详细介绍：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