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CREATE TABLE departments (
department_id SERIAL PRIMARY KEY,
department_name VARCHAR(100) NOT NULL,
location VARCHAR(100)
);
CREATE TABLE employees (
employee_id SERIAL PRIMARY KEY,
first_name VARCHAR(50),
last_name VARCHAR(50),
email VARCHAR(100) UNIQUE,
hire_date DATE,
salary DECIMAL(10,2),
department_id INTEGER REFERENCES departments(department_id)
);
INSERT INTO departments (department_name, location) VALUES
('Engineering', 'New York'),
('Marketing', 'Chicago'),
('Finance', 'Boston');
INSERT INTO employees (first_name, last_name, email, hire_date, salary, department_id) VALUES
('John', 'Smith', 'john@example.com', '2022-01-15', 85000, 1),
('Sarah', 'Johnson', 'sarah@example.com', '2021-03-10', 92000, 1),
('Mike', 'Brown', 'mike@example.com', '2023-05-11', 70000, 2),
('Emma', 'Davis', 'emma@example.com', '2020-07-21', 88000, 3);
SELECT
d.department_name,
COUNT(e.employee_id) AS total_employees,
AVG(e.salary) AS avg_salary,
MAX(e.salary) AS highest_salary
FROM departments d
LEFT JOIN employees e ON d.department_id = e.department_id
GROUP BY d.department_name;
SELECT
first_name,
last_name,
salary,
RANK() OVER (ORDER BY salary DESC) AS salary_rank
FROM employees;
CREATE VIEW employee_summary AS
SELECT
e.employee_id,
e.first_name,
e.last_name,
d.department_name,
e.salary
FROM employees e
JOIN departments d ON e.department_id = d.department_id;
SELECT * FROM employee_summary;Coding works best on desktop or with an external keyboard.
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