There was a question from the op regarding adding a new column to a query output by generating the cyclic sequence numbers from 1 to 3.
Select A =identity(int,1,1),B,C from table_abs
For this output, the 4th column generates the Sequence of numbers from 1 to 3 which is shown below
If you are using SQL 2012 then Sequence would be natural choice for any such operations.
CREATE SEQUENCE Seq AS INTEGER START WITH 1 INCREMENT BY 1 MINVALUE 1 MAXVALUE 3 CYCLE; SELECT table_name,NEXT VALUE FOR Seq New_column FROM information_schema.tables
Using CTE and Modulus operator
;with q as ( select row_number() over (order by (select null)) A, * from sys.objects ) select A, 1+A%3 B, * from q
Loops and Temp table
create table dummyTest ( id int, col1 char(1), col2 int ) insert into dummyTest values(1,'A',410),(2,'B',411),(3,'c',4111),(4,'d',421),(5,'e',441),(6,'f',451),(7,'g',481),(8,'h',401) create table #dummy ( id int, col1 char(1), col2 int, NewColumn int ) declare @n int,@i int,@limit int set @limit=1 set @i=1 select @n=count(*) from dummyTest while @i<=@n begin set @limit=1 while @limit<=3 begin print @limit insert into #dummy select *,NewColumn=@limit from dummyTest where id=@i set @i=@i+1 set @limit=@limit+1 end end select * from #dummy
The simplest of all the above methods
create table dummyTest
insert into dummyTest values(1,’A’,410),(2,’B’,411),(3,’c’,4111),(4,’d’,421),(5,’e’,441),(6,’f’,451),(7,’g’,481),(8,’h’,401)
select id,col1,col2, (row_number() over (order by id)-1) %4+1 as cycle from dummyTest
The same solution can be derived using the cursor and there may be other solutions as well. At many instances, we opt for any solutions without thinking of data volume that may degrade the performance. This is one of the prime examples of why we need to upgrade to newer version.