Reputation: 13091
I have a table
id, date, value
a, 5/22/2019, 22
b, 5/22/2019, 22
c, 5/22/2019, 22
a, 5/21/2019, 21
b, 5/21/2019, 21
c, 5/21/2019, 21
a, 5/20/2019, 20
b, 5/20/2019, 20
c, 5/20/2019, 20
Want to get ID and value of current (latest) and previous date:
id, date, date-1
a, 22, 21
b, 22, 21
c, 22, 21
So far I have some parts but need only 1 ID per row (this bellow give more than 1 row for each ID - it will show for 5/20 date in the next row for the same ID which is wrong):
select
id, value,
lag(value, 1, 0) over (partition by id order by date ) as "date - 1"
from
table1
How to achieve that?
Upvotes: 3
Views: 988
Reputation: 32003
you can try like below
with cte as
(
select
id, value,
lag(value, 1, 0) over (partition by id order by date ) as "date - 1",
lag(value, 2, 0) over (partition by id order by date ) as "date - 2",
row_number() over(partition by id order by date desc) rn
from
table1
) select * from cte where rn=1
Upvotes: 2
Reputation: 1211
How about this? Note that I named the column ddate
instead of date:
create table a_test(id varchar(1), ddate date, value int);
insert into a_test values ('a', '5/22/2019', 22),
('b', '5/22/2019', 22),
('c', '5/22/2019', 22),
('a', '5/21/2019', 21),
('b', '5/21/2019', 21),
('c', '5/21/2019', 21),
('a', '5/20/2019', 20),
('b', '5/20/2019', 20),
('c', '5/20/2019', 20);
with cte as(
select id,
value,
row_number() over(partition by id order by ddate desc) as rn
from a_test)
select id,
MAX(value) FILTER (WHERE rn = 1) as col1,
MAX(value) FILTER (WHERE rn = 2) as col2
from cte
group by id
Upvotes: 2