How Can I Sort A 'Version Number' Column Generically Using a SQL Server Query

Question

I wonder if the SQL geniuses amongst us could lend me a helping hand.

I have a column VersionNo in a table Versions that contains \'version

Answer 1

Another way to do it:

Assuming you only have a,b,c,d only you may as well separate the data out to columns and do an order by a,b,c,d(all desc) and get the top 1 row

If you need to scale to more than d to say e,f,g... just change 1,2,3,4, to 1,2,3,4,5,6,7 and so on in the query

Query : see demo

create table t (versionnumber varchar(255))
insert into t values
('1.0.0.505')
,('1.0.0.506')
,('1.0.0.507')
,('1.0.0.508')
,('1.0.0.509')
,('1.0.1.2')


; with cte as 
(
    select 
    column1=row_number() over (order by (select NULL)) ,
    column2=versionnumber
    from t
    )

select top 1
    CONCAT([1],'.',[2],'.',[3],'.',[4])
from 
(
    select 
        t.column1,
        split_values=SUBSTRING( t.column2, t1.N, ISNULL(NULLIF(CHARINDEX('.',t.column2,t1.N),0)-t1.N,8000)),
        r= row_number() over( partition by column1 order by t1.N) 
    from cte t 
        join
        (
            select 
                t.column2,
                1 as N 
            from cte t  
                UNION ALL
            select 
                t.column2,
                t1.N + 1 as N
            from cte t 
                join
                (
                 select 
                    top 8000
                        row_number() over(order by (select NULL)) as N 
                 from 
                    sys.objects s1 
                        cross join 
                   sys.objects s2 
                ) t1 
            on SUBSTRING(t.column2,t1.N,1) = '.'
         ) t1
          on t1.column2=t.column2
)a
pivot
( 
    max(split_values) for r in ([1],[2],[3],[4])
   )p
  order by [1] desc,[2] desc,[3] desc,[4] desc