高频SQL50题（基础版）-2

主要内容

1. LeetCode–高频·SQL50题（基础版）11-20

一.SQL练习题

1.577-员工奖金

代码如下（示例）:

# Write your MySQL query statement below
select e.name,b.bonus
from employee e 
left join bonus b 
on e.empId = b.empId
where b.bonus<1000 or b.bonus is null;

2.1280-学生们参加各科测试的次数

代码如下（示例）:

# Write your MySQL query statement below
select s.student_id,s.student_name,su.subject_name,
    count(e.subject_name) as attended_exams
from students s 
cross join subjects su
left join examinations e 
on  e.student_id = s.student_id
and e.subject_name = su.subject_name
group by s.student_id,su.subject_name
order by s.student_id,su.subject_name; 

3.570-至少有5名直接下属的经理

代码如下（示例）:

# Write your MySQL query statement below
select name
from employee
where id in (
    select managerId from employee 
    group by managerId
    having count(*) >=5
);

4.1934-确认率

代码如下（示例）:

# Write your MySQL query statement below
select s.user_id,
    round(avg(if(c.action = 'confirmed',1,0)),2)confirmation_rate
from signups s 
left join Confirmations c 
on s.user_id = c.user_id
group by s.user_id;

#confirmation_rate 列使用了 ROUND 函数将平均值保留两位小数。
#IF 函数用于判断 Confirmations 表中的 action 列是否为'confirmed'，如果是则返回1，否则返回0。

5.620-有趣的电影

代码如下（示例）:

# Write your MySQL query statement below
select id,movie,description,rating
from cinema
where description != 'boring' and mod(id,2) = 1
order by rating desc;

6. 1251-平均售价

代码如下（示例）:

# Write your MySQL query statement below
#select p.product_id,round(sum(u.units * p.price)/sum(u.units),2) average_price
#from prices p
#left join UnitsSold u 
#on p.product_id = u.product_id
#where u.purchase_date between p.start_date and p.end_date
#group by p.product_id;

SELECT
    product_id,
    IFNULL(Round(SUM(sales) / SUM(units), 2), 0) AS average_price
FROM (
    SELECT
        Prices.product_id AS product_id,
        Prices.price * UnitsSold.units AS sales,
        UnitsSold.units AS units
    FROM Prices 
    LEFT JOIN UnitsSold ON Prices.product_id = UnitsSold.product_id
    AND (UnitsSold.purchase_date BETWEEN Prices.start_date AND Prices.end_date)
) T
GROUP BY product_id

7.1075-项目员工

代码如下（示例）:

# Write your MySQL query statement below
select project_id,round(avg(experience_years),2) average_years
from project p 
left join employee e 
on p.employee_id=e.employee_id
group by project_id;

8.1633-各赛事的用户注册率

代码如下（示例）:

# Write your MySQL query statement below
select contest_id,
  round(count(distinct user_id)/(select count(*) from users)*100,2) as percentage
from register
group by contest_id
order by percentage desc,contest_id;

9.1211-查询结果的质量和占比

代码如下（示例）:

# Write your MySQL query statement below
SELECT 
    query_name, 
    ROUND(AVG(rating/position), 2) quality,
    ROUND(SUM(IF(rating < 3, 1, 0)) * 100 / COUNT(*), 2) poor_query_percentage
FROM Queries
GROUP BY query_name

10.1193-每月交易

代码如下（示例）:

# Write your MySQL query statement below
 select 
  left(trans_date,7) month,
  country,
  count(state) trans_count,
  count(if(state='approved',1,null)) approved_count,
  sum(amount) trans_total_amount,
  sum(if(state='approved',amount,0)) approved_total_amount
from transactions
group by month,country

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总结

以上是今天要讲的内容，练习了一些 高频SQL题。