✅ SQL Practice Questions with Answers: Part-2 🧠🗃️

🔍 Q7. Find employees who never received a bonus
👤 Table: employees
id | name
1 | Arjun
2 | Riya
3 | Meena

💰 Table: bonus
employee_id | bonus_amount
1 | 3000
3 | 5000

📄 Query:

SELECT e.id, e.name  
FROM employees e  
LEFT JOIN bonus b ON e.id = b.employee_id  
WHERE b.employee_id IS NULL;

📌 Result: Riya

🔍 Q8. Get highest salary employee from each department
🧾 Table: employees
id | name | dept | salary
1 | Arjun | HR | 40000
2 | Riya | IT | 55000
3 | Meena | IT | 62000
4 | Kabir | HR | 45000

📄 Query:

SELECT e.*  
FROM employees e  
JOIN (  
  SELECT department, MAX(salary) AS max_salary  
  FROM employees  
  GROUP BY department  
) t  
ON e.department = t.department  
AND e.salary = t.max_salary;

📌 Result: Kabir (HR), Meena (IT)

🔍 Q9. Count number of employees who joined each year
🗓 Table: employees
id | name | join_date
1 | Arjun | 2021-03-10
2 | Riya | 2022-05-12
3 | Meena | 2021-11-03
4 | Kabir | 2023-01-09

📄 Query:

SELECT YEAR(join_date) AS join_year, COUNT(*) AS total  
FROM employees  
GROUP BY YEAR(join_date)  
ORDER BY join_year;

📌 Result:
2021 → 2
2022 → 1
2023 → 1

🔍 Q10. Find employees earning more than department average
🧾 Table: employees
id | name | dept | salary
1 | Arjun | HR | 40000
2 | Riya | IT | 55000
3 | Meena | IT | 62000
4 | Kabir | HR | 45000

📄 Query:

SELECT e.*  
FROM employees e  
JOIN (  
  SELECT department, AVG(salary) AS avg_salary  
  FROM employees  
  GROUP BY department  
) t  
ON e.department = t.department  
WHERE e.salary > t.avg_salary;

📌 Result: Kabir (HR), Meena (IT)

🔍 Q11. Fetch the 5th highest salary from employee table
🧾 Table: employees
Salaries: 90000, 85000, 78000, 76000, 72000, 70000

📄 Query:

SELECT DISTINCT salary  
FROM employees  
ORDER BY salary DESC  
LIMIT 1 OFFSET 4;

📌 Result: 72000

🔍 Q12. Find employees working on more than one project
📂 Table: project_assignments
employee_id | project_id
1 | 101
1 | 102
2 | 103
3 | 104
3 | 105
3 | 106

📄 Query:

SELECT employee_id, COUNT(*) AS project_count  
FROM project_assignments  
GROUP BY employee_id  
HAVING COUNT(*) > 1;

📌 Result:
1 → 2 projects
3 → 3 projects

💬 Tap ❤️ for more!