SQL Server 2008 - order by strings with number numerically

Question

I have following values in my table:

ABC
ABC1
ABC2
ABC3 and so on...

ABC11
ABC12
ABC13 and so on..

ABC20
ABC21
ABC22 and so on..

So basic

Answer 1

You could adapt the function RemoveNonAlphaCharacters in this answer to filter out everything except numbers, and then use an ORDER BY using that function.

Answer 2

In order by statement, prepend enough zeros with when value contains any number in it to make all alphanumerica value same length

SELECT ColName
FROM TableName
ORDER BY 
 CASE WHEN ColName like '%[0-9]%' 
 THEN Replicate('0', 100 - Len(ColName)) + ColName
 ELSE ColName  END

Answer 3

You could remove the first three characters and cast the rest to int

SELECT Value,
       Num=CAST(RIGHT(Value, LEN(Value) - 3) AS int)
FROM dbo.TableName
ORDER BY Num

Demo

Answer 4

(based on answers from @shenhengbin and @EchO to this question)

The following is what I call a "clean hack". Assuming you are ordering on column Col1:

ORDER BY LEN(Col1), Col1

It is a hack, although I'd personally feel proud using it.

Answer 5

You can do it using PATINDEX() function like below :

select * from Test 
order by CAST(SUBSTRING(Name + '0', PATINDEX('%[0-9]%', Name + '0'), LEN(Name + '0')) AS INT)

SQL Fiddle Demo

If you have numbers in middle of the string then you need to create small user defined function to get number from string and sort data based on that number like below :

CREATE FUNCTION dbo.fnGetNumberFromString (@strInput VARCHAR(255)) 
RETURNS VARCHAR(255) 
AS 
BEGIN 
    DECLARE @intNumber int 
    SET @intNumber = PATINDEX('%[^0-9]%', @strInput)

    WHILE @intNumber > 0
    BEGIN 
        SET @strInput = STUFF(@strInput, @intNumber, 1, '')
        SET @intNumber = PATINDEX('%[^0-9]%', @strInput)
    END 

    RETURN ISNULL(@strInput,0) 
END 
GO

You can sort data by :

select Name from Test order by dbo.fnGetNumberFromString(Name), Name