Reputation: 383
I have following table,
+------+-------------+----------+---------+
| id | date | amount | amount2 |
+------+-------------+----------+---------+
| | | | 500 |
| 1 | 1/1/2020 | 1000 | |
+------+-------------+----------+---------+
| | | | 100 |
| 1 | 1/3/2020 | 1558 | |
+------+-------------+----------+---------+
| | | | 200 |
| 1 | 1/3/2020 | 126 | |
+------+-------------+----------+---------+
| | | | 500 |
| 2 | 2/5/2020 | 4921 | |
+------+-------------+----------+---------+
| | | | 100 |
| 2 | 2/5/2020 | 15 | |
+------+-------------+----------+---------+
| | | | 140 |
| 2 | 1/1/2020 | 5951 | |
+------+-------------+----------+---------+
| | | | 10 |
| 2 | 1/2/2020 | 1588 | |
+------+-------------+----------+---------+
| | | | 56 |
| 2 | 1/3/2020 | 1568 | |
+------+-------------+----------+---------+
| | | | 45 |
| 2 | 1/4/2020 | 12558 | |
+------+-------------+----------+---------+
I need to get each Id's max date and its amount and amount2 summations, how can I do this. according to above data, I need following output.
+------+-------------+----------+---------+
| | | | 300 |
| 1 | 1/3/2020 | 1684 | |
+------+-------------+----------+---------+
| | | | 600 |
| 2 | 2/5/2020 | 4936 | |
+------+-------------+----------+---------+
How can I do this.
Upvotes: 0
Views: 1301
Reputation: 94914
Aggregate and use MAX OVER
to get the IDs' maximum dates:
select id, [date], sum_amount, sum_amount2
from
(
select
id, [date], sum(amount) as sum_amount, sum(amount2) as sum_amount2,
max([date]) over (partition by id) as max_date_for_id
from mytable group by id, [date]
) aggregated
where [date] = max_date_for_id
order by id;
Upvotes: 3
Reputation: 24763
first is to use dense_rank()
to find the row with latest date
dense_rank () over (partition by id order by [date] desc)
after that, just simply group by
with sum()
on the amount
select id, [date], sum(amount), sum(amount2)
from
(
select *,
dr = dense_rank () over (partition by id order by [date] desc)
from your_table
) t
where dr = 1
group by id, [date]
Upvotes: 2