50 SQL QUERY-BASED INTERVIEW QUESTIONS-2026

📘 Beginner 49 Questions 🎯 Fresher Friendly 🕒 Updated Mar 2026

SELECT * FROM employees;

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SELECT DISTINCT department FROM employees;

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SELECT COUNT(*) FROM employees;

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SELECT MAX(salary) FROM employees;

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SELECT name, salary FROM employees;

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SELECT * FROM employees WHERE salary > 50000;

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SELECT * FROM employees WHERE joining_date > '2023-01-01';

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SELECT * FROM employees WHERE name LIKE 'A%';

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SELECT * FROM employees ORDER BY salary DESC;

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SELECT * FROM employees LIMIT 5;

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SELECT * FROM employees WHERE salary BETWEEN 30000 AND 60000;

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SELECT * FROM employees WHERE department IN ('IT','HR');

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SELECT * FROM employees WHERE manager_id IS NULL;

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SELECT department, COUNT(*) FROM employees GROUP BY department;

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SELECT department, SUM(salary) FROM employees GROUP BY department;

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SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees);

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SELECT DISTINCT salary FROM employees ORDER BY salary DESC LIMIT 3;

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SELECT * FROM employees e LEFT JOIN departments d ON e.dept_id = d.id WHERE d.id IS NULL;

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SELECT e.name, d.dept_name FROM employees e INNER JOIN departments d ON e.dept_id = d.id;

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SELECT e.name, d.dept_name FROM employees e LEFT JOIN departments d ON e.dept_id = d.id;

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SELECT e.name, d.dept_name FROM employees e RIGHT JOIN departments d ON e.dept_id = d.id;

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SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees);

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SELECT * FROM employees WHERE joining_date >= CURDATE() - INTERVAL 30 DAY;

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SELECT email, COUNT(*) FROM users GROUP BY email HAVING COUNT(*) > 1;

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DELETE t1 FROM users t1 JOIN users t2 ON t1.email = t2.email AND t1.id > t2.id;

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UPDATE employees SET salary = salary * 1.10;

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INSERT INTO employees (name, salary, dept_id) VALUES ('Rahul',55000,3);

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SELECT emp_id, COUNT(*) FROM tasks GROUP BY emp_id HAVING COUNT(*) > 2;

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SELECT department, MAX(salary) FROM employees GROUP BY department;

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SELECT department, COUNT(*) AS num FROM employees GROUP BY department ORDER BY num DESC LIMIT 1;

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

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SELECT REPLACE(name,'Mr.','') FROM employees;

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SELECT CONCAT(name,' - ',email) AS info FROM employees;

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SELECT name, CASE WHEN salary > 70000 THEN 'High' WHEN salary > 40000 THEN '2' ELSE 'Low' END AS salary_level FROM employees;

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SELECT e1.name, e2.name, e1.salary FROM employees e1 JOIN employees e2 ON e1.salary = e2.salary AND e1.id < e2.id;

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SELECT * FROM employees ORDER BY id LIMIT 5 OFFSET 5;

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SELECT name, salary, RANK() OVER (ORDER BY salary DESC) AS rank_no FROM employees;

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SELECT name, salary, SUM(salary) OVER (ORDER BY id) AS running_total FROM employees;

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SELECT name, salary, DENSE_RANK() OVER (ORDER BY salary DESC) FROM employees;

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SELECT id + 1 AS missing_id FROM employees e WHERE NOT EXISTS (SELECT 1 FROM employees x WHERE x.id = e.id + 1);

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SELECT department, SUM(CASE WHEN gender='M' THEN 1 ELSE 0 END) AS male, SUM(CASE WHEN gender='F' THEN 1 ELSE 0 END) AS female FROM employees GROUP BY department;

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DELETE FROM employees;

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TRUNCATE TABLE employees;

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SELECT emp_id, COUNT(DISTINCT dept_id) FROM emp_dept GROUP BY emp_id HAVING COUNT(DISTINCT dept_id) > 1;

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SELECT a.name AS employee, b.name AS manager FROM employees a JOIN employees b ON a.manager_id = b.id;

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SELECT name FROM employees WHERE manager_id IS NULL;

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SELECT * FROM employees ORDER BY joining_date DESC LIMIT 3;

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SELECT department, COUNT(*) * 100.0 / (SELECT COUNT(*) FROM employees) AS percentage FROM employees GROUP BY department;

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SELECT salary, COUNT(*) FROM employees GROUP BY salary HAVING COUNT(*) > 1;

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50 SQL QUERY-BASED INTERVIEW QUESTIONS-2026

Q1 Write a query to fetch all records from a table.

Q2 Fetch unique values from a column.

Q3 Count total number of records.

Q4 Fetch the highest salary.

Q5 Fetch employee names and salaries.

Q6 Fetch employees with salary greater than 50000.

Q7 Fetch employees who joined after 2023.

Q8 Fetch employees whose name starts with A.

Q9 Sort data by salary descending.

Q10 Limit records to 5.

Q11 Fetch employees between salary range 30000-60000.

Q12 Use IN operator in SQL.

Q13 Fetch NULL values in manager_id.

Q14 Count employees per department.

Q15 Get total salary per department.

Q16 Fetch the second highest salary.

Q17 Fetch top 3 salaries.

Q18 Find employees without department using LEFT JOIN.

Q19 INNER JOIN example.

Q20 LEFT JOIN example.

Q21 RIGHT JOIN example.

Q22 Subquery example: salary > avg salary.

Q23 Find employees hired in last 30 days.

Q24 Find duplicate email records.

Q25 Delete duplicates except latest.

Q26 Update salary by 10%.

Q27 Insert a new employee record.

Q28 Find employees with more than 2 projects.

Q29 Find highest-paid employee per department.

Q30 Find department with max employees.

Q31 Find Nth highest salary.

Q32 Replace a string inside a column.

Q33 Concatenate two columns.

Q34 CASE statement example.

Q35 Find employees with same salary.

Q36 Fetch next 5 records using OFFSET.

Q37 Rank employees by salary (Window Function).

Q38 Running total salary using window.

Q39 Dense rank example.

Q40 Find missing IDs.

Q41 Pivot: convert gender to columns.

Q42 Delete all records from a table.

Q43 Truncate table.

Q44 Employees working in multiple departments.

Q45 Self join example to find managers.

Q46 Find employees with no manager.

Q47 Find 3 most recently joined employees.

Q48 Find department-wise employee percentage.

Q49 Show duplicate salaries.

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