Question

目标：查找过去3个月内前200首曲目。每个轨道的消费者数据为每日。

我用过

RANK () OVER(PARTITION BY report_date ORDER BY SUM(num_streams))

我担心的是，这将使我每天返回前200名不同和我不想要。

我希望前200名每天保持不变。但是每天都有不同的消费者数据。

有什么主意吗？

我正在使用AWS REDSHIFT

我想要这个：

Day       Track_Title       Consumer_counts       Rank
Jul 1      Halo               600                  1
Jul 1      old town road      200                  2
Jul 1      heartbeat          180                  3
Jul 2      Halo               500                  1
Jul 2      old town road      300                  2
Jul 2      heartbeat          400                  3

但是我明白了

 Day       Track_Title       Consumer_counts       Rank
Jul 1      Halo               600                  1
Jul 1      old town road      200                  2
Jul 1      heartbeat          180                  3
Jul 2      Halo               500                  1
Jul 2      wish you were here 400                  2
Jul 2      old town road      300                  3

Answer 1

如果您希望在3个月内排名前200位的曲目，然后获得仅包含所选200条曲目的每日计数，则可以执行以下操作：

SELECT top 200 --To get the top 200 tracks over 3 months
t.track,
RANK () OVER(PARTITION BY t.track ORDER BY SUM(t.num_streams))
INTO #tempTable
FROM 
(SELECT track, num_streams
FROM MyTable
WHERE report_date BETWEEN "2019-01-01" and "2019-04-01") t

SELECT  --To get the top 200 tracks for the selected date.
mt.day,
mt.track,
SUM(mt.num_streams)
RANK () OVER(PARTITION BY mt.report_date ORDER BY SUM(mt.num_streams))
FROM #tempTable tt
INNER JOIN MyTable mt on tt.track = mt.track

Answer 2

我认为您想要一个聚合查询

select Track_Title, sum(Consumer_counts) as total
       rank() over (order by sum(Consumer_counts) desc) as ranking
from t
where report_date >= current_date - interval '3 month'
group by Track_Title
order by total desc
limit 200;

如果您想按日查看详细信息，那么我建议重新加入：

with top200 as (
      select Track_Title, sum(Consumer_counts) as total
             rank() over (order by sum(Consumer_counts) desc) as ranking
      from t
      where report_date >= current_date - interval '3 month'
      group by Track_Title
      order by total desc
      limit 200
     )
select t.*, top200.ranking
from t join
     top200
     on t.Track_Title = top200.Track_Title
order by t.report_date, top200.ranking;

等级函数总是为每个组返回相同的值吗？

2 个答案: