雪花窗函数 last_value 和 max

Question

我有一张这样的桌子：

我想为每个 user_key 获取每个日期/月/工作日/周的 top total_listened

我想我需要使用窗口函数 我可以得到不同的日期格式：

MONTH(stream_date) for months
WEEKDAY(stream_date) for weekday
WEEK(stream_date) for week

我试过这个：

select 
MAX(vals.total_listened_per_day) as RECORD_STREAM_DAY_TIME,
MAX(vals.total_listened_per_month) as RECORD_STREAM_MONTH_TIME,
MAX(vals.total_listened_per_week) as RECORD_STREAM_WEEK_TIME,
MAX(vals.most_active_weekday) as MOST_ACTIVE_WEEKDAY_TIME
 last_value(days.date) over (partition by user_key order by days.total_listened) as RECORD_STREAMDAY,
from
(
select user_key, stream_date as date,
sum(st.length_listened) over (partition by user_key, stream_date) as total_listened_per_day,
sum(st.length_listened) over (partition by user_key, MONTH(stream_date)) as total_listened_per_month,
sum(st.length_listened) over (partition by user_key, WEEK(stream_date)) as total_listened_per_week,
sum(st.length_listened) over (partition by user_key, DAYNAME(stream_date)) as most_active_weekday
group by 1,2
 .....
 )

用于获取金额（以_TIME为结尾的变量），但不适用于获取特定的日期/月份....（末尾没有_TIME的变量，例如RECORD_STREAMDAY），这是因为group by ，它是按 stream_date 而不是按 month(stream_date) 分组的，例如，我不知道如何在没有 doin 子查询的情况下做到这一点

Answer 1

我认为你想要的逻辑是：

select user_key,
    max(total_listened_per_day  ) as max_total_listened_per_day
    max(total_listened_per_week ) as max_total_listened_per_week,
    max(total_listened_per_month) as max_total_listened_per_month,
    max(case when rn_day   = 1 then date_trunc('day',   stream_date) end) as most_active_day,
    max(case when rn_week  = 1 then date_trunc('week',  stream_date) end) as most_active_week,
    max(case when rn_month = 1 then date_trunc('month', stream_date) end) as most_active_month
from (  
    select t.*,
        rank() over(partition by user_key order by total_listened_per_day   desc) as rn_day,
        rank() over(partition by user_key order by total_listened_per_week  desc) as rn_week,
        rank() over(partition by user_key order by total_listened_per_month desc) as rn_month
    from (
        select t.*
            sum(st.length_listened) over (partition by user_key, date_trunc('day',   stream_date)) as total_listened_per_day,
            sum(st.length_listened) over (partition by user_key, date_trunc('week',  stream_date)) as total_listened_per_week
            sum(st.length_listened) over (partition by user_key, date_trunc('month', stream_date)) as total_listened_per_month
        from mytable t
    ) t
) t
group by user_key

最内部的子查询计算每天、每周和每月的收听时间的窗口总和。下一个子查询使用该信息对记录进行排名。最后，外部查询使用条件聚合来带来相应的持续时间和周期。如果有平局，则选择最近的时期。