LeetCode 高频 SQL 50 题（基础版）之 【聚合函数】部分

题目：620. 有趣的电影

题解：

select * from cinema
where description !='boring' and id%2=1
order by rating desc

题目：1251. 平均售价

题解：

select p.product_id product_id,round(ifnull(sum(p.price*u.units)/sum(u.units),0),2) average_price 
from Prices p left join UnitsSold u
on p.product_id=u.product_id
where (p.start_date<=u.purchase_date and u.purchase_date<=p.end_date) or u.units is null
group by p.product_id

题目：1075. 项目员工 I

题解：

select p.project_id project_id, round(avg(e.experience_years),2) average_years from Project p,Employee e
where p.employee_id=e.employee_id
group by p.project_id

题目：1633. 各赛事的用户注册率

题解：

select contest_id,round(count(user_id)*100/(select count(*) from Users),2) as percentage
from Register r 
group by contest_id
order by percentage desc,contest_id asc

题目：1211. 查询结果的质量和占比

题解：

select query_name,round(avg(rating/position),2) quality,
round(100*avg(rating<3),2) poor_query_percentage 
from Queries
group by query_name

题目：1193. 每月交易 I

题解：

select  left(trans_date,7) month,country,
count(*) trans_count,
sum(amount) approved_count,
sum(if(state='approved',1,0)) trans_total_amount,
sum(if(state='approved',amount,0)) approved_total_amount
from Transactions
group by month,country

题目：1174. 即时食物配送 II

题目：

select round(100*avg(order_date=customer_pref_delivery_date),2) immediate_percentage 
from Delivery
where (customer_id,order_date) in(
    select customer_id,min(order_date) 
    from Delivery
    group by customer_id
)

题目：550. 游戏玩法分析 IV

题解：

select round(avg(a2.event_date is not null),2) fraction from (
    select player_id,min(event_date) event_date from Activity
    group by player_id) a1 left join Activity a2
on a1.player_id=a2.player_id and datediff(a2.event_date,a1.event_date)=1