# Write your MySQL query statement below
select id,count(*) num
from (
(
select requester_id id
from requestAccepted
)
union all
(
select accepter_id id
from requestAccepted
)
) tt
group by id
order by num desc
limit 1 ;
# Write your MySQL query statement below
select round(sum(tiv_2016),2) tiv_2016
from(
select *,
count(pid) over (partition by tiv_2015) as num_tiv,
count(pid) over (partition by lat,lon) as num_city
from insurance
)h
where h.num_tiv >1 and num_city =1;
# Write your MySQL query statement below
select Department,Employee,Salary
from (
select d.Name as Department,
e.Name as Employee,
e.Salary as Salary,
dense_rank() over ( partition by DepartmentId order by Salary desc) as rk
from Employee as e, Department as d
where e.DepartmentId = d.Id
) m
where rk <=3;
法2:
select d.Name as Department,e.Name as Employee,e.Salary as Salary
from Employee as e left join Department as d
on e.DepartmentId = d.Id
where e.Id in
(
select e1.Id
from Employee as e1 left join Employee as e2
on e1.DepartmentId = e2.DepartmentId and e1.Salary < e2.Salary
group by e1.Id
having count(distinct e2.Salary) <= 2
)
and e.DepartmentId in (select Id from Department)
order by d.Id asc,e.Salary desc
# Write your MySQL query statement below
select user_id,
concat(upper(substr(name,1,1)),lower(substr(name,2))) name
from users
order by user_id;
# Write your MySQL query statement below
select patient_id,patient_name,conditions
from patients
where conditions like '%DIAB1'
or conditions like 'DIAB1%'
or conditions like '% DIAB1%';
# Write your MySQL query statement below
delete u
from Person u , Person v
where v.id < u.id and u.email = v.email
# Write your MySQL query statement below
select ifNull((
select distinct Salary
from Employee
order by Salary desc limit 1,1),null) as SecondHighestSalary;
要想获取第二高,需要排序,使用 order by(默认是升序 asc,即从小到大),若想降序则使用关键字 desc
去重,如果有多个相同的数据,使用关键字 distinct 去重
判断临界输出,如果不存在第二高的薪水,查询应返回 null,使用 ifNull(查询,null)方法
起别名,使用关键字 as ...
因为去了重,又按顺序排序,使用 limit()方法,查询第二大的数据,即第二高的薪水,即 limit(1,1) (因为默认从0开始,所以第一个1是查询第二大的数,第二个1是表示往后显示多少条数据,这里只需要一条)
为什么最外层没有加上 from employee?如果没有 ifNull() 函数,单纯只是嵌套SQL,加上 from employee 没有问题;但问题是 select ifNull 是搭配使用的,并不属于嵌套查询
# Write your MySQL query statement below
select sell_date,
count(distinct product) num_sold,
group_concat(distinct product) products
from Activities
group by sell_date
order by sell_date;
# Write your MySQL query statement below
select product_name, sum(unit) unit
from Products join Orders using (product_id)
where order_date like "2020-02%"
group by product_name
having unit >= 100;
# Write your MySQL query statement below
select *
from users
where mail regexp '^[a-zA-A]+[a-zA-Z0-9_\\./\\-]*@leetcode\\.com$'
以上是今天要讲的内容,练习了一些SQL题。