在一周的范围内每天获得最频繁的选择

Question

我还有一个问题，但这有点令人困惑，所以这次我将尝试简化。

我有一个表“ option”：

id  |  option_name
1   "Option 1"
2   "Option 2"
3   "Option 3"
4   "Option 4"
5   "Option 5"
6   "Option 6"
7   "Option 7"

和另一个表“ user”，其中包含一些用户信息。然后，我创建了一个名为“ daily_voting”的表，每个用户每天都可以选择一个选项。应该是这样的：

user_id | option_id | timestamp
1  1  2018-12-24 00:01:00.091055-03
2  1  2018-12-24 01:01:00.091055-03
3  2  2018-12-24 02:01:00.091055-03
1  2  2018-12-25 00:01:00.091055-03
2  2  2018-12-25 00:02:00.091055-03
3  5  2018-12-26 00:02:00.091055-03

如果我这样做：

select count(*) as votes, option_id, date_trunc('day', daily_voting.timestamp) as day
from daily_voting
where date_trunc('day', daily_voting.timestamp) = '2018-12-24 00:00:00-03'
group by option_id, day
order by votes desc
limit 1

我会得到：

votes | option_id | day
2    1    2018-12-24 00:00:00-03

我需要的是，从当前日期过去一周，可能是：

date_trunc('week', current_date) and date_trunc('week', current_date+7)

查询的输出将是：

votes | option_id | day
2    1    2018-12-24 00:00:00-03
2    2    2018-12-25 00:00:00-03
1    5    2018-12-26 00:00:00-03

Answer 1

ROW_NUMBER()将为您排名。

SELECT Votes, Option_ID, Day
FROM (
SELECT Votes, Option_id, day, row_number() over (partition by day order by votes DESC) AS VoteRank
FROM (
select count(*) as votes, option_id, date_trunc('day', daily_voting.timestamp) as day
from daily_voting
where date_trunc('day', daily_voting.timestamp) BETWEEN '2018-12-24' AND '2018-12-31'
group by option_id, day
)Aux ) T
WHERE VoteRank = 1

您可以随意使用其他方法定义日期范围。