Chapter IX sub query (key)

Chapter nine Subquery

# Query salary ratio of employees Abel Many people
# Subquery
SELECT e2.last_name,e2.salary
FROM employees e1,employees e2
WHERE e1.last_name=‘Abel’ AND e1.salary<e2.salary;

# Subquery
SELECT last_name,salary
FROM employees
WHERE salary>(
SELECT salary
FROM employees
WHERE last_name=‘Abel’
);

# Queries can be divided into single line sub queries and multi line sub queries , It can also be divided into related sub queries and unrelated sub queries

# Correlation subquery ： Query employee information whose salary is greater than the average salary of the Department
SELECT last_name,department_id,salary
FROM employees
WHERE salary>(
SELECT AVG(salary)
FROM employees
)
GROUP BY department_id;

# Uncorrelated subqueries ： Query the information of employees whose salary is greater than the average salary of the company
SELECT last_name,salary
FROM employees
WHERE salary>(
SELECT AVG(salary)
FROM employees
)

# Query salary greater than 149 Information of the employee whose salary is No
SELECT employee_id,last_name,salary
FROM employees
WHERE salary>(
SELECT salary
FROM employees
WHERE employee_id=149
);

# return job_id And 141 Same as employee No ,salary Than 143 The name of the employee with a large number of employees ,job_id And wages
SELECT last_name,job_id,salary
FROM employees
WHERE job_id=(
SELECT job_id
FROM employees
WHERE employee_id=141
) AND salary>(
SELECT salary
FROM employees
WHERE employee_id=143
);

# Return to the lowest paid employees of the company last_name,job_id and salary
SELECT last_name,job_id,salary
FROM employees
WHERE salary=(
SELECT MIN(salary)
FROM employees
);

# Query and 141 Number (employee_id) Staff manager_id and department_id Of the same other employees employee_id,manager_id,department_id

# Mode one ： Unpaired comparison , Compare them separately
SELECT employee_id,manager_id,department_id
FROM employees
WHERE manager_id=(
SELECT manager_id
FROM employees
WHERE employee_id=141
)
AND department_id=(
SELECT department_id
FROM employees
WHERE employee_id=141
)
AND employee_id<>141;

# Mode two ： Compare in pairs
SELECT employee_id,manager_id,department_id
FROM employees
WHERE (manager_id,department_id)=(
SELECT manager_id,department_id
FROM employees
WHERE employee_id=141
)
AND employee_id <>141;

# Query and 141 Number or 174 Staff manager_id and department_id Of the same other employees employee_id,manager_id,department_id

# Mode one ： Unpaired comparison
SELECT employee_id,manager_id,department_id
FROM employees
WHERE manager_id IN
(SELECT manager_id
FROM employees
WHERE employee_id IN(141,174))
AND department_id IN
(SELECT department_id
FROM employees
WHERE employee_id IN(141,174))
AND employee_id NOT IN(141,174);

# Compare in pairs
SELECT employee_id,manager_id,department_id
FROM employees
WHERE (manager_id,department_id)IN(
SELECT manager_id,department_id
FROM employees
WHERE employee_id IN(141,174))
AND employee_id NOT IN(141,174);

#having Subqueries in
# The minimum wage is greater than 50 The Department of minimum wage id And its minimum wage
SELECT department_id,MIN(salary)
FROM employees
GROUP BY department_id
HAVING MIN(salary)>(
SELECT MIN(salary)
FROM employees
WHERE department_id=50
);

#case Subqueries in

# Displays the employee's employee_id,last_name and location, among , If employee department_id And location_id by 1800 Of department_id identical , be lacation by ’Canada’, Others are ’USA’
SELECT employee_id,last_name,
(CASE department_id
WHEN
(SELECT department_id FROM departmentsemployees
WHERE location_id=1800)
THEN ‘Canada’ ELSE ‘USA’ END)
AS location
FROM employees;

# Check the title and type of the book , among note The value is novel Show novel ,law Show legal ,
medicine Show medicine ,cartoon Show cartoon ,joke Show jokes
SELECT name’ Title ’,note,
CASE note
WHEN’novel’ THEN’ A novel ’
WHEN’law’THEN’ law ’
WHEN’medicine’THEN’ medicine ’
WHEN’cartoon’THEN’ cartoon ’
WHEN’joke’THEN’ joke ’
ELSE’ other ’
END
AS ‘ type ’
FROM books;

# Check the title of the book 、 stock , among num Value exceeds 30 Ben's , Show unsalable , Greater than 0 And below 10 Of , Show best sellers , by 0 The display needs to be out of stock
SELECT nameAS’ Title ’,num AS’ stock ’,
CASE WHEN num>30 THEN’ Unsalable ’
WHEN 0<num<10 THEN’ Sell well ’
WHEN num=0 THEN’ No goods ’
ELSE ‘ normal ’
END
AS ‘ Display state ’
FROM books;

# Null value of subquery : Because the subquery inside is empty , No call Haas People who , All eventually did not return any rows
SELECT last_name,job_id
FROM employees
WHERE job_id=(
SELECT job_id
FROM employees
WHERE last_name=‘Haas’
)

# Illegal use of subquery : Using equal to will report an error , Because this sub query will return multiple rows of data , Change the equal sign to in that will do
SELECT employee_id,last_name
FROM employees
WHERE salary =(
SELECT MIN(salary)
FROM employees
GROUP BY department_id
)

# Because different employees receive the same minimum wage
SELECT employee_id,last_name
FROM employees
WHERE salary IN(
SELECT MIN(salary)
FROM employees
GROUP BY department_id
)

# Multi line sub query

# Operator of multi row subquery ：in() any all some( yes any Another name for , It's usually used any)

# Return to other job_id Middle ratio job_id by ’IT_PROG’ The employee number of any employee in the Department with low salary 、 full name 、job_id as well as salary
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<ANY(
SELECT salary
FROM employees
WHERE job_id=‘IT_PROG’)
AND job_id!=‘IT_PROG’

SELECT* FROM employees
WHERE employee_id!=101;

# Return to other job_id Middle ratio job_id by ’IT_PROG’ Employee number with low salary in the Department 、 full name 、job_id as well as salary
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<(
SELECT MIN(salary)
FROM employees
WHERE job_id=‘IT_PROG’)
AND job_id!=‘IT_PROG’
# perhaps
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<ALL(
SELECT salary
FROM employees
WHERE job_id=‘IT_PROG’)
AND job_id!=‘IT_PROG’

# Query the Department with the lowest average wage id
# Mode one : Or is it a way to use any,all This is easy to understand
SELECT department_id
FROM employees
GROUP BY department_id
HAVING AVG(salary)<=ALL(
SELECT AVG(salary)
FROM employees
GROUP BY department_id
)

# Mode two : The three field aliases must have , Otherwise, the report will be wrong ,every derived table must have its own itias
# Each reserved table should have its own alias
SELECT department_id
FROM employees
GROUP BY department_id
HAVING AVG(salary)=(
SELECT MIN(avg_sal)
FROM (SELECT AVG(salary)avg_sal
FROM employees
GROUP BY department_id)dept_avg_sal
)

# Look at the first mistake below , The second correct : Nested functions should be named by their aliases
SELECT MIN(AVG(salary))
FROM (
SELECT AVG(salary)
FROM employees
GROUP BY department_id
)

SELECT MIN(ll)
FROM (
SELECT AVG(salary)ll
FROM employees
GROUP BY department_id
)kkk

# Correlation subquery : More important , Every time, you need to make an internal query
# review ： Query employees whose salary is greater than the average salary of the Department last_name,salary And its department_id
SELECT last_name,salary,department_id
FROM employees
WHERE salary>(
SELECT AVG(salary)
FROM employees
)

# Query employees whose salary is greater than 60 Of the average salary of department No last_name,salary And its department_id
SELECT last_name,salary,department_id
FROM employees
WHERE salary>(
SELECT AVG(salary)
FROM employees
WHERE department_id=60
)

# Correlation subquery ： Query employees whose salary is greater than 60 Of the average salary of department No last_name,salary And its department_id Query employees whose salary is greater than the average salary of the Department last_name,salary And its department_id
SELECT last_name,salary,department_id
FROM employees e1
WHERE salary>(
SELECT AVG(salary)
FROM employees
WHERE department_id=e1.department_id
)

# stay from Use subqueries in
SELECT last_name,salary,e1.department_id
FROM employees e1,(
SELECT department_id,AVG(salary)dept_avg_sal # It appears as a field of the table instead of a function , Alias is required
FROM employees
GROUP BY department_id)e2
WHERE e1.department_id=e2.department_id
AND e2.dept_avg_sal<e1.salary;

# Check the employee's id,salary, according to department_name Sort
SELECT e1.department_id,employee_id,salary
FROM employees e1
ORDER BY(
SELECT department_name
FROM departments e2
WHERE e1.department_id=e2.department_id
)ASC;

# Check the employee's id,salary, according to department_name Sort
# If there are fields in the table of the query and between the two tables join, The fields of the table should be specified
SELECT employee_id,salary
FROM employees e1
ORDER BY(
SELECT department_id
FROM departments e2
WHERE e1.department_id=e2.department_id
);

# if employees In the table employee_id And job_history In the table employee_id The same number is not less than 2, Output these same
#id The employees' employee_id,last_name,job_id
# The following is wrong
SELECT employees.employee_id,last_name,employees.job_id
FROM employees JOIN job_history
ON employees.employee_id=job_history.department_id;

# The correct ones are as follows ,count(*) Can be written as count( Field name )
SELECT e.employee_id,last_name,e.job_id
FROM employees e
WHERE 2<=(
#select count() // Sure count() perhaps count( Field )
SELECT COUNT(employee_id)
FROM job_history
WHERE employee_id=e.employee_id
);

#exists And not exists keyword
# The following two questions are self connected
# Query the company employees employee_id,last_name,job_id,department_id Information
SELECT employee_id,last_name,job_id,department_id
FROM employees
WHERE employee_id NOT IN(
SELECT manager_id
FROM employees
WHERE manager_id IS NOT NULL
);

# Query company managers employee_id,last_name,job_id,department_id Information : Self join , want distinct De emphasis on managers
SELECT DISTINCT e2.employee_id,e2.last_name,e2.job_id,e2.department_id
FROM employees e1
JOIN employees e2
ON e1.manager_id=e2.employee_id

# Query company managers employee_id,last_name,job_id,department_id Information ： Subquery used
SELECT employee_id,last_name,job_id,department_id
FROM employees
WHERE employee_id IN(
SELECT manager_id
FROM employees
);

# Query company managers employee_id,last_name,job_id,department_id Information ： use exists Realization
SELECT employee_id,last_name,job_id,department_id
FROM employees e1
WHERE EXISTS(
SELECT * FROM employees e2
WHERE e1.employee_id=e2.manager_id
)

# Inquire about departments In the table , There is no such thing as employees Of the departments in the table department_id and department_name
# That is, find out that there are no employees in some departments
# Mode one ：
SELECT d.department_id,d.department_name
FROM employees e RIGHT JOIN departments d
ON e.department_id=d.department_id
WHERE e.department_id IS NULL;

# Mode two
SELECT department_id,department_name
FROM departments d
WHERE EXISTS(
SELECT* FROM employees e
WHERE d.department_id=e.department_id
);