Create TASK_1.sql

TASK_1.sql

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+CREATE TABLE departments (
+    emp_id INT PRIMARY KEY,
+    dept_name VARCHAR(50),
+    dept_id INT
+);
+INSERT INTO departments (emp_id, dept_name, dept_id) VALUES
+(17679, 'Operations', 13),
+(26650, 'Marketing', 14),
+(30840, 'Operations', 13),
+(49823, 'Technology', 12),
+(51821, 'Operations', 13),
+(67323, 'Marketing', 14),
+(71119, 'Administrative', 11),
+(76589, 'Operations', 13),
+(97927, 'Technology', 12);
+CREATE TABLE employees (
+    emp_id INT PRIMARY KEY,
+    first_name VARCHAR(50),
+    last_name VARCHAR(50),
+    salary INT,
+    job_title VARCHAR(100),
+    gender VARCHAR(10),
+    hire_date DATE
+);
+INSERT INTO employees (emp_id, first_name, last_name, salary, job_title, gender, hire_date) VALUES
+(17679, 'Robert', 'Gilmore', 110000, 'Operations Director', 'Male', '2018-09-04'),
+(26650, 'Elvis', 'Ritter', 86000, 'Sales Manager', 'Male', '2017-11-24'),
+(30840, 'David', 'Barrow', 85000, 'Data Scientist', 'Male', '2019-12-02'),
+(49714, 'Hugo', 'Forester', 55000, 'IT Support Specialist', 'Male', '2019-11-22'),
+(51821, 'Linda', 'Foster', 95000, 'Data Scientist', 'Female', '2019-04-29'),
+(67323, 'Lisa', 'Wiener', 75000, 'Business Analyst', 'Female', '2018-08-09'),
+(70950, 'Rodney', 'Weaver', 87000, 'Project Manager', 'Male', '2018-12-20'),
+(71329, 'Gayle', 'Meyer', 77000, 'HR Manager', 'Female', '2019-06-28'),
+(76589, 'Jason', 'Christian', 99000, 'Project Manager', 'Male', '2019-01-21'),
+(97927, 'Billie', 'Lanning', 67000, 'Web Developer', 'Female', '2018-06-25');
+
+-- 1. Find the employees who get paid more than Rodney Weaver.
+SELECT first_name, last_name, salary
+FROM employees
+WHERE salary > (
+    SELECT salary FROM employees WHERE first_name = 'Rodney' AND last_name = 'Weaver'
+);
+
+-- 2. Find the average, min and max salaries.
+SELECT AVG(salary) AS avg_salary, MIN(salary) AS min_salary, MAX(salary) AS max_salary
+FROM employees;
+
+-- 3. Find the employees whose salary is more than 87000. Our query should return first name, last name, and salary info of the employees.
+SELECT first_name, last_name, salary
+FROM employees
+WHERE salary > 87000;
+
+-- 4. Find the employees (first name, last name from employees table) who work under the Operations department (departments table). Our query should return first name and last name info.
+SELECT e.first_name, e.last_name
+FROM employees e
+JOIN departments d ON e.emp_id = d.emp_id
+WHERE d.dept_name = 'Operations';
+
+-- 5. Find the employees (first name, last name from employees table) who work under the Technology department (departments table). Our query should return first name and last name info.
+SELECT e.first_name, e.last_name
+FROM employees e
+JOIN departments d ON e.emp_id = d.emp_id
+WHERE d.dept_name = 'Technology';
+
+-- 6. Find the average salary of female employees.
+SELECT AVG(salary) AS avg_salary_female
+FROM employees
+WHERE gender = 'Female';
+
+-- 7. Find the average salaries of each department.
+SELECT d.dept_name, AVG(e.salary) AS avg_salary
+FROM employees e
+JOIN departments d ON e.emp_id = d.emp_id
+GROUP BY d.dept_name;
+
+-- 8. Find the oldest and newest employees.
+-- Oldest employee
+SELECT first_name, last_name, hire_date
+FROM employees
+ORDER BY hire_date ASC
+LIMIT 1;
+
+-- Newest employee
+SELECT first_name, last_name, hire_date
+FROM employees
+ORDER BY hire_date DESC
+LIMIT 1;
+
+-- 9. Find the hiring date and department of the highest paid employee.
+SELECT e.hire_date, d.dept_name
+FROM employees e
+JOIN departments d ON e.emp_id = d.emp_id
+WHERE e.salary = (SELECT MAX(salary) FROM employees);
+
+-- 10. Find the hiring date and department of the lowest paid employee.
+SELECT e.hire_date, d.dept_name
+FROM employees e
+JOIN departments d ON e.emp_id = d.emp_id
+WHERE e.salary = (SELECT MIN(salary) FROM employees);