SQL Interview Questions for Freshers (2026): 60 Questions and Answers

Oct 28, 2026

What SQL questions are asked to freshers? Interviewers usually ask the difference between DELETE, TRUNCATE and DROP, primary key vs foreign key, WHERE vs HAVING, INNER JOIN vs LEFT JOIN, GROUP BY with aggregate functions, subqueries, normalisation basics and then ask you to write queries live, such as the second highest salary, duplicate records, employees earning more than their manager and department-wise counts. This guide gives you both: 60 concept questions and 25 query problems on one sample database so you can practise exactly like the interview.

This guide from VibrantMinds Technologies in Pune is written for anyone searching SQL interview questions for freshers, SQL interview questions and answers, SQL query interview questions for freshers, SQL joins interview questions, second highest salary in SQL, SQL practice questions with answers, SQL window functions interview questions, difference between WHERE and HAVING in SQL, SQL interview questions for Java developers or a free IT job WhatsApp group for SQL and Java freshers in Pune.

Takeaway	Detail
Interviews mix theory and live queries	Know definitions, but practise writing queries from memory.
Joins and GROUP BY decide most rounds	Master INNER, LEFT, GROUP BY and HAVING.
The classic queries repeat	Second highest salary, duplicates, department counts, Nth highest.
Say the logic aloud	Explain tables, join keys and filters before typing.
Know NULL behaviour	`NULL = NULL` is not true. Use `IS NULL`.
Name your dialect	Mention MySQL, PostgreSQL or SQL Server when syntax differs.
Practise on a real database	Install MySQL or PostgreSQL and run every query yourself.

HOW SQL INTERVIEW ROUNDS USUALLY FLOW

THE SAMPLE DATABASE USED IN THIS GUIDE

Create these two tables in MySQL or PostgreSQL and run every query yourself.

sql

CREATE TABLE departments (
    dept_id   INT PRIMARY KEY,
    dept_name VARCHAR(50) NOT NULL
);

CREATE TABLE employees (
    emp_id     INT PRIMARY KEY,
    emp_name   VARCHAR(50) NOT NULL,
    dept_id    INT,
    salary     INT,
    manager_id INT,
    city       VARCHAR(50),
    FOREIGN KEY (dept_id) REFERENCES departments(dept_id)
);

INSERT INTO departments VALUES (1,'IT'),(2,'HR'),(3,'Testing'),(4,'Finance');

INSERT INTO employees VALUES
(1,'Amit',   1, 60000, NULL, 'Pune'),
(2,'Divya',  1, 50000, 1,    'Pune'),
(3,'Farhan', 2, 40000, 1,    'Mumbai'),
(4,'Gauri',  3, 45000, 1,    'Pune'),
(5,'Harsh',  1, 50000, 2,    'Nashik'),
(6,'Isha',   3, 35000, 4,    'Pune'),
(7,'Jay',    2, 42000, 3,    'Mumbai'),
(8,'Kiran',  NULL, 30000, 1, 'Pune')

CREATE TABLE departments (
    dept_id   INT PRIMARY KEY,
    dept_name VARCHAR(50) NOT NULL
);

CREATE TABLE employees (
    emp_id     INT PRIMARY KEY,
    emp_name   VARCHAR(50) NOT NULL,
    dept_id    INT,
    salary     INT,
    manager_id INT,
    city       VARCHAR(50),
    FOREIGN KEY (dept_id) REFERENCES departments(dept_id)
);

INSERT INTO departments VALUES (1,'IT'),(2,'HR'),(3,'Testing'),(4,'Finance');

INSERT INTO employees VALUES
(1,'Amit',   1, 60000, NULL, 'Pune'),
(2,'Divya',  1, 50000, 1,    'Pune'),
(3,'Farhan', 2, 40000, 1,    'Mumbai'),
(4,'Gauri',  3, 45000, 1,    'Pune'),
(5,'Harsh',  1, 50000, 2,    'Nashik'),
(6,'Isha',   3, 35000, 4,    'Pune'),
(7,'Jay',    2, 42000, 3,    'Mumbai'),
(8,'Kiran',  NULL, 30000, 1, 'Pune')

CREATE TABLE departments (
    dept_id   INT PRIMARY KEY,
    dept_name VARCHAR(50) NOT NULL
);

CREATE TABLE employees (
    emp_id     INT PRIMARY KEY,
    emp_name   VARCHAR(50) NOT NULL,
    dept_id    INT,
    salary     INT,
    manager_id INT,
    city       VARCHAR(50),
    FOREIGN KEY (dept_id) REFERENCES departments(dept_id)
);

INSERT INTO departments VALUES (1,'IT'),(2,'HR'),(3,'Testing'),(4,'Finance');

INSERT INTO employees VALUES
(1,'Amit',   1, 60000, NULL, 'Pune'),
(2,'Divya',  1, 50000, 1,    'Pune'),
(3,'Farhan', 2, 40000, 1,    'Mumbai'),
(4,'Gauri',  3, 45000, 1,    'Pune'),
(5,'Harsh',  1, 50000, 2,    'Nashik'),
(6,'Isha',   3, 35000, 4,    'Pune'),
(7,'Jay',    2, 42000, 3,    'Mumbai'),
(8,'Kiran',  NULL, 30000, 1, 'Pune')

departments

dept_id	dept_name
1	IT
2	HR
3	Testing
4	Finance

employees

emp_id	emp_name	dept_id	salary	manager_id	city
1	Amit	1	60000	NULL	Pune
2	Divya	1	50000	1	Pune
3	Farhan	2	40000	1	Mumbai
4	Gauri	3	45000	1	Pune
5	Harsh	1	50000	2	Nashik
6	Isha	3	35000	4	Pune
7	Jay	2	42000	3	Mumbai
8	Kiran	NULL	30000	1	Pune

Notice the deliberate details: Kiran has no department (NULL), the Finance department has no employees, two people share the salary 50000 and Jay earns more than his manager Farhan. These traps are exactly what interviewers use.

MODULE 1: SQL AND DATABASE BASICS (Q1 TO Q12)

Q1. What is SQL?
Answer: Structured Query Language, the standard language to create, read, update and delete data in relational databases.

Q2. DBMS vs RDBMS?
Answer: A DBMS stores data in any structure (files, for example). An RDBMS stores data in related tables with keys and supports SQL and constraints. MySQL, PostgreSQL, SQL Server and Oracle are RDBMS.

Q3. What are the types of SQL commands?

Type	Full form	Commands
DDL	Data Definition Language	CREATE, ALTER, DROP, TRUNCATE
DML	Data Manipulation Language	INSERT, UPDATE, DELETE
DQL	Data Query Language	SELECT
DCL	Data Control Language	GRANT, REVOKE
TCL	Transaction Control Language	COMMIT, ROLLBACK, SAVEPOINT

Q4. DELETE vs TRUNCATE vs DROP

Point	DELETE	TRUNCATE	DROP
Removes	Selected rows (or all)	All rows	The whole table
WHERE clause	Yes	No	No
Type	DML	DDL	DDL
Rollback	Usually possible inside a transaction	Depends on the database	Depends on the database
Table structure	Kept	Kept	Removed

Q5. What is a table, a row and a column?
Answer: A table stores data. A row (record or tuple) is one entry. A column (field or attribute) is one data property.

Q6. What are the common data types?
Answer: Numeric (INT, DECIMAL), text (VARCHAR, CHAR, TEXT), date and time (DATE, DATETIME, TIMESTAMP), and boolean.

Q7. CHAR vs VARCHAR?
Answer: CHAR(n) is fixed length. VARCHAR(n) is variable length and stores only what’s needed.

Q8. What is NULL?
Answer: The absence of a value. It isn’t zero or an empty string. Comparisons with NULL need IS NULL or IS NOT NULL.

Q9. What are aliases?
Answer: Temporary names for columns or tables to make queries readable.

sql

SELECT e.emp_name AS employee, d.dept_name AS department
FROM employees e JOIN departments d ON

SELECT e.emp_name AS employee, d.dept_name AS department
FROM employees e JOIN departments d ON

SELECT e.emp_name AS employee, d.dept_name AS department
FROM employees e JOIN departments d ON

Q10. What is a schema?
Answer: A logical container for database objects such as tables, views and indexes.

Q11. What is a database transaction?
Answer: A group of SQL operations treated as one unit: all succeed or all fail.

Q12. What are ACID properties?

Property	Meaning
Atomicity	All steps happen or none do
Consistency	Data stays valid after the transaction
Isolation	Concurrent transactions don’t interfere
Durability	Committed data survives failures

MODULE 2: KEYS, CONSTRAINTS AND NORMALISATION (Q13 TO Q22)

Q13. What is a primary key?
Answer: A column (or set of columns) that uniquely identifies each row. It cannot be NULL or duplicate, and a table has only one.

Q14. What is a foreign key?
Answer: A column that refers to the primary key of another table, enforcing referential integrity.

Q15. Primary key vs unique key?

Point	Primary key	Unique key
NULL	Not allowed	Usually allowed (rules vary by database)
Count per table	One	Many
Purpose	Row identity	Prevent duplicate values

Q16. What is a composite key?
Answer: A primary key made of two or more columns.

Q17. What are constraints?
Answer: Rules on data: NOT NULL, UNIQUE, PRIMARY KEY, FOREIGN KEY, CHECK, DEFAULT.

Q18. What is normalisation?
Answer: Organising data to reduce redundancy and improve integrity by splitting data into related tables.

Q19. 1NF, 2NF, 3NF in simple words

Form	Rule
1NF	Each cell holds one value, no repeating groups
2NF	1NF, and every non-key column depends on the whole primary key
3NF	2NF, and non-key columns don’t depend on other non-key columns

Q20. What is denormalisation?
Answer: Deliberately adding redundancy to speed up reads, often used in reporting.

Q21. What is referential integrity?
Answer: Ensuring a foreign key value always matches an existing primary key value (or is NULL).

Q22. What are ON DELETE CASCADE and ON DELETE SET NULL?
Answer: Rules for what happens to child rows when the parent row is deleted. CASCADE deletes the child rows. SET NULL sets the foreign key to NULL.

MODULE 3: SELECT, WHERE, ORDER BY AND FUNCTIONS (Q23 TO Q32)

Q23. Basic SELECT, WHERE and ORDER BY

sql

SELECT emp_name, salary
FROM employees
WHERE city = 'Pune' AND salary > 40000
ORDER BY salary DESC

SELECT emp_name, salary
FROM employees
WHERE city = 'Pune' AND salary > 40000
ORDER BY salary DESC

SELECT emp_name, salary
FROM employees
WHERE city = 'Pune' AND salary > 40000
ORDER BY salary DESC

Q24. DISTINCT

sql

SELECT DISTINCT city FROM employees;   -- Pune, Mumbai, Nashik

SELECT DISTINCT city FROM employees;   -- Pune, Mumbai, Nashik

SELECT DISTINCT city FROM employees;   -- Pune, Mumbai, Nashik

Q25. IN, BETWEEN, LIKE

sql

SELECT * FROM employees WHERE city IN ('Pune','Mumbai');
SELECT * FROM employees WHERE salary BETWEEN 40000 AND 50000;   -- inclusive
SELECT * FROM employees WHERE emp_name LIKE 'A%';               -- starts with A

SELECT * FROM employees WHERE city IN ('Pune','Mumbai');
SELECT * FROM employees WHERE salary BETWEEN 40000 AND 50000;   -- inclusive
SELECT * FROM employees WHERE emp_name LIKE 'A%';               -- starts with A

SELECT * FROM employees WHERE city IN ('Pune','Mumbai');
SELECT * FROM employees WHERE salary BETWEEN 40000 AND 50000;   -- inclusive
SELECT * FROM employees WHERE emp_name LIKE 'A%';               -- starts with A

Q26. How do you handle NULL?

sql

SELECT * FROM employees WHERE dept_id IS NULL;           -- Kiran
SELECT emp_name, COALESCE(dept_id, 0) FROM employees;    -- replace NULL with 0

SELECT * FROM employees WHERE dept_id IS NULL;           -- Kiran
SELECT emp_name, COALESCE(dept_id, 0) FROM employees;    -- replace NULL with 0

SELECT * FROM employees WHERE dept_id IS NULL;           -- Kiran
SELECT emp_name, COALESCE(dept_id, 0) FROM employees;    -- replace NULL with 0

Q27. LIMIT, TOP and FETCH FIRST
Answer: They return the first N rows. MySQL and PostgreSQL use LIMIT, SQL Server uses TOP, and Oracle (12c+) and standard SQL use FETCH FIRST n ROWS ONLY.

sql

SELECT * FROM employees ORDER BY salary DESC LIMIT 3

SELECT * FROM employees ORDER BY salary DESC LIMIT 3

SELECT * FROM employees ORDER BY salary DESC LIMIT 3

Q28. Common aggregate functions?
Answer: COUNT, SUM, AVG, MIN, MAX.

Q29. COUNT(*) vs COUNT(column)?
Answer: COUNT(*) counts all rows. COUNT(column) counts only non-NULL values in that column.

sql

SELECT COUNT(*) FROM employees;        -- 8
SELECT COUNT(dept_id) FROM employees;  -- 7 (Kiran's NULL is ignored)

SELECT COUNT(*) FROM employees;        -- 8
SELECT COUNT(dept_id) FROM employees;  -- 7 (Kiran's NULL is ignored)

SELECT COUNT(*) FROM employees;        -- 8
SELECT COUNT(dept_id) FROM employees;  -- 7 (Kiran's NULL is ignored)

Q30. Common string and date functions?
Answer: UPPER, LOWER, LENGTH, SUBSTRING, CONCAT, TRIM, NOW, CURRENT_DATE, DATEDIFF. Names vary a little by database.

Q31. What is CASE?

sql

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

Q32. UNION vs UNION ALL?
Answer: Both combine results of two queries with the same column structure. UNION removes duplicates, while UNION ALL keeps them and is faster.

MODULE 4: GROUP BY, HAVING AND AGGREGATES (Q33 TO Q40)

Q33. What does GROUP BY do?
Answer: It groups rows with the same values so you can apply aggregate functions per group.

sql

SELECT dept_id, COUNT(*) AS emp_count, AVG(salary) AS avg_salary
FROM employees
GROUP BY

SELECT dept_id, COUNT(*) AS emp_count, AVG(salary) AS avg_salary
FROM employees
GROUP BY

SELECT dept_id, COUNT(*) AS emp_count, AVG(salary) AS avg_salary
FROM employees
GROUP BY

Q34. WHERE vs HAVING

Point	WHERE	HAVING
Filters	Rows, before grouping	Groups, after grouping
Aggregates	Cannot use them	Can use them
Example	`WHERE city = 'Pune'`	`HAVING COUNT(*) > 2`

sql

SELECT dept_id, COUNT(*) AS emp_count
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING COUNT(*) > 2;       -- returns dept 1 (IT, 3 employees)

SELECT dept_id, COUNT(*) AS emp_count
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING COUNT(*) > 2;       -- returns dept 1 (IT, 3 employees)

SELECT dept_id, COUNT(*) AS emp_count
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING COUNT(*) > 2;       -- returns dept 1 (IT, 3 employees)

Q35. Can you use a column in SELECT that isn’t in GROUP BY?
Answer: Not unless it’s inside an aggregate function (strict SQL modes enforce this). Otherwise the result is ambiguous.

Q36. How do you count employees per city, sorted by count?

sql

SELECT city, COUNT(*) AS total
FROM employees
GROUP BY city
ORDER BY total DESC

SELECT city, COUNT(*) AS total
FROM employees
GROUP BY city
ORDER BY total DESC

SELECT city, COUNT(*) AS total
FROM employees
GROUP BY city
ORDER BY total DESC

Q37. How do you find the highest salary in each department?

sql

SELECT dept_id, MAX(salary) AS max_salary
FROM employees
GROUP BY

SELECT dept_id, MAX(salary) AS max_salary
FROM employees
GROUP BY

SELECT dept_id, MAX(salary) AS max_salary
FROM employees
GROUP BY

Q38. Can GROUP BY have multiple columns?
Answer: Yes. Groups are formed by the combination, for example GROUP BY dept_id, city.

Q39. What does COUNT(DISTINCT column) do?

sql

SELECT COUNT(DISTINCT city) FROM employees;   -- 3

SELECT COUNT(DISTINCT city) FROM employees;   -- 3

SELECT COUNT(DISTINCT city) FROM employees;   -- 3

Q40. How do aggregates treat NULL?
Answer: They ignore NULL values (except COUNT(*)). AVG divides by the number of non-NULL values.

MODULE 5: JOINS (Q41 TO Q48)

Q41. INNER JOIN

sql

SELECT e.emp_name, d.dept_name
FROM employees e
INNER JOIN departments d ON e.dept_id = d.dept_id;
-- 7 rows. Kiran (no dept) and the Finance department are excluded.

SELECT e.emp_name, d.dept_name
FROM employees e
INNER JOIN departments d ON e.dept_id = d.dept_id;
-- 7 rows. Kiran (no dept) and the Finance department are excluded.

SELECT e.emp_name, d.dept_name
FROM employees e
INNER JOIN departments d ON e.dept_id = d.dept_id;
-- 7 rows. Kiran (no dept) and the Finance department are excluded.

Q42. LEFT JOIN

sql

SELECT e.emp_name, d.dept_name
FROM employees e
LEFT JOIN departments d ON e.dept_id = d.dept_id;
-- 8 rows. Kiran appears with dept_name = NULL.

SELECT e.emp_name, d.dept_name
FROM employees e
LEFT JOIN departments d ON e.dept_id = d.dept_id;
-- 8 rows. Kiran appears with dept_name = NULL.

SELECT e.emp_name, d.dept_name
FROM employees e
LEFT JOIN departments d ON e.dept_id = d.dept_id;
-- 8 rows. Kiran appears with dept_name = NULL.

Q43. INNER JOIN vs LEFT JOIN: the difference

Point	INNER JOIN	LEFT JOIN
Returns	Only matching rows	All left-table rows, with NULLs where no match
Kiran (no department)	Missing	Present
Use when	You need matched data only	You must keep every row from the left table

Q44. How do you find departments with no employees?

sql

SELECT d.dept_name
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
WHERE e.emp_id IS NULL;      -- Finance

SELECT d.dept_name
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
WHERE e.emp_id IS NULL;      -- Finance

SELECT d.dept_name
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
WHERE e.emp_id IS NULL;      -- Finance

Q45. What is a self join? Show employee and manager names.

sql

SELECT e.emp_name AS employee, m.emp_name AS manager
FROM employees e
LEFT JOIN employees m ON

SELECT e.emp_name AS employee, m.emp_name AS manager
FROM employees e
LEFT JOIN employees m ON

SELECT e.emp_name AS employee, m.emp_name AS manager
FROM employees e
LEFT JOIN employees m ON

Q46. What is a CROSS JOIN?
Answer: It pairs every row of one table with every row of another (the Cartesian product). 8 employees × 4 departments gives 32 rows.

Q47. RIGHT JOIN and FULL OUTER JOIN?
Answer: RIGHT JOIN keeps all rows from the right table. FULL OUTER JOIN keeps all rows from both. MySQL has no direct FULL OUTER JOIN, so emulate it with LEFT JOIN UNION RIGHT JOIN.

Q48. Does the position of a condition matter in a LEFT JOIN?
Answer: Yes. A filter on the right table in WHERE can remove the NULL rows and turn the LEFT JOIN into an INNER JOIN. Put such a filter in the ON clause to keep the unmatched rows.

MODULE 6: SUBQUERIES, VIEWS, INDEXES AND TRANSACTIONS (Q49 TO Q60)

Q49. What is a subquery?
Answer: A query nested inside another query.

sql

SELECT emp_name, salary
FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);   -- average is 44000
-- Amit, Divya, Gauri, Harsh

SELECT emp_name, salary
FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);   -- average is 44000
-- Amit, Divya, Gauri, Harsh

SELECT emp_name, salary
FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);   -- average is 44000
-- Amit, Divya, Gauri, Harsh

Q50. Correlated vs non-correlated subquery?
Answer: A non-correlated subquery runs once independently. A correlated subquery depends on the outer query and runs for each outer row.

Q51. IN vs EXISTS?
Answer: IN compares against a list of values. EXISTS checks whether a subquery returns any row, and often performs well on large tables. Watch NULL behaviour with NOT IN, because a NULL in the list can make the result empty. NOT EXISTS is safer.

Q52. What is a view?
Answer: A saved query that behaves like a virtual table.

sql

CREATE VIEW pune_employees AS
SELECT emp_id, emp_name, salary FROM employees WHERE city = 'Pune'

CREATE VIEW pune_employees AS
SELECT emp_id, emp_name, salary FROM employees WHERE city = 'Pune'

CREATE VIEW pune_employees AS
SELECT emp_id, emp_name, salary FROM employees WHERE city = 'Pune'

Q53. What is an index? Pros and cons?
Answer: A structure that speeds up searches on a column. It makes reads faster but uses storage and slows inserts and updates.

sql

CREATE INDEX idx_emp_city ON employees(city)

CREATE INDEX idx_emp_city ON employees(city)

CREATE INDEX idx_emp_city ON employees(city)

Q54. Clustered vs non-clustered index?
Answer: A clustered index decides the physical order of rows (one per table). A non-clustered index is a separate structure pointing to rows (many per table). Exact behaviour varies by database.

Q55. What is a stored procedure?
Answer: A saved set of SQL statements stored in the database that you can call by name, with parameters.

Q56. What is a trigger?
Answer: Code that runs automatically on events such as INSERT, UPDATE or DELETE.

Q57. COMMIT, ROLLBACK and SAVEPOINT

sql

START TRANSACTION;
UPDATE employees SET salary = salary + 1000 WHERE dept_id = 1;
-- check the result
ROLLBACK;   -- undo, or use COMMIT to save permanently

START TRANSACTION;
UPDATE employees SET salary = salary + 1000 WHERE dept_id = 1;
-- check the result
ROLLBACK;   -- undo, or use COMMIT to save permanently

START TRANSACTION;
UPDATE employees SET salary = salary + 1000 WHERE dept_id = 1;
-- check the result
ROLLBACK;   -- undo, or use COMMIT to save permanently

Answer: COMMIT saves changes, ROLLBACK undoes them and SAVEPOINT marks a point to roll back to.

Q58. What is SQL injection and how do you prevent it?
Answer: An attack where malicious input changes a query. Prevent it with parameterised queries or prepared statements and input validation, never string concatenation.

Q59. What is a CTE?
Answer: A Common Table Expression, a named temporary result defined with WITH that makes complex queries readable.

sql

WITH dept_avg AS (
    SELECT dept_id, AVG(salary) AS avg_sal FROM employees GROUP BY dept_id
)
SELECT * FROM dept_avg WHERE avg_sal > 40000

WITH dept_avg AS (
    SELECT dept_id, AVG(salary) AS avg_sal FROM employees GROUP BY dept_id
)
SELECT * FROM dept_avg WHERE avg_sal > 40000

WITH dept_avg AS (
    SELECT dept_id, AVG(salary) AS avg_sal FROM employees GROUP BY dept_id
)
SELECT * FROM dept_avg WHERE avg_sal > 40000

Q60. What is the difference between a function and a stored procedure?
Answer: A function returns a value and can be used inside queries. A stored procedure performs actions and may return results through output parameters or result sets. Details differ by database.

QUERY PRACTICE: 25 PROBLEMS WITH ANSWERS (USE THE SAMPLE TABLES)

Try each problem yourself before reading the answer.

P1. Find the second highest salary.

sql

SELECT MAX(salary) AS second_highest
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
-- 50000

SELECT MAX(salary) AS second_highest
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
-- 50000

SELECT MAX(salary) AS second_highest
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);
-- 50000

P2. Second highest salary using LIMIT and OFFSET (MySQL or PostgreSQL).

sql

SELECT DISTINCT salary
FROM employees
ORDER BY salary DESC
LIMIT 1 OFFSET 1;
-- 50000

SELECT DISTINCT salary
FROM employees
ORDER BY salary DESC
LIMIT 1 OFFSET 1;
-- 50000

SELECT DISTINCT salary
FROM employees
ORDER BY salary DESC
LIMIT 1 OFFSET 1;
-- 50000

P3. Nth highest salary (N = 3, using DENSE_RANK).

sql

SELECT salary
FROM (
    SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rnk
    FROM employees
) t
WHERE rnk = 3;
-- 45000

SELECT salary
FROM (
    SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rnk
    FROM employees
) t
WHERE rnk = 3;
-- 45000

SELECT salary
FROM (
    SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rnk
    FROM employees
) t
WHERE rnk = 3;
-- 45000

P4. Find duplicate salaries.

sql

SELECT salary, COUNT(*) AS cnt
FROM employees
GROUP BY salary
HAVING COUNT(*) > 1;
-- 50000, 2

SELECT salary, COUNT(*) AS cnt
FROM employees
GROUP BY salary
HAVING COUNT(*) > 1;
-- 50000, 2

SELECT salary, COUNT(*) AS cnt
FROM employees
GROUP BY salary
HAVING COUNT(*) > 1;
-- 50000, 2

P5. Employees earning more than the average salary.

sql

SELECT emp_name, salary FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees)

SELECT emp_name, salary FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees)

SELECT emp_name, salary FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees)

P6. Department-wise employee count including departments with no employees.

sql

SELECT d.dept_name, COUNT(e.emp_id) AS emp_count
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
GROUP BY d.dept_name;
-- IT 3, HR 2, Testing 2, Finance 0

SELECT d.dept_name, COUNT(e.emp_id) AS emp_count
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
GROUP BY d.dept_name;
-- IT 3, HR 2, Testing 2, Finance 0

SELECT d.dept_name, COUNT(e.emp_id) AS emp_count
FROM departments d
LEFT JOIN employees e ON d.dept_id = e.dept_id
GROUP BY d.dept_name;
-- IT 3, HR 2, Testing 2, Finance 0

Note: COUNT(e.emp_id) gives 0 for Finance, while COUNT(*) would wrongly give 1.

P7. Departments with more than 2 employees.

sql

SELECT dept_id, COUNT(*) AS cnt
FROM employees
GROUP BY dept_id
HAVING COUNT(*) > 2;   -- IT

SELECT dept_id, COUNT(*) AS cnt
FROM employees
GROUP BY dept_id
HAVING COUNT(*) > 2;   -- IT

SELECT dept_id, COUNT(*) AS cnt
FROM employees
GROUP BY dept_id
HAVING COUNT(*) > 2;   -- IT

P8. Highest-paid employee in each department.

sql

SELECT e.dept_id, e.emp_name, e.salary
FROM employees e
WHERE e.salary = (
    SELECT MAX(salary) FROM employees WHERE dept_id = e.dept_id
)

SELECT e.dept_id, e.emp_name, e.salary
FROM employees e
WHERE e.salary = (
    SELECT MAX(salary) FROM employees WHERE dept_id = e.dept_id
)

SELECT e.dept_id, e.emp_name, e.salary
FROM employees e
WHERE e.salary = (
    SELECT MAX(salary) FROM employees WHERE dept_id = e.dept_id
)

P9. Employees who earn more than their manager.

sql

SELECT e.emp_name
FROM employees e
JOIN employees m ON e.manager_id = m.emp_id
WHERE e.salary > m.salary;
-- Jay

SELECT e.emp_name
FROM employees e
JOIN employees m ON e.manager_id = m.emp_id
WHERE e.salary > m.salary;
-- Jay

SELECT e.emp_name
FROM employees e
JOIN employees m ON e.manager_id = m.emp_id
WHERE e.salary > m.salary;
-- Jay

P10. Employees without a department.

sql

SELECT emp_name FROM employees WHERE dept_id IS NULL;   -- Kiran

SELECT emp_name FROM employees WHERE dept_id IS NULL;   -- Kiran

SELECT emp_name FROM employees WHERE dept_id IS NULL;   -- Kiran

P11. Departments with no employees.

sql

SELECT dept_name FROM departments d
WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.dept_id = d.dept_id);
-- Finance

SELECT dept_name FROM departments d
WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.dept_id = d.dept_id);
-- Finance

SELECT dept_name FROM departments d
WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.dept_id = d.dept_id);
-- Finance

P12. Employees and their manager names.

sql

SELECT e.emp_name, m.emp_name AS manager
FROM employees e LEFT JOIN employees m ON

SELECT e.emp_name, m.emp_name AS manager
FROM employees e LEFT JOIN employees m ON

SELECT e.emp_name, m.emp_name AS manager
FROM employees e LEFT JOIN employees m ON

P13. Employees whose name starts with a vowel.

sql

SELECT emp_name FROM employees
WHERE UPPER(SUBSTRING(emp_name, 1, 1)) IN ('A','E','I','O','U');
-- Amit, Isha

SELECT emp_name FROM employees
WHERE UPPER(SUBSTRING(emp_name, 1, 1)) IN ('A','E','I','O','U');
-- Amit, Isha

SELECT emp_name FROM employees
WHERE UPPER(SUBSTRING(emp_name, 1, 1)) IN ('A','E','I','O','U');
-- Amit, Isha

P14. Total salary per city.

sql

SELECT city, SUM(salary) AS total_salary FROM employees GROUP BY

SELECT city, SUM(salary) AS total_salary FROM employees GROUP BY

SELECT city, SUM(salary) AS total_salary FROM employees GROUP BY

P15. Top 3 highest-paid employees.

sql

SELECT emp_name, salary FROM employees ORDER BY salary DESC LIMIT 3

SELECT emp_name, salary FROM employees ORDER BY salary DESC LIMIT 3

SELECT emp_name, salary FROM employees ORDER BY salary DESC LIMIT 3

P16. Employees in the same department as Divya (excluding Divya).

sql

SELECT emp_name FROM employees
WHERE dept_id = (SELECT dept_id FROM employees WHERE emp_name = 'Divya')
  AND emp_name <> 'Divya';
-- Amit, Harsh

SELECT emp_name FROM employees
WHERE dept_id = (SELECT dept_id FROM employees WHERE emp_name = 'Divya')
  AND emp_name <> 'Divya';
-- Amit, Harsh

SELECT emp_name FROM employees
WHERE dept_id = (SELECT dept_id FROM employees WHERE emp_name = 'Divya')
  AND emp_name <> 'Divya';
-- Amit, Harsh

P17. Add a salary band label to every employee.

sql

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

SELECT emp_name, salary,
       CASE WHEN salary >= 50000 THEN 'High'
            WHEN salary >= 40000 THEN 'Medium'
            ELSE 'Low' END AS band
FROM

P18. Department-wise average salary, only above 40000.

sql

SELECT dept_id, AVG(salary) AS avg_sal
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING AVG(salary) > 40000;
-- IT (53333.33), HR (41000)

SELECT dept_id, AVG(salary) AS avg_sal
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING AVG(salary) > 40000;
-- IT (53333.33), HR (41000)

SELECT dept_id, AVG(salary) AS avg_sal
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY dept_id
HAVING AVG(salary) > 40000;
-- IT (53333.33), HR (41000)

P19. Delete duplicate rows keeping the lowest id (concept, for a table with an id column).

sql

DELETE e1 FROM employees e1
JOIN employees e2
  ON e1.emp_name = e2.emp_name AND e1.emp_id > e2.emp_id;   -- MySQL style

DELETE e1 FROM employees e1
JOIN employees e2
  ON e1.emp_name = e2.emp_name AND e1.emp_id > e2.emp_id;   -- MySQL style

DELETE e1 FROM employees e1
JOIN employees e2
  ON e1.emp_name = e2.emp_name AND e1.emp_id > e2.emp_id;   -- MySQL style

Practise this on a copy of a table, never on live data. Syntax differs by database.

P20. Rank employees by salary within each department.

sql

SELECT emp_name, dept_id, salary,
       RANK() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS dept_rank
FROM

SELECT emp_name, dept_id, salary,
       RANK() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS dept_rank
FROM

SELECT emp_name, dept_id, salary,
       RANK() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS dept_rank
FROM

P21. Running total of salary ordered by emp_id.

sql

SELECT emp_id, salary,
       SUM(salary) OVER (ORDER BY emp_id) AS running_total
FROM

SELECT emp_id, salary,
       SUM(salary) OVER (ORDER BY emp_id) AS running_total
FROM

SELECT emp_id, salary,
       SUM(salary) OVER (ORDER BY emp_id) AS running_total
FROM

P22. Employees hired in the current month (for a table with a join_date column).

sql

SELECT * FROM employees
WHERE EXTRACT(YEAR FROM join_date)  = EXTRACT(YEAR FROM CURRENT_DATE)
  AND EXTRACT(MONTH FROM join_date) = EXTRACT(MONTH FROM CURRENT_DATE)

SELECT * FROM employees
WHERE EXTRACT(YEAR FROM join_date)  = EXTRACT(YEAR FROM CURRENT_DATE)
  AND EXTRACT(MONTH FROM join_date) = EXTRACT(MONTH FROM CURRENT_DATE)

SELECT * FROM employees
WHERE EXTRACT(YEAR FROM join_date)  = EXTRACT(YEAR FROM CURRENT_DATE)
  AND EXTRACT(MONTH FROM join_date) = EXTRACT(MONTH FROM CURRENT_DATE)

P23. Swap values using CASE (for example, change city names).

sql

UPDATE employees
SET city = CASE WHEN city = 'Pune' THEN 'Mumbai'
                WHEN city = 'Mumbai' THEN 'Pune'
                ELSE city END

UPDATE employees
SET city = CASE WHEN city = 'Pune' THEN 'Mumbai'
                WHEN city = 'Mumbai' THEN 'Pune'
                ELSE city END

UPDATE employees
SET city = CASE WHEN city = 'Pune' THEN 'Mumbai'
                WHEN city = 'Mumbai' THEN 'Pune'
                ELSE city END

Run inside a transaction and roll back after testing.

P24. Find the manager who has the most direct reports.

sql

SELECT manager_id, COUNT(*) AS reports
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
ORDER BY reports DESC
LIMIT 1;
-- manager 1 (Amit), 4 reports

SELECT manager_id, COUNT(*) AS reports
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
ORDER BY reports DESC
LIMIT 1;
-- manager 1 (Amit), 4 reports

SELECT manager_id, COUNT(*) AS reports
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
ORDER BY reports DESC
LIMIT 1;
-- manager 1 (Amit), 4 reports

P25. List cities with employees in more than one department.

sql

SELECT city
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY city
HAVING COUNT(DISTINCT dept_id) > 1;
-- Pune (IT and Testing), Mumbai has only HR so it's excluded

SELECT city
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY city
HAVING COUNT(DISTINCT dept_id) > 1;
-- Pune (IT and Testing), Mumbai has only HR so it's excluded

SELECT city
FROM employees
WHERE dept_id IS NOT NULL
GROUP BY city
HAVING COUNT(DISTINCT dept_id) > 1;
-- Pune (IT and Testing), Mumbai has only HR so it's excluded

WINDOW FUNCTIONS FOR FRESHERS

Function	What it does	Behaviour on ties
ROW_NUMBER()	Unique sequence number	Always unique
RANK()	Rank with gaps	Ties share a rank, next rank skips (1, 1, 3)
DENSE_RANK()	Rank without gaps	Ties share a rank, no skip (1, 1, 2)
SUM() OVER	Running or partition total	n/a
LAG() / LEAD()	Previous or next row’s value	n/a

sql

SELECT emp_name, salary,
       ROW_NUMBER() OVER (ORDER BY salary DESC) AS rn,
       RANK()       OVER (ORDER BY salary DESC) AS rnk,
       DENSE_RANK() OVER (ORDER BY salary DESC) AS drnk
FROM

SELECT emp_name, salary,
       ROW_NUMBER() OVER (ORDER BY salary DESC) AS rn,
       RANK()       OVER (ORDER BY salary DESC) AS rnk,
       DENSE_RANK() OVER (ORDER BY salary DESC) AS drnk
FROM

SELECT emp_name, salary,
       ROW_NUMBER() OVER (ORDER BY salary DESC) AS rn,
       RANK()       OVER (ORDER BY salary DESC) AS rnk,
       DENSE_RANK() OVER (ORDER BY salary DESC) AS drnk
FROM

With the two salaries of 50000, you’ll see RANK jump (2, 2, 4) while DENSE_RANK doesn’t (2, 2, 3). Window functions need MySQL 8.0+ or another modern database.

SQL ORDER OF EXECUTION (WHY WHERE CAN’T USE ALIASES OR AGGREGATES)

This explains why WHERE can’t use aggregates (groups don’t exist yet) and why a SELECT alias can be used in ORDER BY but usually not in WHERE.

PERFORMANCE BASICS FRESHERS SHOULD KNOW

Practice	Why
Select only needed columns	Avoid `SELECT *` in production code
Index columns used in WHERE, JOIN and ORDER BY	Faster lookups
Avoid functions on indexed columns in WHERE	`WHERE YEAR(date) = 2026` can stop index use. Use a date range.
Use `EXISTS` over `IN` for large subqueries (test both)	Often more efficient
Filter early	Reduce rows before joining and grouping
Use `EXPLAIN`	See how the database runs your query

SQL WITH JAVA: QUESTIONS FOR JAVA FULL STACK FRESHERS

Question	Short answer
Statement vs PreparedStatement?	PreparedStatement is precompiled, safer against SQL injection and faster for repeated queries.
How do you connect Java to a database?	Load the JDBC driver, get a connection, create a statement, execute, process the ResultSet and close resources (try-with-resources).
What is a connection pool?	A cache of reusable connections that avoids the cost of opening a new one for each request.
How does JPA map to SQL?	Entities map to tables, fields to columns, relationships to foreign keys.
What is the N+1 problem?	One query loads parents and then N more queries load children. Fix with joins or fetch strategies.

java

String sql = "SELECT emp_name FROM employees WHERE city = ?";
try (PreparedStatement ps = conn.prepareStatement(sql)) {
    ps.setString(1, "Pune");
    try (ResultSet rs = ps.executeQuery()) {
        while (rs.next()) System.out.println(rs.getString("emp_name"));
    }
}

String sql = "SELECT emp_name FROM employees WHERE city = ?";
try (PreparedStatement ps = conn.prepareStatement(sql)) {
    ps.setString(1, "Pune");
    try (ResultSet rs = ps.executeQuery()) {
        while (rs.next()) System.out.println(rs.getString("emp_name"));
    }
}

String sql = "SELECT emp_name FROM employees WHERE city = ?";
try (PreparedStatement ps = conn.prepareStatement(sql)) {
    ps.setString(1, "Pune");
    try (ResultSet rs = ps.executeQuery()) {
        while (rs.next()) System.out.println(rs.getString("emp_name"));
    }
}

COMMON MISTAKES FRESHERS MAKE IN SQL INTERVIEWS

Mistake	Why it hurts	Better approach
Using `= NULL`	Returns nothing	Use `IS NULL`
Using `COUNT()` after a LEFT JOIN*	Counts unmatched rows as 1	Count a column from the right table
Mixing WHERE and HAVING	Wrong or error	Row filter in WHERE, group filter in HAVING
Forgetting the join condition	Accidental CROSS JOIN	Always write `ON`
Memorising only the answer	Fails on a variation	Learn the logic
Ignoring ties	Wrong top-N results	Know RANK vs DENSE_RANK
Typing without explaining	Interviewer can’t follow	Narrate tables, joins, filters
Not testing queries	Syntax mistakes	Run every query on the sample database
Using `NOT IN` with NULLs	Empty results	Use `NOT EXISTS`

14-DAY SQL PREPARATION PLAN

Day	Focus	Task
1	Setup	Install MySQL or PostgreSQL, create the sample tables
2	Basics	Q1 to Q12, DDL vs DML, DELETE vs TRUNCATE vs DROP
3	Keys and normalisation	Q13 to Q22, design a small 3-table schema
4	SELECT practice	Q23 to Q32, plus P10, P13, P15
5	Aggregates	Q33 to Q40, plus P7, P14, P18
6	Joins 1	INNER and LEFT joins, P6, P12
7	Joins 2	Self join, P9, P11 and P24
8	Subqueries	Q49 to Q51, P1, P5, P8, P16
9	Nth highest and duplicates	P2, P3, P4, P19
10	Window functions	P20, P21, RANK vs DENSE_RANK
11	Views, indexes, transactions	Q52 to Q60, run a transaction with ROLLBACK
12	SQL with Java	Write a JDBC PreparedStatement example
13	Mock interview	Solve 8 random problems aloud in 45 minutes
14	Review	Redo the problems you got wrong and explain each aloud

MYTH VS REALITY

Myth	Reality
“SQL is only for data roles.”	Developers, testers and analysts all use SQL.
“Joins are too hard.”	Practise on the sample tables and they become pattern recognition.
“Memorising 100 queries is enough.”	Variations test whether you understand the logic.
“ORMs like JPA mean I don’t need SQL.”	Interviewers still test SQL, and debugging needs it.
“Only the correct answer matters.”	Explanation, edge cases (NULL, ties) and clarity count too.
“Window functions are for experts only.”	They are now common even in fresher rounds.

HOW VIBRANTMINDS HELPS FRESHERS CRACK SQL AND JAVA INTERVIEWS

VibrantMinds Technologies Pvt. Ltd. is a Pune-based IT training and placement support company. If you searched for VibrantMinds or VibrantMinds Technologies, here is what it offers:

Full Stack Java Development (about 4 to 6 months): Core Java, OOP, JDBC, SQL, Spring Boot, REST APIs, front-end basics, projects, aptitude and interview preparation. Please check the candidate portal for the latest syllabus and tools.
Software Testing (Manual + Automation) (SQL is useful for data validation), Core Java and Python & Data Science for other career paths
Interview preparation: Technical rounds, SQL and coding practice, project rounds, HR rounds, aptitude, communication and mock interviews
Placement support: Screening, drive coordination and interview support. Check https://candidates.vibrantminds.in/ for current openings.
Free WhatsApp job alert groups for fresher IT jobs, campus drives and walk-in alerts
Honesty first: VibrantMinds offers placement support, not a guaranteed job, and no legitimate recruiter should ask you to pay to be interviewed.

VibrantMinds works as a talent ecosystem connecting candidates, colleges and companies, with 1900+ partner colleges and 600+ hiring companies. More fresher job guidance: https://candidates.vibrantminds.in/blog/fresher-jobs-pune-2026

VIBRANTMINDS COURSES AT A GLANCE

Course	Duration	Key topics
Full Stack Java Development	4 to 6 months	Core Java, OOP, JDBC, SQL, Spring Boot, REST APIs, front-end basics, projects, aptitude and interview preparation
Software Testing (Manual + Automation)	2 to 5 months	Test cases, STLC, bug tracking, SQL, Selenium basics
Core Java	1 to 2 months	Syntax, OOP, collections, exceptions, strings
Python & Data Science	2 to 4 months	Python basics, data handling, analysis fundamentals

Fees, syllabus and batch dates are on https://candidates.vibrantminds.in/. Please confirm them directly before enrolling.

THE VIBRANTMINDS PLACEMENT PIPELINE: SIX STEPS

REAL OUTCOMES: WHAT TO EXPECT AND WHAT NOT TO

Past students have shared positive feedback about the training and support, and have gone on to roles at companies such as Aress Software and Education Technologies, Saama Technologies and Three M Software. Results vary by candidate, skill level, effort and market conditions, so please confirm details directly with the VibrantMinds team. No one can honestly promise a guaranteed job.

JOIN FREE VIBRANTMINDS WHATSAPP GROUPS: FREE IT JOB UPDATES FOR SQL, JAVA AND PYTHON FRESHERS IN PUNE AND INDIA

Looking for free WhatsApp groups, free IT job WhatsApp groups, a Java WhatsApp group for freshers or a job updates WhatsApp group in India? The VibrantMinds groups share fresher IT openings, campus drives, walk-in interview alerts and interview preparation updates, free of charge. Join whichever group has space. All three carry the same type of updates.

Group	What it offers	Link
VibrantMinds Fresher IT Jobs, Group 1	Fresher openings, campus drives, walk-in alerts	https://chat.whatsapp.com/CUCS1derlmTFee2c9dyUHd
VibrantMinds Fresher IT Jobs, Group 2	Fresher openings, campus drives, walk-in alerts	https://chat.whatsapp.com/FvMf1fGYQYnHppQ8AwFX40
VibrantMinds Fresher IT Jobs, Group 3	Fresher openings, campus drives, walk-in alerts	https://chat.whatsapp.com/I7RNnGMkb2L8S8HibdyWpQ

Never pay to apply. Legitimate recruiters do not charge candidates for tests, interviews or offer letters. Do not share sensitive personal or financial details in any group chat.

FREQUENTLY ASKED QUESTIONS

1. What are the most asked SQL interview questions for freshers?
DELETE vs TRUNCATE vs DROP, primary key vs foreign key, WHERE vs HAVING, INNER JOIN vs LEFT JOIN, GROUP BY with aggregates, normalisation basics, subqueries, and live queries such as the second highest salary, duplicate records, department-wise counts and employees earning more than their manager.

2. How do I find the second highest salary in SQL?
The simplest portable way is SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees);. You can also use SELECT DISTINCT salary FROM employees ORDER BY salary DESC LIMIT 1 OFFSET 1; in MySQL or PostgreSQL, or DENSE_RANK() to find any Nth highest salary, which handles ties correctly.

3. What is the difference between INNER JOIN and LEFT JOIN with an example?
INNER JOIN returns only rows that match in both tables. LEFT JOIN returns every row from the left table and fills the right side with NULL where nothing matches. In the sample tables, an employee with no department appears in the LEFT JOIN result but not in the INNER JOIN result.

4. What is the difference between WHERE and HAVING in SQL?
WHERE filters individual rows before grouping and can’t use aggregate functions. HAVING filters groups after GROUP BY and can use aggregates such as COUNT(*) > 2.

5. Is there a free WhatsApp group for SQL and Java freshers in Pune?
Yes. VibrantMinds runs free fresher IT job WhatsApp groups that share Java, SQL-related, testing and other IT openings, campus drives and walk-in alerts. Join through the links above, and always verify a job post and never pay to apply.

CONCLUSION: PRACTISE QUERIES, NOT JUST DEFINITIONS

The fastest way to master SQL interview questions for freshers is to run every query on a real database. Learn the 60 concepts, then solve the 25 problems on the sample tables until joins, GROUP BY, HAVING, subqueries and window functions feel natural. Watch the NULL and tie traps, explain your logic aloud and follow the 14-day plan.

If you’re in Pune and want structured Java and SQL training with practice projects, interview preparation and placement support, VibrantMinds Technologies is here to help.

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