每台机器的进程平均运行时间
题目
1661. 每台机器的进程平均运行时间 - 力扣(LeetCode)
思路
首先应该了解下面几个函数
- round:把数值字段舍入为指定的小数位数
- avg:用于计算一组值或表达式的平均值
然后将activity根据题目要求连接起来
代码
sql
# Write your MySQL query statement below
select a.machine_id,
round(avg(a.timestamp-b.timestamp),3) as processing_time
from Activity as a
join Activity as b
on a.machine_id=b.machine_id
and a.process_id=b.process_id
and b.activity_type = 'start'
and a.activity_type = 'end'
group by a.machine_id
员工奖金
题目
思路
注意点在于不能写成b.bonus = null而要写成b.bonus is null
代码
sql
# Write your MySQL query statement below
select e.name,b.bonus
from employee as e
left join bonus as b
on e.empId=b.empId
where b.bonus<1000
or b.bonus is null
学生们参加各科测试的次数
题目
1280. 学生们参加各科测试的次数 - 力扣(LeetCode)
思路
- Student表和Subjects表进行笛卡尔积连接(在第一点的基础上拼接Examinations中的每个学生参加每门科目的数量。
- 学生名单必须完整,在Examinations表中不存在则为0。所以使用左连接LEFT JOIN进行连接
代码
sql
# 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 as s
join subjects as su
left join examinations as e
on s.student_id=e.student_id
and e.subject_name = su.subject_name
group by s.student_name,su.subject_name
order by s.student_id,su.subject_name
至少有五名直接下属的经理
题目
570. 至少有5名直接下属的经理 - 力扣(LeetCode)
思路
使用子查询先找出有五个直接下属的经理id,再找出具体的员工名字
了解having函数
mysql中,当我们用到聚合函数,如sum,count后,又需要筛选条件时,having就派上用场了,因为WHERE是在聚合前筛选记录的,having和group by是组合着用的
注意where之后要用in,不能写等于号
代码
sql
# Write your MySQL query statement below
select name from employee
where id in
(select managerId from employee
group by managerId
having count(managerId)>=5)
确认率
题目
思路
主要在于avg函数的应用
ifnull函数:把null值转化成对应的值
代码
sql
# Write your MySQL query statement below
select
s.user_id,
ROUND(IFNULL(AVG(c.action='confirmed'), 0), 2) AS confirmation_rate
from
Signups as s
left join
Confirmations as c
on
s.user_id = c.user_id
group by
s.user_id