SQL training 2

12 Information about the highest salary of current employees in all departments

, give dept_no/emp_no And corresponding salary

select d.dept_no,d.emp_no,max(salary)
from dept_emp d
inner join salaries s
on d.emp_no=s.emp_no
where d.to_date=s.to_date
group by d.dept_no

13 title The table is obtained according to title Grouping

The number of each group is greater than or equal to 2, give titile And the corresponding number

select title,count(*) t
from titles
group by title
having count(*) >=2;

15 lookup employees Table all emp_no It's odd

And last_name Not for mary Employee information , according to hire_date The reverse

select * 
from employees e
where e.emp_no%2=1
//where e.emp_no mod 2 =1
and e.last_name is not 'mary'
//and e.last_name <> 'mary'
order by e.hire_date desc

16 Count the current title type

The current average salary of the corresponding employee ; The results show that title And the average wage avg

select t.title,avg(s.salary) avg
from titles t
inner join salaries s
on t.emp_no=s.emp_no
where t.to_date='9999-01-01'
and s.to_date='9999_01_01'
group by t.title;

17 Get the second highest salary at present

Of our employees dept_no And employee salary

select emp_no,salary
from salaries
where to_date='9999-01-01'
and salary=(
	select distinct salary
    from salaries
    order by salary desc
    limit 1,1
);

18 The number of the employee with the second highest salary

emp_no, salary salary,last_name as well as first_name, Do not use order by

select e.emp_no,max(s.salary),e.last_name,e.first_name
from employees e
inner join salaries s
on e.emp_no =s.emp_no
where s.to-date='lll'
and s.salary not in(
    select max(salary)
    from salaries
    where to_date='jjj'
);

19 All employees last_name/first_name/dept_name

Including employees who have not been assigned departments for the time being （ Less departments , The employee is on the left ）

incomprehension , What does three tables mean ;

select e.last_name.e.first_name,d.dept_name
from employees
left join dept_emp de on e.dept_no=de.dept_no
left join departments d on d.dept_no=de.dept_no;

20 Find the employee number emp_no=100001 The salary increase value of employees since their entry growth

select 
(
    (select salary from salaries where emp_no=10001 order by salary)
    -(select salary from salaries where emp_no=10001 order by salary)
) growth;

21 Find out the salary increase of all employees since their entry

select emp_no, (s1.salary-s2.salary) as growth
from employees e
inner join salaries s1
on e.emp_no =s1.emp_no and s1.to_date='9999-01-01'
inner join salaries s2
on e.emp_no=s2.emp_no and e.hire_date=s2.from_date
order by growth asc;

22 The total number of employee increases corresponding to each department

dept_no dept_name frequency sum

select d.dept_no,d.dept_name,count(s.salary) 'sum'
from departments d
inner join dept_emp de on d.dept_no=de.dept_no
inner join salaries s on de.emp_no=s.emp_no
group by d.dept_no;

23 The current salary of all employees shall be in accordance with salary Sort

select emp_no,salary,count(distinct s2.salary) rank
from salaries s1,salaries s2
where s1.salary<=s2.salary
and s1.to_date='9999-01-01'
and s2.to_date='9999-01-01'
group by s1.emp_no
order by s1.salary desc , s1.emp_no asc

24 All non manager Current salary of employees

Ideas 1 Inquire about all employees

2 Filter out manager staff

select de.depts_no,e.emp_no,s.salary
from employees e
inner join salaries s
on e.emp_no=s.emp_no
and s.to_date=''
inner join dept_emp de
on e.emp_no=de.emp_no
where de.emp_no not in(
    select emp_no
    from dept_manager dm
    where dm.to_date=''
)