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add SQL 30 Questions
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Diff for: ‎SQL 30 Questions/30_Simple_ SQL_Queries.sql

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/****************************************/
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/* 30 Simple SQL Interview Queries */
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/****************************************/
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/*1. Delete table Employee, Department and Company.*/
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DROP TABLE IF EXISTS Employee;
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DROP TABLE IF EXISTS Department;
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DROP TABLE IF EXISTS Company;
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/*
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2. Create tables:
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Employee with attributes (id, name, city, department, salary)
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Department with attributes (id, name)
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Company with attributes (id, name, revenue)
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*/
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CREATE TABLE department(
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id INT PRIMARY KEY AUTO_INCREMENT,
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name VARCHAR(50) NOT NULL
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);
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CREATE TABLE company(
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id INT PRIMARY KEY AUTO_INCREMENT,
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name VARCHAR(100) NOT NULL,
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revenue INT
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);
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CREATE TABLE Employee(
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id INT PRIMARY KEY AUTO_INCREMENT,
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name VARCHAR(150) NOT NULL,
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city VARCHAR(150) NOT NULL,
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department_id INT NOT NULL,
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salary INT NOT NULL,
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FOREIGN KEY (department_id) REFERENCES department(id)
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);
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/*
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4. Add rows into Department table
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(1, 'IT'),
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(2, 'Management'),
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(3, 'IT'),
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(4, 'Support');
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*/
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INSERT INTO department(name)
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VALUES
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('IT'),
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('Management'),
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('IT'),
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('Support');
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/*
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5. Add rows into Company table
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(1, 'IBM', 2000000),
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(2, 'GOOGLE', 9000000),
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(3, 'Apple', 10000000);
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*/
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INSERT INTO company(name,revenue)
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VALUES
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('IBM', 2000000),
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('GOOGLE', 9000000),
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('Apple', 10000000);
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/*
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3.Add rows into employee table:
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(1, 'David', 'London', 'IT', 80000),
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(2, 'Emily', 'London', 'IT', 70000),
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(3, 'Peter', 'Paris', 'IT', 60000),
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(4, 'Ava', 'Paris', 'IT', 50000),
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(5, 'Penny', 'London', 'Management', 110000),
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(6, 'Jim', 'London', 'Management', 90000),
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(7, 'Amy', 'Rome', 'Support', 30000),
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(8, 'Cloe', 'London', 'IT', 110000);
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*/
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INSERT INTO employee (name,city,department_id,salary)
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VALUES
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('David', 'London', 3, 80000),
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('Emily', 'London', 3, 70000),
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('Peter', 'Paris', 3, 60000),
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('Ava', 'Paris', 3, 50000),
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('Penny', 'London', 2, 110000),
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('Jim', 'London', 2, 90000),
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('Amy', 'Rome', 4, 30000),
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('Cloe', 'London', 3, 110000);
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/*
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6. Query all rows from Department table
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*/
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SELECT * FROM department;
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/*
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7. Change the name of department with id = 1 to 'Management'
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*/
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UPDATE department
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SET name = 'Management'
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WHERE id = 1;
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/*
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8. Delete employees with salary greater than 100 000
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*/
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DELETE FROM employee
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WHERE salary > 100000;
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/*
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9. Query the names of companies
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*/
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SELECT name FROM company;
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/*
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10. Query the name and city of every employee
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*/
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SELECT name, city
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FROM employee;
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/*
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11. Query all companies with revenue greater than 5 000 000
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*/
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SELECT * FROM company
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WHERE revenue > 5000000;
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/*
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12. Query all companies with revenue smaller than 5 000 000
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*/
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SELECT * FROM company
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WHERE revenue < 5000000;
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/*
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13. Query all companies with revenue smaller than 5 000 000, but you cannot use the '<' operator
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*/
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SELECT * FROM company
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ORDER BY revenue
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LIMIT 1;
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/*version 2*/
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SELECT * FROM company
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WHERE NOT revenue >= 5000000;
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/*
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14. Query all employees with salary greater than 50 000 and smaller than 70 000
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*/
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SELECT * FROM employee
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WHERE salary BETWEEN 50000 AND 70000;
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/*
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15. Query all employees with salary greater than 50 000 and smaller than 70 000, but you cannot use BETWEEN
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*/
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SELECT * FROM employee
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WHERE salary >= 50000 AND salary <= 70000;
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/*
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16. Query all employees with salary equal to 80 000
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*/
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SELECT * FROM employee
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WHERE salary = 80000;
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/*
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17. Query all employees with salary not equal to 80 000
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*/
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SELECT * FROM employee
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WHERE salary <> 80000;
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/*
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18. Query all names of employees with salary greater than 70 000 together with employees who work on the 'IT' department.
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*/
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SELECT name FROM employee
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WHERE salary > 70000
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OR department_id IN (
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SELECT id FROM department
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WHERE name = 'IT'
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);
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/*
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19. Query all employees that work in city that starts with 'L'
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*/
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SELECT * FROM employee
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WHERE city LIKE 'L%';
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/*
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20. Query all employees that work in city that starts with 'L' or ends with 's'
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*/
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SELECT * FROM employee
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WHERE city LIKE 'L%' OR city LIKE '%s';
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/*
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21. Query all employees that work in city with 'o' somewhere in the middle
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*/
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SELECT * FROM employee
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WHERE city LIKE '%o%';
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/*
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22. Query all departments (each name only once)
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*/
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SELECT DISTINCT name FROM department;
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/*
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22. Query names of all employees together with id of department they work in, but you cannot use JOIN
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*/
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SELECT emp.name,dep.id,dep.name
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FROM employee emp, department dep
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WHERE emp.department_id = dep.id
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ORDER BY emp.name, dep.id;
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/*
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23. Query names of all employees together with id of department they work in, using JOIN
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*/
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SELECT emp.name,dep.id,dep.name
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FROM employee emp
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JOIN department dep
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ON emp.department_id = dep.id
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ORDER BY emp.name, dep.id;
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/*
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24. Query name of every company together with every department
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Personal thoughts: It is kinda weird question, as there is no relationship between company and departement
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*/
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SELECT com.name,dep.name
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FROM company com, department dep
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ORDER BY com.name;
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/*
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25. Query name of every company together with departments without the 'Support' department
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*/
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SELECT com.name,dep.name
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FROM company com, department dep
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WHERE dep.name NOT LIKE 'Support'
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ORDER BY com.name;
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/*
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26. Query employee name together with the department name that they are not working in
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*/
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SELECT emp.name, dep.name
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FROM employee emp, department dep
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WHERE emp.department_id <> dep.id;
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/*
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27. Query company name together with other companies names
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LIKE:
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GOOGLE Apple
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GOOGLE IBM
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Apple IBM
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...
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*/
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SELECT com1.name, com2.name
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FROM company com1, company com2
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WHERE com1.name <> com2.name
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ORDER BY com1.name,com2.name;
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/*
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28. Query employee names with salary smaller than 80 000 without using NOT and <
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NOTE: for POSTGRESQL only. Mysql doesn't support except
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*/
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SELECT e1.name FROM employee e1
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EXCEPT
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SELECT e2.name FROM employee e2 WHERE e2.salary >= 80000;
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/*
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29.Query names of every company and change the name of column to 'Company'
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*/
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SELECT name AS Company
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FROM company;
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/*
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30. Query all employees that work in same department as Peter
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*/
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SELECT * FROM employee
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WHERE department_id IN(
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SELECT department_id FROM employee
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WHERE name LIKE 'Peter'
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)
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AND name NOT LIKE 'Peter';

Diff for: ‎readme.md

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**Instagram Clone Project**: MySQL project which is a cloned mimic version of Instagram database. It is used to perform analysis for real world business related questions.
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## Challenges
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**Ultimate MySQL** : challenges are from Udemy course [The Ultimate MySQL Bootcamp: Go from SQL Beginner to Expert'](https://www.udemy.com/course/the-ultimate-mysql-bootcamp-go-from-sql-beginner-to-expert/) which is an amazing course.
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**Ultimate MySQL** : challenges are from Udemy course [The Ultimate MySQL Bootcamp: Go from SQL Beginner to Expert'](https://www.udemy.com/course/the-ultimate-mysql-bootcamp-go-from-sql-beginner-to-expert/) which is an amazing course.
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**SQL 30 Questions**: challenges from Skillshare course [Skill Share - 30 Simple SQL Interview Queries](https://www.skillshare.com/classes/SQL-Interview-30-Simple-SQL-Interview-Queries-in-2019/809081836).

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