我需要进行查询,但是我找不到有效的方法。
客户每次听一首歌,就会用他的键和track_id,日期...记录一行。
我想检查每个客户,如果他上个月收听的曲目是新曲目(本月之前从未收听过)
一行如下:
key | track_id | date
asd | 12312 | 12/02/2020
fds | 12323 | 12/05/2020
等
我认为我可以使用窗口函数执行某些操作,但似乎找不到一种好的方法。
然后,我还需要从此列表中获得前3名听众最多的歌曲,我可以通过窗口功能来实现。
如果有人可以帮助我?非常感谢。
答案 0 :(得分:1)
使用聚合和窗口功能:
select key, count(*) as num_rows,
count(*) as num_tracks,
sum( case when first_yyyymm = yyyymm then 1 else 0 end) as num_new_tracks,
sum( case when first_yyyymm < yyyymm then 1 else 0 end) as num_prev_tracks
from (select t.key, track_id, date_trunc('month', date) as yyyymm,
min(date_trunc('month', date) ) over (partition by key, track_id) as first_yyyymm
from t
group by key, track_id, yyyymm
) t
where yyyymm >= date_trunc('month', current_date)
group by key
答案 1 :(得分:0)
一条建议。如果您需要别人的帮助,请始终在示例数据集中包含DDL。期望人们为您做这是不公平的。 (在这种情况下,我做到了)
create table sandbox.songs
(key varchar,
track_id number,
dt date);
insert into sandbox.songs
values
('asd', 12312, '12/02/2020'),
('fds', 12323, '12/05/2020'),
('asd', 11000, '04/05/2020'),
('fds', 92823, '04/07/2020'),
('asd', 11000, '3/05/2020'),
('fds', 12323, '12/05/2020'),
('asd', 92834, '3/05/2020');
问题1
select
key,
count(0),
sum(case when dt < current_date - 30 then 1 else 0 end) as older_than_30,
sum(case when dt >= current_date - 30 then 1 else 0 end) as within_30
from sandbox.songs group by key;
[在此处输入图片描述] [1]
问题2
track_id,
count(0)
from sandbox.songs
group by track_id
order by count(0) desc
limit 3;
[enter image description here][2]
[1]: https://i.stack.imgur.com/CeDqw.png
[2]: https://i.stack.imgur.com/sdKV0.png