一、高频SQL50题(基础版)
| 题目考查类型 | 题号 |
|---|---|
| 查询 | 1-5 |
| 连接 | 6-14 |
| 聚合函数 | 15-22 |
| 排序和分组 | 23-29 |
| 高级查询和连接 | 30-36 |
| 子查询 | 37-43 |
| 高级字符串函数 / 正则表达式 / 子句 | 44-50 |
SELECT product_id
FROM Products
WHERE low_fats='Y' and recyclable='Y';
select name
from Customer
where referee_id != 2 or referee_id is null
select name,population,area
from World
where area>=3000000 or population>=25000000
select distinct author_id as id
from Views
where author_id=viewer_id
order by id
select tweet_id
from Tweets
where length(content)>15
select unique_id,name
from Employees left join EmployeeUNI
on EmployeeUNI.ID=Employees.ID
select product_name,year,price
from Sales join Product
on Sales.product_id = Product.product_id
select customer_id, count(*) as count_no_trans
from Visits v left join Transactions t on
v.visit_id = t.visit_id
where transaction_id is null
group by customer_id
select w2.id id
from Weather w1 join Weather w2
on w1.recordDate = w2.recordDate-interval 1 day
where w1.Temperature < w2.Temperature
# 运行时间17%
select start.machine_id, round(avg(end.timestamp-start.timestamp),3) processing_time
from
(select *
from Activity
where activity_type ='start') as start
join
(select *
from Activity
where activity_type ='end') as end
on
start.machine_id=end.machine_id and start.process_id=end.process_id
group by start.machine_id
# 运行时间45%
select machine_id,
round((2*sum(timestamp*(case when activity_type = 'start' then -1 else 1 end)))/count(activity_type),3) as processing_time
from Activity
group by machine_id
11.577. 员工奖金
select name,bonus
from Employee left join Bonus
on Employee.empId = Bonus.empid
where bonus<1000 or bonus is null
select a.student_id,a.student_name,a.subject_name,ifnull(attended_exams,0) attended_exams
from
(select *
from subjects join students) a
left join
(select *,count(e.student_id) as attended_exams
from Examinations e
group by e.student_id,e.subject_name) b
on a.student_id = b.student_id
and a.subject_name = b.subject_name
order by a.student_id,a.subject_name
select e2.name name
from Employee e2
left join Employee e1
on e1.managerId=e2.id
group by e2.id
having count(*)>=5
14.1934. 确认率
select s.user_id,round(sum(if(action='confirmed',1,0))/count(*),2) confirmation_rate
from Signups s
left join Confirmations c
on s.user_id = c.user_id
group by s.user_id
15.620. 有趣的电影
select *
from cinema
where description!='boring' and id%2!=0
order by rating desc
16.1251. 平均售价
select p.product_id,ifnull(round((sum(price*units)/sum(units)),2),0) as average_price
from Prices p left join UnitsSold u
on p.product_id = u.product_id
and u.purchase_date between p.start_date and p.end_date
group by p.product_id
17.1075. 项目员工 I
select project_id,round(avg(experience_years),2) as average_years
from Project p left join Employee e
on p.employee_id = e.employee_id
group by project_id
select contest_id,round(count(contest_id)/(select count(*) from Users)*100,2) as percentage
from Register r left join Users u
on r.user_id = u.user_id
group by contest_id
order by percentage desc,contest_id
select query_name,round(avg(rating/position),2) as quality,
round((100*sum(case when rating<3 then 1 else 0 end)/count(*)),2) as poor_query_percentage
from Queries
group by query_name
having query_name is not null
20.1193. 每月交易 I
select left(trans_date,7) as month,
country,count(*) as trans_count,
sum(case when state='approved' then 1 else 0 end) as approved_count,
sum(amount) as trans_total_amount,
sum((case when state='approved' then 1 else 0 end)*amount) as approved_total_amount
from Transactions
group by month,country
select round((sum(case when customer_pref_delivery_date=order_date then 1 else 0 end)*100/count(*)),2) as immediate_percentage
from
(select customer_id,min(order_date) as order_date,min(customer_pref_delivery_date) as customer_pref_delivery_date
from Delivery
group by customer_id) as first_order
# 卡了很久最小时间
select round(count(*)/(select count(distinct player_id) from Activity),2) as fraction
from
((select player_id,min(event_date) as event_date
from Activity
group by player_id) as a1
join Activity a2
on a1.player_id=a2.player_id
and a1.event_date=a2.event_date - interval 1 day)
select teacher_id,count(distinct subject_id) as cnt
from teacher
group by teacher_id
select activity_date as day,count(distinct user_id) as active_users
from Activity
group by activity_date
having activity_date between ("2019-07-27"- interval 29 day) and "2019-07-27"
# 注意sum=count的用法，用于“所有都是……”的场景
select s.product_id,product_name
from Sales s left join Product p
on s.product_id=p.product_id
group by s.product_id
having sum(s.sale_date between "2019-01-01" and "2019-03-31")=count(*)
select class
from Courses
group by class
having count(*)>=5
select user_id,count(*) as followers_count
from Followers
group by user_id
order by user_id
select max(num) num
from
(select num
from MyNumbers
group by num
having count(*)=1) num1
select customer_id
from Customer
group by customer_id
having count(distinct product_key)= (select count(*) from Product)
select e2.employee_id,e2.name,count(*) as reports_count,
round(avg(e1.age),0) as average_age
from Employees e1 left join Employees e2
on e1.reports_to = e2.employee_id
group by e2.employee_id
having e2.employee_id is not null
order by employee_id
(select employee_id,department_id
from Employee
where primary_flag ='Y')
UNION
(select employee_id,department_id
from Employee
group by employee_id
having count(*)=1)
order by employee_id
32.610. 判断三角形
select *,
(case when (x+y>z and x+z>y and z+y>x) then "Yes"
else "No"
end) as triangle
from Triangle
33.180. 连续出现的数字
select distinct L1.num as ConsecutiveNums
from Logs L1
join Logs L2 on L1.id=L2.id-1
join Logs L3 on L2.id=L3.id-1
where L1.num=L2.num and L2.num=L3.num
select product_id, new_price as price
from Products
where (product_id,change_date) in
(select product_id,max(change_date)
from Products
where change_date<="2019-08-16"
group by product_id)
union
select product_id,10 as price
from Products
where product_id not in (select product_id from Products where change_date<="2019-08-16")
select q1.person_name
from Queue q1
join Queue q2 on q1.turn>=q2.turn
group by q1.person_id
having sum(q2.weight)<=1000
order by q1.turn desc limit 1
select "Low Salary" category,count(*) accounts_count
from Accounts
where income<20000
union
select "Average Salary" category,count(*) accounts_count
from Accounts
where income between 20000 and 50000
union
select "High Salary" category,count(*) accounts_count
from Accounts
where income>50000
select employee_id
from Employees
where salary<30000 and manager_id not in (select employee_id from Employees)
order by employee_id
38.626. 换座位
select
(case when id%2!=0 and id!=(select count(*) from Seat) then id+1
when id%2=0 then id-1
else id end) as id,student
from Seat
order by id
39.1341. 电影评分
(select name as results
from Users u
join MovieRating r1 on u.user_id = r1.user_id
group by u.user_id
order by count(*) desc,name limit 1)
union all
(select title as results
from Movies m
join MovieRating r2 on m.movie_id = r2.movie_id and left(r2.created_at,7) = "2020-02"
group by r2.movie_id
order by avg(rating) desc,title limit 1)
⭐⭐40.1321. 餐馆营业额变化增长
#注意join时where的用法以及分组之后avg函数的使用
select a.visited_on,sum(c.amount) as amount,round((sum(c.amount))/7,2) as average_amount
from
(select distinct visited_on
from Customer) as a
left join customer c
on (c.visited_on>=a.visited_on - interval 6 day) and (c.visited_on<=a.visited_on)
where a.visited_on>=(select min(visited_on) from customer)+6
group by a.visited_on
order by a.visited_on
select a.id,count(*) as num
from
(select requester_id as id from RequestAccepted r1
union all
select accepter_id as id from RequestAccepted r2) as a
group by id
order by num desc limit 1
select round(sum(tiv_2016),2) tiv_2016
from Insurance
where tiv_2015 in
(select tiv_2015 from Insurance
group by tiv_2015 having count(*)>1)
and concat(lat, lon) in
(select concat(lat, lon)
from Insurance
group by concat(lat, lon)
having count(*)=1)
⭐⭐⭐43.185. 部门工资前三高的所有员工
select d.name as Department,e.name as Employee,e.salary
from Employee e left join Department d
on e.departmentId=d.id
where e.id in
(select e1.id
from Employee e1 left join Employee e2
on e1.departmentId=e2.departmentId and e1.salary<e2.salary
group by e1.id
having count(distinct e2.salary)<=2)
select user_id,concat(upper(left(name,1)),lower(SUBSTRING(name,2))) name
from Users
order by user_id
select *
from Patients
where conditions like "DIAB1%" or conditions like "% DIAB1%"
delete from
Person
where id not in
(select id from(select min(id) id
from Person
group by email) as a)
47.176. 第二高的薪水
select ifnull((
select distinct salary
from Employee
order by salary desc
limit 1 offset 1),null) as SecondHighestSalary
select sell_date,count(distinct product) as num_sold,
group_concat(distinct product order by product SEPARATOR ',') as products
from Activities
group by sell_date
order by sell_date
select product_name,sum(unit) as unit
from Products p join Orders o
on p.product_id=o.product_id and left(o.order_date,7)="2020-02"
group by product_name
having sum(unit)>=100
SELECT user_id, name, mail
FROM Users
WHERE mail REGEXP '^[a-zA-Z][a-zA-Z0-9_.-]*\\@leetcode\\.com$';
二、高频SQL50题（进阶版）
| 题目考查类型 | 题号 |
|---|---|
| 查询 | 1-5 |
| 连接 | 6-11 |
| 聚合函数 | 12-19 |
| 排序和分组 | 20-26 |
| 高级查询和连接 | 27-35 |
| 子查询 | 36-43 |
| 高级字符串函数 / 正则表达式 / 子句 | 44-50 |
select customer_id
from Customers
where year=2021 and revenue>0
2.183. 从不订购的客户
select name as Customers
from Customers
where id not in
(select customerId from Orders)
- 1873. 计算特殊奖金
select employee_id,salary*(case when employee_id%2!=0 and left(name,1)!="M" then 1 else 0 end) as bonus
from Employees
order by employee_id
4.1398. 购买了产品 A 和产品 B 却没有购买产品 C 的顾客
# 方法一
select customer_id,customer_name
from Customers
where customer_id in
(select customer_id
from Orders
where product_name="A" or product_name="B"
group by customer_id
having count(distinct product_name)=2)
and customer_id not in
(select customer_id
from Orders
where product_name="C"
group by customer_id)
# 方法二：巧用sum
select customer_id,customer_name
from Customers
where customer_id in
(select customer_id
from Orders
group by customer_id
having sum(product_name="A")*sum(product_name="B")>0
and sum(product_name="C")=0)
5.1112. 每位学生的最高成绩
select student_id,min(course_id) course_id,grade
from Enrollments
where (student_id,grade) in
(
select student_id,max(grade) grade
from Enrollments
group by student_id)
# 注意这里的group by，因为取了min，所有要group by
group by student_id
order by student_id,course_id
6.175. 组合两个表
select firstName,lastName,city,state
from Person p left join Address a on p.PersonId = a.personId
select seller_name
from Seller s
where seller_id not in
(select seller_id
from Orders
where left(sale_date,4)="2020")
order by seller_name
- 1407. 排名靠前的旅行者
select name,ifnull(sum(r.distance),0) travelled_distance
from Users u
left join Rides r on u.id=r.user_id
group by u.id
order by travelled_distance desc,name
9.607. 销售员
select name
from SalesPerson
where sales_id not in
(select sales_id
from Orders o join Company c
on o.com_id = c.com_id and c.name="RED" )
10.1440. 计算布尔表达式的值
select e.*,
(case when operator=">" and v1.value>v2.value then "true"
when operator="<" and v1.value<v2.value then "true"
when operator="=" and v1.value=v2.value then "true"
else "false"
end) as value
from Expressions e
left join Variables v1 on v1.name=e.left_operand
left join Variables v2 on v2.name=e.right_operand
11.1212. 查询球队积分
select t.team_id,t.team_name,
sum(case when m.host_team = t.team_id and host_goals>guest_goals then 3
when m.guest_team = t.team_id and host_goals<guest_goals then 3
when host_goals=guest_goals then 1
else 0
end) as num_points
from Matches m
right join Teams t on m.host_team = t.team_id or m.guest_team=t.team_id
group by t.team_id
order by num_points desc,team_id
12.1890. 2020年最后一次登录
select user_id,max(time_stamp) as last_stamp
from Logins
where left(time_stamp,4)='2020'
group by user_id
13.511. 游戏玩法分析 I
select player_id,min(event_date) as first_login
from Activity
group by player_id
14.1571. 仓库经理
select name as warehouse_name,sum(units*Width*Length*Height) as volume
from Warehouse w left join Products p
on w.product_id=p.product_id
group by w.name
15.586. 订单最多的客户
select customer_number
from Orders
group by customer_number
order by count(customer_number) desc limit 1
16.1741. 查找每个员工花费的总时间
select event_day as day,emp_id,sum(out_time-in_time) as total_time
from Employees
group by event_day,emp_id
17.1173. 即时食物配送 I
select round(100*sum(case when order_date=customer_pref_delivery_date then 1
else 0
end)/(count(delivery_id)),2) as immediate_percentage
from Delivery
18.1445. 苹果和桔子
select sale_date,sum(case when fruit="apples" then sold_num
else -sold_num
end ) as diff
from Sales
group by sale_date
19.1699. 两人之间的通话次数
# 先找到适合的顺序
select
(case when from_id<to_id then from_id else to_id end)as person1,
(case when from_id<to_id then to_id else from_id end)as person2,
count(*) as call_count,
sum(duration) as total_duration
from Calls
group by person1,person2
20.1587. 银行账户概要 II
select u.name as NAME,sum(t.amount) as BALANCE
from Transactions t left join Users u on t.account=u.account
group by t.account
having sum(amount)>10000
21.182. 查找重复的电子邮箱
select email Email
from Person
group by email
having count(*)>=2
22.1050. 合作过至少三次的演员和导演
select actor_id,director_id
from ActorDirector
group by actor_id,director_id
having count(*)>=3
23.1511. 消费者下单频率
select o.customer_id,c.name
from Orders o join Product p on o.product_id=p.product_id
join Customers c on o.customer_id=c.customer_id
group by customer_id
having sum(case when left(o.order_date,7)="2020-06" then quantity*price else 0 end)>=100
and sum(case when left(o.order_date,7)="2020-07" then quantity*price else 0 end)>=100
24.1693. 每天的领导和合伙人
select date_id,make_name,count(distinct lead_id) unique_leads,count(distinct partner_id) unique_partners
from DailySales
group by date_id,make_name
25.1495. 上月播放的儿童适宜电影
select distinct title
from Content c left join TVProgram t on c.content_id =t.content_id
where c.Kids_content='Y' and left(t.program_date,7)="2020-06" and content_type="Movies"
26.1501. 可以放心投资的国家
select co.name as country
from Person p join Country co on left(p.phone_number,3)=co.country_code
join Calls ca on p.id=ca.caller_id or p.id=ca.callee_id
group by co.country_code
having avg(duration)>(select avg(duration) from Calls)
27.603. 连续空余座位
select distinct c1.seat_id
from Cinema c1 join Cinema c2
on abs(c1.seat_id - c2.seat_id)=1
where c1.free=1 and c2.free=1
order by c1.seat_id
28.1795. 每个产品在不同商店的价格
select product_id,"store1" as store,store1 price
from Products
where store1 is not null
union all
select product_id,"store2" as store,store2 price
from Products
where store2 is not null
union all
select product_id,"store3" as store,store3 price
from Products
where store3 is not null
29.613. 直线上的最近距离
select min(abs(p1.x-p2.x)) as shortest
from Point p1 join Point p2 on p1.x!=p2.x
30.1965. 丢失信息的雇员
select employee_id
from Employees
where employee_id not in (select employee_id from Salaries)
union
select employee_id
from Salaries
where employee_id not in (select employee_id from Employees)
order by employee_id
31.1264. 页面推荐
select distinct page_id as recommended_page
from Likes
where user_id in(
select user2_id user_id
from Friendship
where user1_id ="1"
union
select user1_id user_id
from Friendship
where user2_id ="1")
and page_id not in (select page_id from Likes where user_id=1)
32.608. 树节点
select id,(case when p_id is null then "Root"
when id not in (select p_id from Tree where p_id is not null)then "Leaf"
else "Inner"
end )type
from Tree
33.534. 游戏玩法分析 III
SELECT a2.player_id,a2.event_date,sum(a1.games_played) as games_played_so_far
FROM Activity a1 left join Activity a2 on a1.player_id=a2.player_id and a2.event_date>=a1.event_date
group by a2.event_date,player_id
34.1783. 大满贯数量
select p.player_id,p.player_name,sum(c.Wimbledon=p.player_id)+sum(c.Fr_open=p.player_id)+sum(c.US_open=p.player_id)+sum(c.Au_open=p.player_id)grand_slams_count
from Players p,Championships c
group by p.player_id
having grand_slams_count>0
35.1747. 应该被禁止的 Leetflex 账户
select account_id
from(select account_id,login,logout,lead (login,1) over() as ll,nums
from(select *,row_number() over (partition by account_id order by login) as nums
from loginfo)aaa)bbb
where nums=1 and ll between login and logout
select id,name
from Students s
where s.department_id not in (select id from Departments)
37.1303. 求团队人数
select e1.employee_id,count(*) as team_size
from Employee e1 left join Employee e2 on e1.team_id=e2.team_id
group by e1.employee_id
select player_id,device_id
from Activity
where (player_id,event_date) in
(select player_id,min(event_date)
from Activity
group by player_id)
select d.name Department,e.name Employee,e.salary
from Employee e left join Department d on e.departmentId=d.id
where (e.departmentId,e.salary) in (select departmentId,max(salary)
from Employee
group by departmentId )
select p.product_name,o.product_id,o.order_id,o.order_date
from Orders o left join Products p on o.product_id=p.product_id
where (o.product_id,o.order_date) in
(select product_id,max(order_date)
from Orders
group by product_id)
order by p.product_name,o.product_id,o.order_id
select c.name as customer_name,c.customer_id,o2.order_id,o2.order_date
from Orders o1 left join Orders o2 on o1.customer_id=o2.customer_id and o1.order_date>=o2.order_date
left join Customers c on o1.customer_id=c.customer_id
group by o2.order_id
#比我日期更近的order_date不超过3个，所以是前三
having count(o1.order_date)<=3
order by c.name,c.customer_id,o2.order_date desc
select transaction_id
from Transactions
where (day,amount) in(select day,max(amount)
from Transactions
group by day)
order by transaction_id
43.1077. 项目员工 III
# 解题思路同39题
select p.project_id,p.employee_id
from Project p left join Employee e on p.employee_id=e.employee_id
where (p.project_id,e.experience_years) in
(select p.project_id,max(experience_years)
from Project p left join Employee e on p.employee_id=e.employee_id
group by p.project_id)
44.1285. 找到连续区间的开始和结束数字
#开窗函数，还看不太懂
select min(log_id) as start_id,max(log_id) as end_id
from
(SELECT
log_id,
log_id - row_number() over() diff
FROM logs)as t
group by diff
45.1596. 每位顾客最经常订购的商品
select o.customer_id,o.product_id,p.product_name
from
(select customer_id,product_id,
rank() over(partition by customer_id order by count(product_id) desc)rnk
from Orders
group by customer_id,product_id)o
join products p on o.product_id=p.product_id
where rnk=1
46.1709. 访问日期之间最大的空档期
select user_id,max(datediff(next_day,visit_date)) as biggest_window
from
(select user_id,visit_date,LEAD(visit_date,1,'2021-1-1') over(partition by user_id order by visit_date)as next_day
from UserVisits)as tmp
group by user_id
order by user_id
47.1270. 向公司 CEO 汇报工作的所有人
select distinct employee_id
from
(select employee_id from Employees where manager_id=1
union all
(select employee_id from Employees
where manager_id in (select employee_id from Employees where manager_id=1 ))
union all
select employee_id from Employees
where manager_id in (select employee_id from Employees
where manager_id in (select employee_id from Employees where manager_id=1 )))e
where employee_id!=1
SELECT e1.employee_id
FROM Employees e1
JOIN Employees e2 ON e1.manager_id = e2.employee_id
JOIN Employees e3 ON e2.manager_id = e3.employee_id
WHERE e1.employee_id != 1 AND e3.manager_id = 1
48.1412. 查找成绩处于中游的学生
select tmp.student_id,s.student_name
from
(select *,
if(dense_rank() over(partition by exam_id order by score desc)=1,1,0) d_rank,
if(dense_rank() over(partition by exam_id order by score )=1,1,0) a_rank
from Exam)tmp
left join Student s on tmp.student_id=s.student_id
group by tmp.student_id
having sum(d_rank)=0 and sum(a_rank)=0
order by tmp.student_id
⭐⭐⭐49.1767. 寻找没有被执行的任务对
with recursive table1 as(select task_id,subtasks_count subtask_id
from Tasks
union all
select task_id,subtask_id-1 from table1 where subtask_id > 1)
select task_id,subtask_id
from table1
left join Executed E using(task_id, subtask_id)
where E.task_id is null
⭐⭐⭐1225. 报告系统状态的连续日期
select type as period_state,min(date) as start_date,max(date) as end_date
from
(select type,date,subdate(date,row_number() over(partition by type order by date))as diff
from
(select "failed" as type,fail_date as date from Failed
union all
select "succeeded" as type,success_date as date from Succeeded)tmp1)tmp2
where date between "2019-01-01" and "2019-12-31"
group by type,diff
order by start_date