如何获取最大日期及其行的总和 SQL
How to get Max date and sum of its rows SQL
我关注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 | |
+------+-------------+----------+---------+
我需要获取每个 ID 的最大日期及其金额和金额 2 的总和,我该怎么做。根据以上数据,我需要以下输出。
+------+-------------+----------+---------+
| | | | 300 |
| 1 | 1/3/2020 | 1684 | |
+------+-------------+----------+---------+
| | | | 600 |
| 2 | 2/5/2020 | 4936 | |
+------+-------------+----------+---------+
我该怎么做。
首先是使用dense_rank()
查找最新的行date
dense_rank () over (partition by id order by [date] desc)
之后,只需 group by
和 amount
上的 sum()
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]
汇总并使用 MAX OVER
获取 ID 的最大日期:
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;
我关注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 | |
+------+-------------+----------+---------+
我需要获取每个 ID 的最大日期及其金额和金额 2 的总和,我该怎么做。根据以上数据,我需要以下输出。
+------+-------------+----------+---------+
| | | | 300 |
| 1 | 1/3/2020 | 1684 | |
+------+-------------+----------+---------+
| | | | 600 |
| 2 | 2/5/2020 | 4936 | |
+------+-------------+----------+---------+
我该怎么做。
首先是使用dense_rank()
查找最新的行date
dense_rank () over (partition by id order by [date] desc)
之后,只需 group by
和 amount
sum()
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]
汇总并使用 MAX OVER
获取 ID 的最大日期:
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;