这是一个创建于 1687 天前的主题，其中的信息可能已经有所发展或是发生改变。
看到一个面试题想请假一下
two tables:
friending (sender_id, receiver_id, send_date, accept_date, sender_country)
age (user_id, age_group)
1.1 How many friend-requests sent on each day of the week for the last 4 weeks?
1.2 What is the average number of friend-requests sent per user over the last past week by age group? // 前 7 天内每个 age group 的平均好友请求数
第 1 个是直接 select sender_id 然后 where send_data < 28 吧？
主要是想请问一下第 2 个怎么写的？ 这 2 个 tables union，然后 COUNT(sender_id) OVER (PARTITION BY age_group) 这种？
不好意思 sql 不太懂临时抱佛脚学的比较乱七八糟，希望指教
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1
hand515 2020-03-23 08:46:04 +08:00
Join
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2
a87965028 2020-03-23 09:31:05 +08:00  1
select a.age_group, count(1) / count(distinct a.user_id) as avg
from friending as f inner join age as a on f.sender_id = a.user_id where send_date < datediff(day, -7, getdate()) group by age_group |
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