The Top MySQL Query Interview Questions for 2023

Mastering MySQL queries is a must for any aspiring data analyst or data engineer. With many companies like Google, Facebook, Amazon etc using MySQL as their database of choice, you can expect MySQL query based questions in tech interviews.

In this comprehensive guide, I will be covering the most commonly asked MySQL interview questions and their answers in detail. Whether you are a fresher just starting your tech career or a seasoned professional, this guide will help you prepare and ace your next MySQL interview.

Why MySQL Queries are Important for Interviews

Here are some key reasons why MySQL query based questions are so popular in interviews

• MySQL is one of the most widely used relational database management systems Knowledge of MySQL is a must for any data related role

• Writing efficient queries to extract and manipulate data is an essential skill required in real world projects. Interviewers want to test this skill.

• How you structure your queries gives insights into your SQL skills and database concepts clarity.

• Query questions test your logical thinking and problem solving abilities.

• Senior roles require expertise in complex SQL operations like joins, subqueries, triggers etc. Query questions help assess this.

Now let’s look at some of the most frequently asked MySQL query interview questions and their ideal responses.

Common MySQL Interview Questions and Answers

Q1. Fetch the first name from the student table and display it as ‘STUDENT_NAME’

SELECT first_name AS STUDENT_NAME FROM student;

This simple query demonstrates aliasing using the AS keyword to rename a column in the result set.

Q2. Get unique values of subject from the student table

SELECT DISTINCT subject FROM student;

We use the DISTINCT keyword to eliminate duplicates and get unique values.

Q3. Get the first 3 characters of students’ first name

SELECT SUBSTRING(first_name, 1, 3) FROM student; 

The SUBSTRING function is used to extract part of a string by specifying start position and length.

Q4. Replace occurrence of ‘a’ with ‘A’ in first name

SELECT REPLACE(first_name, 'a', 'A') FROM student;

The REPLACE function substitutes all occurrences of a substring in a string.

Q5. Display first and last name together as full name

SELECT CONCAT(first_name, ' ', last_name) AS full_name FROM student;

The CONCAT function joins strings together.

Q6. Sort students by first name and subject in descending order

SELECT * FROM student ORDER BY first_name, subject DESC;

The ORDER BY clause sorts rows based on specified columns. DESC sorts in descending order.

Q7. Get students with first names ‘Tony’ and ‘John’

SELECT * FROM studentWHERE first_name IN ('Tony', 'John'); 

The IN operator allows checking if a value exists in a list of values.

Q8. Get students whose first name ends with ‘a’

SELECT * FROM student WHERE first_name LIKE '%a';

The LIKE operator with wildcard ‘%’ matches patterns in strings.

Q9. Get students with GPA between 9.0 and 9.5

SELECT * FROM studentWHERE gpa BETWEEN 9.0 AND 9.5;

The BETWEEN operator selects values within a range.

Q10. Get the number of students for each subject

SELECT subject, COUNT(*) FROM studentGROUP BY subject; 

The COUNT() aggregate with GROUP BY gives count grouped by a column.

Q11. Get subjects having less than 3 students

SELECT subject, COUNT(*) as student_count  FROM studentGROUP BY subjectHAVING COUNT(*) < 3;

HAVING filters grouped rows matched by the condition.

Q12. Get the 5th highest GPA without using TOP/LIMIT

SELECT * FROM student s1WHERE 4 = (  SELECT COUNT(DISTINCT gpa)   FROM student s2  WHERE s2.gpa >= s1.gpa);

This uses a correlated subquery to determine the 5th highest GPA row.

Q13. Join students and scholarship tables

SELECT s.first_name, s.last_name, sc.amountFROM student sJOIN scholarship sc   ON s.student_id = sc.student_id; 

JOINs combine rows from multiple tables based on a condition.

Q14. Show running total of scholarship amounts

SELECT first_name,   SUM(amount) OVER (ORDER BY first_name) AS running_totalFROM student sJOIN scholarship scON s.student_id = sc.student_id;

The SUM() OVER() window function calculates running totals partitioned by first_name.

Q15. List students with no scholarships

SELECT s.student_idFROM student sLEFT JOIN scholarship scON s.student_id = sc.student_idWHERE sc.amount IS NULL;

A LEFT JOIN with IS NULL condition finds missing values.

So these were some of the most common MySQL query based interview questions. Practicing questions on joins, subqueries, window functions, wildcards etc. thoroughly will help you master this skill and stand out in your next data interview.

In addition to writing queries, you should also focus on explaining the logic, efficiency, and use cases of your queries to interviewers. This demonstrates your deeper understanding of database concepts.

With regular practice and solid grasp of basic and advanced SQL concepts, you can crack any MySQL interview questions that come your way! All the best!