get the most two recent dates for each customer - sql

Question

basically, I need to retrieve the last two dates for customers who purchased in at least two different dates, implying there are some customer who had purchased only in one date, the data has the following form
client_id date
1 2016-07-02
1 2016-07-02
1 2016-06-01
2 2015-06-01
and I would like to get it in the following form
client_id previous_date last_date
1 2016-06-01 2016-07-02
remarques:
a client can have multiple entries for the same date
a client can have entries only for one date, such customer should be discarded

Answer 1

Rank your dates with DENSE_RANK. Then group by client_id and show the last dates (ranked #1 and #2).
select
client_id,
max(case when rn = 2 then date end) as previous_date,
max(case when rn = 1 then date end) as last_date
from
(
select
client_id,
date,
dense_rank() over (partition by client_id order by date desc) as rn
from mytable
)
group by client_id
having max(rn) > 1;

Answer 2

build up:
t=# create table s153 (c int, d date);
CREATE TABLE
t=# insert into s153 values (1,'2016-07-02'), (1,'2016-07-02'),(1,'2016-06-01'),(2,'2016-06-01');
INSERT 0 4
query:
t=# with a as (
select distinct c,d from s153
)
, b as (
select c,nth_value(d,1) over (partition by c order by d) last_date, nth_value(d,2) over (partition by c order by d) prev_date
from a
)
select * from b where prev_date is not null
;
c | last_date | prev_date
---+------------+------------
1 | 2016-06-01 | 2016-07-02
(1 row)

Answer 3

UNTESTED:
We use a common table expression to assign a row number based on the date in descending order and then only include those records having a row number <=2 and then ensure that those having 1 row are excluded by the having.
WITH CTE AS (
SELECT Distinct Client_ID
, Date
, row_number() over (partition by clientID order by date desc) rn
FROM Table)
SELECT Client_ID, min(date) previous_date, max(date) last_date)
FROM CTE
WHERE RN <=2
GROUP BY Client_ID
HAVING max(RN) > 1

Answer 4

All you need is a group by...
--test date
declare #tablename TABLE
(
client_id int,
[date] datetime
);
insert into #tablename
values( 1 , '2016-07-02'),
(1 , '2016-07-02'),
(1 , '2016-06-01'),
(2 , '2015-06-01');
--query
SELECT client_id,MIN([DATE]) AS [PREVIOUS_DATE], MAX([DATE]) AS [LAST_DATE]
FROM #tablename
GROUP BY client_id
Updated
-- create data
create table myTable
(
client_id integer,
given_date date
);
insert into myTable
values( 1 , '2016-07-02'),
(1 , '2016-07-02'),
(1 , '2016-06-01'),
(1 , '2016-06-03'),
(1 , '2016-06-09'),
(2 , '2015-06-01'),
(3 , '2016-06-03'),
(3 , '2016-06-09');
-- query
SELECT sub.client_id, sub.PREVIOUS_DATE, sub.LAST_DATE
FROM
(select
ROW_NUMBER() OVER (PARTITION BY a.client_id order by b.given_date desc,(MAX(b.given_date) - a.given_date)) AS ROW_NUMBER,
a.client_id,a.given_date AS PREVIOUS_DATE, MAX(b.given_date) - a.given_date AS diff, (b.given_date) AS LAST_DATE
FROM myTable AS a
JOIN myTable AS b
ON b.client_id = a.client_id
WHERE a.given_date <> b.given_date
group by a.client_id, a.given_date, b.given_date) AS sub
WHERE sub.ROW_NUMBER = 1