Getting two rows after nth row for each user - sql

Question

My table structure:
consumer_id, signup_date, plan_id, subscription_date
It has multiple subscription_dates for the same consumer_id.
I wish to get results only for those users who have atleast two rows of data
For each user I need to get a result which gives me the top two rows ordered by subscription_date..
Then I want another set of result of all users who have atleast three rows of data..
For each user I need to get a result which gives me the top second and third rows..
and so on...
I have a feeling its something similar to this but could not get it to work in my case..
Update:
Sample table data:
1 1/1/2015 1 3/1/2015
2 1/1/2015 1 3/1/2015
2 1/1/2015 1 4/1/2015
3 1/1/2015 1 6/1/2015
2 1/1/2015 1 6/1/2015
3 1/1/2015 1 7/1/2015
Sample Output1:
2 1/1/2015 1 3/1/2015
2 1/1/2015 1 4/1/2015
3 1/1/2015 1 6/1/2015
3 1/1/2015 1 7/1/2015
Sample Output2:
2 1/1/2015 1 4/1/2015
2 1/1/2015 1 6/1/2015

Answer 1

This is should give you answer to your first part,But to get entire answer please let us know more details of the query (will be great if we have some output)
Note: I am assuming #test as your main table.
select * from
(
SELECT a.consumer_id, a.signup_date, a.plan_id, a.subscription_date
,RANK() OVER
(PARTITION BY a.consumer_id ORDER BY a.subscription_date ASC) AS Rank1
FROM #test a
where a.consumer_id in
(
select consumer_id as count from #test
group by consumer_id
having count(consumer_id)>=2
)
) as b
where b.Rank1<=2
For the second part.
select * from #test
select * from
(
SELECT a.consumer_id, a.signup_date, a.plan_id, a.subscription_date
,RANK() OVER
(PARTITION BY a.consumer_id ORDER BY a.subscription_date ASC) AS Rank1
FROM #test a
where a.consumer_id in
(
select consumer_id as count from #test
group by consumer_id
having count(consumer_id)>=3
)
) as b
where b.Rank1 between 2 and 3