Cattle off network SQL combat two brush | Day4

"Cattle off network SQL combat two brush" is a series of learning notes Bowen, SQL parsing six topics every day ~ ~ Today is the first title 19-24

The format of each note roughly, three major sections:

• Outline
• Title (Title Description, ideas, codes, related reference materials / Q)
• review

First, outline

Question number	Knowledge Point
19	LEFT JOIN
20	Nested SELECT
21	Nested SELECT
22	GROUP BY， COUNT()
23	Table reuse, DISTINCT, ORDER BY, GROUP BY
24	NOT IN

Second, the title

19. Find all the staff and first_name last_name and corresponding dept_name

• Title Description

Find last_name and first_name and corresponding dept_name all employees, including temporary no employees assigned sector of
the CREATE TABLE departments(
dept_nochar (4) the NOT NULL,
dept_nameVARCHAR (40) the NOT NULL,
PRIMARY KEY ( dept_no));
the CREATE TABLE dept_emp(
emp_noint (11) the NOT NULL,
dept_nochar (. 4) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no, dept_no));
the CREATE TABLE employees(
emp_noint (. 11) the NOT NULL,
birth_dateDATE the NOT NULL,
first_nameVARCHAR (14) the NOT NULL,
last_nameVARCHAR (16) NULL the NOT,
genderchar (. 1) the NOT NULL,
hire_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no));

• Output Description

last_name	first_name	dept_name
Facello	Georgi	Marketing
Omission	Omission	Omission
Lock	Mary	NULL

• Thinking

1. The first table LEFT JOIN employees and dept_emp table, and obtain last_name FIRST_NAME;
2. The second table LEFT JOIN departments give dept_name,;
3. Because Find all employees, "not including temporary staff distribution sector", so use LEFT JOIN.

• Code

SELECT e.last_name, e.first_name, d.dept_name
FROM employees AS e LEFT JOIN dept_emp AS de ON e.emp_no = de.emp_no
LEFT JOIN departments AS d ON de.dept_no = d.dept_no

20. Find emp_no employee number 10001 from their salary salary increase since the entry value growth

• Title Description

Find emp_no employee number 10001 from their salary salary increase since the entry value Growth
the CREATE TABLE salaries(
emp_noint (11) the NOT NULL,
salaryint (11) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
PRIMARY KEY ( emp_no, from_date));

• Output Description

growth
28841

• Ideas
  were found emp_no 10001 employees current salary and the entry salary, do difference is the growth

• Code

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 from_date ASC LIMIT 1)
)AS growth

21. Find all the staff since the entry salary increase case

• Title Description

Find a salary increase of all employees from entry since, given the employee number emp_no and its corresponding salary increase growth, and in ascending order according to Growth
the CREATE TABLE employees(
emp_noint (11) the NOT NULL,
birth_dateDATE the NOT NULL,
first_nameVARCHAR (14) the NOT NULL,
last_nameVARCHAR (16) the NOT NULL,
genderchar (. 1) the NOT NULL,
hire_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no));
the CREATE TABLE salaries(
emp_noint (. 11) the NOT NULL,
salaryint (. 11) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no, from_date));

• Output Description

emp_no	growth
10011	0
Omission	Omission
54496	10004

• Thinking

• Produced two tables, a record current salary, a record entry salary;

• Emp_no connected with two tables, corresponding to the deviation to make salary growth.

• Code

SELECT sCurrent.emp_no, (sCurrent.salary - sStart.salary) AS growth
FROM (SELECT e.emp_no, s.salary FROM employees AS e, salaries AS s WHERE e.emp_no = s.emp_no AND s.to_date = '9999-01-01') AS sCurrent,
(SELECT e.emp_no, s.salary FROM employees AS e, salaries AS s WHERE e.emp_no = s.emp_no AND e.hire_date = s.from_date) AS sStart
WHERE sCurrent.emp_no = sStart.emp_no
ORDER BY growth

22. The statistical sum corresponding to the number of employees rose in all sectors

• Title Description

The sum of the number of times corresponding to various departments of employees statistical gains, coding dept_no given sector, the department name and the number dept_name SUM
the CREATE TABLE departments(
dept_nochar (. 4) the NOT NULL,
dept_nameVARCHAR (40) the NOT NULL,
a PRIMARY KEY ( dept_no));
the CREATE TABLE dept_emp(
emp_noint ( . 11) the NOT NULL,
dept_nochar (. 4) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no, dept_no));
the CREATE TABLE salaries(
emp_noint (. 11) the NOT NULL,
salaryint (. 11) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no, from_date));

• Output Description

dept_no	dept_name	sum
d001	Marketing	24
d002	Finance	14
d003	Human Resources	13
d004	Production	24
d005	Development	25
d006	Quality Management	25

• Thinking
• First connection table and salaries dept_emp table, and to obtain, emp_no, dept_no;
• Reconnection departments table, to give dept_name,;
• According dept_no packet, () with the count COUNT with GROUP BY.
• Code

SELECT de.dept_no, d.dept_name, COUNT(s.emp_no) AS sum
FROM (dept_emp AS de INNER JOIN salaries AS s ON de.emp_no = s.emp_no)
    INNER JOIN departments AS d ON de.dept_no = d.dept_no
GROUP BY de.dept_no

23. The salary of all employees are ranked according to 1-N in accordance with the salary

• Title Description

All current employees (to_date = '9999-01-01') according to the salary paid in accordance with the ranking of 1-N, the same salary in parallel and arranged in ascending order emp_no
the CREATE TABLE salaries(
emp_noint (. 11) the NOT NULL,
salaryint (. 11) the NOT NULL,
from_dateDATE the NOT NULL,
to_dateDATE the NOT NULL,
a PRIMARY KEY ( emp_no, from_date));

• Output Description

emp_no	salary	rank
10005	94692	1
10009	94409	2
10010	94409	2
10001	88958	3
10007	88070	4
10004	74057	5
10002	72527	6
10003	43311	7
10006	43311	7
10011	25828	8

• Thinking

1. Re-rank comparison with salaries table;
2. It is the second rank in the table equal to the number greater than the number of the current salary, since the ranking parallel, should be used to re DISTINCT;
3. According emp_no packet;
4. Emp_no salary and arranged in reverse order.

• Code

SELECT s1.emp_no, s1.salary, COUNT(DISTINCT s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01' AND s2.to_date = '9999-01-01' AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no

24. Get the current salaries of all non-manager employees

• Title Description

获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date=‘9999-01-01’
CREATE TABLE dept_emp (
emp_no int(11) NOT NULL,
dept_no char(4) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,dept_no));
CREATE TABLE dept_manager (
dept_no char(4) NOT NULL,
emp_no int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,dept_no));
CREATE TABLE employees (
emp_no int(11) NOT NULL,
birth_date date NOT NULL,
first_name varchar(14) NOT NULL,
last_name varchar(16) NOT NULL,
gender char(1) NOT NULL,
hire_date date NOT NULL,
PRIMARY KEY (emp_no));
CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,from_date));

• 输出描述

dept_no	emp_no	salary
d001	10001	88958
d004	10003	43311
d005	10007	88070
d006	10009	95409

• 思路

1. 连接dept_emp表和salaries表得到dept_no，emp_no和salary
2. 「所有非manager员工」，因此要排除dept_manager表中的员工emp_no

• 代码

SELECT de.dept_no, de.emp_no, s.salary
FROM dept_emp AS de INNER JOIN salaries AS s ON de.emp_no = s.emp_no AND s.to_date = '9999-01-01'
WHERE de.emp_no NOT IN (SELECT emp_no FROM dept_manager)

三、回顾

知识点	题号
LEFT JOIN	19
ORDER BY	23
表的复用	23
DISTINCT	23
NOT IN	24
函数	22「COUNT()」
SELECT 嵌套	20, 21
GROUP BY	22, 23