MySQL Practice Summary -

Case 1:

Known Information
Employee Table Employee

Name	Type	Description
EmployeeId	Int	Employee id (unique)
Name	Varchar(50)	Employee Name
Age	Int	Employee Age
Email	Varchar(50)	Staff Email
Level	Int	Employee Level

Salary salary

Name	Type	Description
SalaryId	Int	salary id (unique)
EmployeeId	Int	employee id
Money	Double	Wage Amount
Date	Datetime	release date

Questions and Answers
Query the names and email addresses of occasional employees, and sort by age in descending order and name in ascending order.

select Name,Emailfrom Employeeorder by Age asc,Name desc;

Query the names of all employees with employee level 1 and the sum of their corresponding salary amounts.

①select e.Name,s.Moneyfrom Employee einner join Salary son e.EmployeeId = s.EmployeeIdwhere e.Level = 1

Query all salary payment information higher than average salary and corresponding employee information (payment date, salary amount, employee id, employee name, employee age, employee email, employee level)

select s.Date,s.Money,s.EmployeeId,e.Name,e,Age,e.Email,e.Levelfrom Employee einner join Salary son e.EmployeeId = s.EmployeeIdwhere s.Money >(select avg(Money)from Salary);