Selecting the second result for each client...

Question

Hi,
I have a table containing all the contacts we had with our clients.
We consider a file an active file from the moment there have been 2 contacts for that client.
Today somebody asked me "can you tell me how many new clients with an active file we have this year"
So what I actualy want to do now, is make a view containing the DATE of the SECOND CONTACT for each CLIENT_ID.
the table looks like this:
CONTACT_ID
CLIENT_ID
DATE
CONTACT_FORM (doesn't matter in this story)
REMARKS (doesn't matter in this story)
Can somebody assist me with this please?
Thanks in advance for your replies.
L
edit I do know how to make a view, that's not the problem :) in this case only the select query to make it is the problem.
edit 2 I made a little example
If I start with this table data:
CLIENT_ID || DATE           .... other columns here...
436     || 01-02-2011
328     || 05-04-2011
319     || 13-01-2011
220     || 03-09-2011
319     || 17-12-2011
220     || 25-11-2011
436     || 01-04-2011
436     || 01-04-2011than the view would have to make me end up with:
CLIENT_ID  || DATE_OF_SECOND_CONTACT
436     || 01-04-2011
220     || 25-11-2011
319     || 17-12-2011(working with version 11.2.0.1.0)
Edited by: 901981 on 13-dec-2011 2:16

Answer 1

I hope this helps:
with t as
select 436 client_id, to_date('01-02-2011','DD-MM-YYYY') d from dual union all
select 328 client_id, to_date('05-04-2011','DD-MM-YYYY') d from dual union all
select 319 client_id, to_date('13-01-2011','DD-MM-YYYY') d from dual union all
select 220 client_id, to_date('03-09-2011','DD-MM-YYYY') d from dual union all
select 319 client_id, to_date('17-12-2011','DD-MM-YYYY') d from dual union all
select 220 client_id, to_date('25-11-2011','DD-MM-YYYY') d from dual union all
select 436 client_id, to_date('01-04-2011','DD-MM-YYYY') d from dual union all
select 436 client_id, to_date('01-04-2011','DD-MM-YYYY') d from dual
select client_id,
       to_char(d,'DD-MM-YYYY') d
from
  select client_id,
         d,
         row_number() over (partition by client_id order by d) rn
  from t
where rn=2;  Edited by: hm on 13.12.2011 02:17
please try out this to see how it works:
with t as
select 436 client_id, to_date('01-02-2011','DD-MM-YYYY') d from dual union all
select 328 client_id, to_date('05-04-2011','DD-MM-YYYY') d from dual union all
select 319 client_id, to_date('13-01-2011','DD-MM-YYYY') d from dual union all
select 220 client_id, to_date('03-09-2011','DD-MM-YYYY') d from dual union all
select 319 client_id, to_date('17-12-2011','DD-MM-YYYY') d from dual union all
select 220 client_id, to_date('25-11-2011','DD-MM-YYYY') d from dual union all
select 436 client_id, to_date('01-04-2011','DD-MM-YYYY') d from dual union all
select 436 client_id, to_date('01-04-2011','DD-MM-YYYY') d from dual
select client_id,
       d,
       row_number() over (partition by client_id order by d) rn
from t;(By the way: in my opinion it is not a good idea to choose a keyword "DATE" as a column name. That the reason why I choose "d" in my example.)