SQL cattle off topic

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Find all the information at the latest recruits

select * 
from 
employees
where 
hire_date = (select max(hire_date) from employees);

Find recruits ranked No. 3 time employees all the information

SELECT  *  
from  
the Employees 
Order  by  
the hire_date 
desc  - descending 
limit 2 , . 1 ; - from 3, an offset, i.e., item 3

Find current salary details of the current (to_date = '9999-01-01') leadership of the various departments and their corresponding department number dept_no

select s.*,d.dept_no
from salaries as s
join dept_manager as d
on d.emp_no=s.emp_no
where s.to_date='9999-01-01'
and d.to_date='9999-01-01';

Find all the staff have been assigned sectors last_name and first_name

select e.last_name,e.first_name,d.dept_no
from employees as e
join dept_emp as d
on e.emp_no = d.emp_no;

Find all employees last_name and first_name and the corresponding department number dept_no, but also show employees are not assigned specific sectors

SELECT e.LAST_NAME, e.first_name, d.dept_no
 from the Employees AS E
 left  the Join dept_emp AS D - sitting on the main table, the right table supplement, complement empty NULL 
ON e.emp_no = d.emp_no;

Find all employees recruited when the salary situation, given emp_no and salary, and in reverse order according to emp_no

select e.emp_no,s.salary
from salaries as s
join employees as e
where e.emp_no = s.emp_no
and e.hire_date = s.from_date
order by
e.emp_no
desc;

Find the number rose rose more than 15 times the salary of the employee number and its corresponding t emp_no

count (): number of statistical records

Also you need to group by emp_no records each employee's record in a display in
```
select emp_no,count(emp_no) as t
from salaries
group by
emp_no
having t > 15;
```
Find all current employees (to_date = '9999-01-01') specific salary salary, the same salary for the show only once, and are displayed in reverse order

distinct: remove duplicate
```
select distinct salary-- 去除重复
from salaries
where to_date = '9999-01-01'
order by
salary
desc;
```

Get the current salary situation in all sectors of the current manager, gives dept_no, emp_no and salary, currently represents to_date = '9999-01-01'

select d.dept_no,d.emp_no,s.salary
from dept_manager as d
join salaries as s
on d.emp_no=s.emp_no
where d.to_date='9999-01-01'
and s.to_date='9999-01-01';

Get all non-manager employees emp_no

select emp_no
from employees
where emp_no not in (select emp_no from dept_manager);

Get all the current staff of manager, if the current manager is not in their own words the results show that the current representation to_date = '9999-01-01'.

SELECT e.emp_no, m.emp_no AS manager_no
 from dept_emp AS E
 the Join dept_manager AS m
 ON e.dept_no = m.dept_no
 WHERE e.emp_no <> m.emp_no - this is that they do not display manager, <> Not equal 
and E .to_date = ' 9999-01-01 ' 
and m.to_date = ' 9999-01-01 ' ;

Get the current staff of the highest salaries in all sectors relevant information given dept_no, emp_no and their corresponding salary

select d.dept_no,d.emp_no,max(s.salary)
from dept_emp as d
join salaries as s
on d.emp_no = s.emp_no
group by
d.dept_no;

Obtained from the title titles table grouped, in groups of more than 2, and gives the corresponding title number t.
```
select title,count(title) as t
from titles
group by
title
having t >= 2;
```
Find all emp_no odd employees table, and last_name employee information is not Mary's, and arranged in reverse order hire_date
```
select * from employees
where emp_no%2 =1
and last_name <> 'Mary'
order by
hire_date
desc;
```
Current statistics of each title type corresponds to the current average wage of employees (to_date = '9999-01-01') corresponding salary. The results are given title, and the average wage avg.
```
select t.title,avg(s.salary)
from titles as t
join salaries as s
on t.emp_no = s.emp_no
where s.to_date='9999-01-01'
and t.to_date='9999-01-01'
group by
t.title;
```

Obtaining a second plurality of current emp_no (to_date = '9999-01-01') and its corresponding salary employees salary salary

SELECT , emp_no,, the salary from ual relations
 WHERE TO_DATE = ' 9999-01-01 ' 
Order  by 
the salary 
desc 
limit . 1 , . 1 ;   - in descending order, the second to take

Find the current salary (to_date = '9999-01-01') ranked second and more employees number emp_no, salary salary, last_name and first_name, are not allowed to use order by

select e.emp_no,s.salary,e.last_name,e.first_name
from employees as e
join salaries as s
on e.emp_no=s.emp_no
where s.to_date='9999-01-01'
order by
s.salary
desc
limit 1,1;

Without using order by: nested need to combine and select max

SELECT e.emp_no, max (the salary), e.LAST_NAME, e.first_name - where a max (the salary) 
from   ual relations AS S, the Employees AS E
 WHERE s.emp_no = e.emp_no
 and the salary <   - less than the maximum salary the largest is the second largest 
(
 SELECT  max (the salary)
 from ual relations
 WHERE TO_DATE =  ' 9999-01-01 ' 
);

Find last_name and first_name and corresponding dept_name all employees, including temporary employees are not assigned sector

Double left connection

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

Find employee number 10001 emp_no since its entry salary salary increase value growth

By entry time to sort the query:

If the direct use of the salary maximum - minimum, it may be the last time a pay cut.

SELECT ( 
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date DESC LIMIT 1) -
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date ASC LIMIT 1)
) AS growth

Find all the staff since the entry salary increase, the number of employees is given emp_no and its corresponding salary increase growth, and growth in ascending order

- the two results of the query, as two tables, for the Join 
SELECT S2.emp_no, (S2.salary - S1.salary) AS Growth
 from  
( SELECT e.emp_no, s.salary 
  from the Employees AS E 
  the Join ual relations AS S 
  ON e.emp_no = s.emp_no 
  and e.hire_date = s.from_date) AS S1   - all employees of the workers owned 
the Join  
( the SELECT e.emp_no, s.salary 
  from the employees AS E 
  the Join salaries ASS 
  ON e.emp_no = s.emp_no 
  and s.to_date = ' 9999-01-01 ' ) AS S2    - all employees of current wages 
ON S1.emp_no = S2.emp_no
 the Order  by Growth;

SQL cow brush off questions (continually updated)

SQL cattle off topic

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