SQL Server 2008 - order by strings with number numerically

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

(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.

SQL Server 2008 - order by strings with number numerically

Related

Recent Posts