3:

Try this:

select substring('test string', 1, (len('test string') - 1))

回答4:

If your string is empty,

DECLARE @String VARCHAR(100)
SET @String = ''
SELECT LEFT(@String, LEN(@String) - 1)

then this code will cause error message 'Invalid length parameter passed to the substring function.'

You can handle it this way:

SELECT LEFT(@String, NULLIF(LEN(@String)-1,-1))

It will always return result, and NULL in case of empty string.

回答5:

If your coloumn is text and not varchar, then you can use this:

SELECT SUBSTRING(@String, 1, NULLIF(DATALENGTH(@String)-1,-1))

回答6:

select left('TEST STRING', len('TEST STRING')-1)

回答7:

If you want to do this in two steps, rather than the three of REVERSE-STUFF-REVERSE, you can have your list separator be one or two spaces. Then use RTRIM to trim the trailing spaces, and REPLACE to replace the double spaces with ','

select REPLACE(RTRIM('a  b  c  d  '),'  ', ', ')

However, this is not a good idea if your original string can contain internal spaces.

Not sure about performance. Each REVERSE creates a new copy of the string, but STUFF is a third faster than REPLACE.

also see this

回答8:

@result = substring(@result, 1, (LEN(@result)-1))

回答9:

I can suggest this -hack- ;).

select 
    left(txt, abs(len(txt + ',') - 2))
from 
    t;

SQL Server Fiddle Demo

回答10:

you can create function

CREATE FUNCTION [dbo].[TRUNCRIGHT] (@string NVARCHAR(max), @len int = 1)
RETURNS NVARCHAR(max)
AS
BEGIN
    IF LEN(@string)<@len
        RETURN ''
    RETURN LEFT(@string, LEN(@string) - @len)
END

回答11:

Try this

DECLARE @String VARCHAR(100)
SET @String = 'TEST STRING'
SELECT LEFT(@String, LEN(@String) - 1) AS MyTrimmedColumn

回答12:

My answer is similar to the accepted answer, but it also check for Null and Empty String.

DECLARE @String VARCHAR(100)

SET @String = 'asdfsdf1'

-- If string is null return null, else if string is empty return as it is, else chop off the end character
SET @String = Case @String when null then null else (case LEN(@String) when 0 then @String else LEFT(@String, LEN(@String) - 1) end ) end

SELECT @String

回答13:

Get the last character

Right(@string, len(@String) - (len(@String) - 1))

回答14:

I love @bill-hoenig 's answer; however, I was using a subquery and I got caught up because the REVERSE function needed two sets of parentheses. Took me a while to figure that one out!

SELECT
   -- Return comma delimited list of all payment reasons for this Visit
   REVERSE(STUFF(REVERSE((
        SELECT DISTINCT
               CAST(CONVERT(varchar, r1.CodeID) + ' - ' + c.Name + ', ' AS VARCHAR(MAX))
          FROM VisitReason r1
          LEFT JOIN ReasonCode c        ON c.ID = r1.ReasonCodeID
         WHERE p.ID = r1.PaymentID
         FOR XML PATH('')
              )), 1, 2, ''))                        ReasonCode
  FROM Payments p

回答15:

To update the record by trimming the last N characters of a particular column:

UPDATE tablename SET columnName = LEFT(columnName , LEN(columnName )-N) where clause

回答16:

Try It :

  DECLARE @String NVARCHAR(100)
    SET @String = '12354851'
    SELECT LEFT(@String, NULLIF(LEN(@String)-1,-1))

回答17:

declare @string varchar(20)= 'TEST STRING'
Select left(@string, len(@string)-1) as Tada

output:

Tada
--------------------
TEST STRIN

回答18:

Try this,

DECLARE @name NVARCHAR(MAX) SET @name='xxxxTHAMIZHMANI****'SELECT Substring(@name, 5, (len(@name)-8)) as UserNames

And the output will be like, THAMIZHMANI

回答19:

declare @x varchar(20),@y varchar(20)
select @x='sam'
select 
case when @x is null then @y
      when @y is null then @x
      else @x+','+@y
end


go

declare @x varchar(20),@y varchar(20)
select @x='sam'
--,@y='john'
DECLARE @listStr VARCHAR(MAX)   

SELECT @listStr = COALESCE(@x + ', ' ,'') +coalesce(@y+',','')
SELECT left(@listStr,len(@listStr)-1)