如何计算mysql中重复的project_no

Question

想象一下，我有 3 个表员工详细信息、项目详细信息和 Works on（其中设置了哪个员工将在哪个项目上工作）。 问题是要显示那些从事超过三名员工的项目的所有员工的平均工资。

create table Employee
(
empid varchar(5) primary key,
ename varchar(20),
address varchar(50),
phone bigint(10),
job varchar(10),
salary varchar(10)
);

insert into Employee values ('e101','Ankita','Delhi',9987654321,'Manager','12000');
insert into Employee values ('e102','Rahul','Banglore',9219849816,'Developer','80000');
insert into Employee values ('e103','Aditya','Noida',9517536842,'Chef','16000');
insert into Employee values ('e104','Rachna','Mumbai',7539561682,'Teacher','45000');
insert into Employee values ('e105','Karan','Delhi',8529517463,'Driver','18000');


create table Project
(
project_no varchar(5) primary key,
pname varchar(20),
no_of_hours int(3)
);

Insert into Project values('p101', 'Data Handling', 5);
Insert into Project values('p102', 'Game Develop', 3);
Insert into Project values('p103', 'New Recipe', 6);
Insert into Project values('p104', 'Question Paper', 9);
Insert into Project values('p105', 'Ride to Shimla', 12);


create table Works_on
(
empid varchar(5),
project_no varchar(5),
start_date date,
end_date date,
manager_name varchar(20),
primary key(empid, project_no),
foreign key(empid) references Employee(empid),
foreign key(project_no) references Project(project_no)
);

Insert into Works_on value('e101', 'p101', '2021-02-24', '2021-03-25', 'Anshika');
Insert into Works_on value('e102', 'p101', '2021-03-4', '2021-04-2', 'Anshika');
Insert into Works_on value('e103', 'p102', '2021-02-4', '2021-03-5', 'Roshan');
Insert into Works_on value('e104', 'p101', '2021-05-20', '2021-05-29', 'Anshika');
Insert into Works_on value('e105', 'p103', '2021-02-4', '2021-02-5', 'Dharmesh');
Insert into Works_on value('e103', 'p101', '2021-02-4', '2021-02-5', 'Anshika');
Insert into Works_on value('e105', 'p104', '2021-02-4', '2021-02-5', 'Ritika');
Insert into Works_on value('e105', 'p105', '2021-02-4', '2021-02-5', 'Sid');

Answer 1

select w.project_no project_number, 
       count(w.project_no) number_of_employees,
       avg(e.salary) average_salary 
from 
    Employee e, Project p, Works_on w 
where 
    w.project_no= p.project_no 
    and w.empid=e.empid  
    group by w.project_no 
    having(count(w.project_no))>3;

DBFiddle