Top 50 SQL Query Interview Questions and Answers for Practice

Hello friends! in this post, we will see some of the most common SQL queries asked in interviews. Whether you are a DBA, developer, tester, or data analyst, these SQL query interview questions and answers are going to help you.

In fact, I have been asked most of these questions during interviews in the different phases of my career.

If you want to skip the basic questions and start with some tricky SQL queries then you can directly move to our SQL queries interview questions for the experienced section. 

Consider the below two tables for reference while trying to solve the SQL queries for practice.

Table – EmployeeDetails

Table – EmployeeSalary

For your convenience, I have compiled the top 10 questions for you. You can try solving these questions and click on the links to go to their respective answers.

1. SQL Query to fetch records that are present in one table but not in another table.
2. SQL query to fetch all the employees who are not working on any project.
3. SQL query to fetch all the Employees from EmployeeDetails who joined in the Year 2020.
4. Fetch all employees from EmployeeDetails who have a salary record in EmployeeSalary.
5. Write an SQL query to fetch a project-wise count of employees.
6. Fetch employee names and salaries even if the salary value is not present for the employee.
7. Write an SQL query to fetch all the Employees who are also managers.
8. Write an SQL query to fetch duplicate records from EmployeeDetails.
9. Write an SQL query to fetch only odd rows from the table.
10. Write a query to find the 3rd highest salary from a table without top or limit keyword.
    
    

Or, you can also jump to our below two sections on SQL query interview questions for freshers and experienced professionals.

SQL Query Interview Questions for Freshers

Here is a list of top SQL query interview questions and answers for fresher candidates that will help them in their interviews. In these queries, we will focus on the basic SQL commands only.

1. Write an SQL query to fetch the EmpId and FullName of all the employees working under the Manager with id – ‘986’.
We can use the EmployeeDetails table to fetch the employee details with a where clause for the manager-

SELECT  EmpId, FullName
FROM EmployeeDetails
WHERE ManagerId = 986;

2. Write an SQL query to fetch the different projects available from the EmployeeSalary table.
While referring to the EmployeeSalary table, we can see that this table contains project values corresponding to each employee, or we can say that we will have duplicate project values while selecting Project values from this table.

So, we will use the distinct clause to get the unique values of the Project.

SELECT DISTINCT(Project)
FROM EmployeeSalary;

3. Write an SQL query to fetch the count of employees working in project ‘P1’.
Here, we would be using aggregate function count() with the SQL where clause-

SELECT COUNT(*) 
FROM EmployeeSalary 
WHERE Project = 'P1';

4. Write an SQL query to find the maximum, minimum, and average salary of the employees.
We can use the aggregate function of SQL to fetch the max, min, and average values-

SELECT Max(Salary), 
Min(Salary), 
AVG(Salary) 
FROM EmployeeSalary;

5. Write an SQL query to find the employee id whose salary lies in the range of 9000 and 15000.
Here, we can use the ‘Between’ operator with a where clause.

SELECT EmpId, Salary
FROM EmployeeSalary
WHERE Salary BETWEEN 9000 AND 15000;

6. Write an SQL query to fetch those employees who live in Toronto and work under the manager with ManagerId – 321.
Since we have to satisfy both the conditions – employees living in ‘Toronto’ and working in Project ‘P2’. So, we will use the AND operator here-

SELECT EmpId, City, ManagerId
FROM EmployeeDetails
WHERE City='Toronto' AND ManagerId='321';

7. Write an SQL query to fetch all the employees who either live in California or work under a manager with ManagerId – 321.
This interview question requires us to satisfy either of the conditions – employees living in ‘California’ and working under Manager with ManagerId – 321. So, we will use the OR operator here-

SELECT EmpId, City, ManagerId
FROM EmployeeDetails
WHERE City='California' OR ManagerId='321';

8. Write an SQL query to fetch all those employees who work on Projects other than P1.
Here, we can use the NOT operator to fetch the rows which are not satisfying the given condition.

SELECT EmpId
FROM EmployeeSalary
WHERE NOT Project='P1';

Or using the ‘not equal to’ operator-

SELECT EmpId
FROM EmployeeSalary
WHERE Project <> 'P1';

For the difference between NOT and <> SQL operators, check this link – Difference between the NOT and != operators.

9. Write an SQL query to display the total salary of each employee adding the Salary with Variable value.
Here, we can simply use the ‘+’ operator in SQL.

SELECT EmpId,
Salary+Variable as TotalSalary 
FROM EmployeeSalary;

10. Write an SQL query to fetch the employees whose name begins with any two characters, followed by a text “hn” and ends with any sequence of characters.
For this question, we can create an SQL query using like operator with ‘_’ and ‘%’ wild card characters, where ‘_’ matches a single character and ‘%’ matches ‘0 or multiple characters.

SELECT FullName
FROM EmployeeDetails
WHERE FullName LIKE ‘__hn%’;

11. Write an SQL query to fetch all the EmpIds which are present in either of the tables – ‘EmployeeDetails’ and ‘EmployeeSalary’.
In order to get unique employee ids from both tables, we can use the Union clause which can combine the results of the two SQL queries and return unique rows.

SELECT EmpId FROM EmployeeDetails
UNION 
SELECT EmpId FROM EmployeeSalary;

12. Write an SQL query to fetch common records between two tables.
SQL Server – Using INTERSECT operator-

SELECT * FROM EmployeeSalary
INTERSECT
SELECT * FROM ManagerSalary;

MySQL – Since MySQL doesn’t have INTERSECT operator so we can use the subquery-

SELECT *
FROM EmployeeSalary
WHERE EmpId IN 
(SELECT EmpId from ManagerSalary);

13. Write an SQL query to fetch records that are present in one table but not in another table.
SQL Server – Using MINUS- operator-

SELECT * FROM EmployeeSalary
MINUS
SELECT * FROM ManagerSalary;

MySQL – Since MySQL doesn’t have a MINUS operator so we can use LEFT join-

SELECT EmployeeSalary.*
FROM EmployeeSalary
LEFT JOIN
ManagerSalary USING (EmpId)
WHERE ManagerSalary.EmpId IS NULL;

14. Write an SQL query to fetch the EmpIds that are present in both the tables –  ‘EmployeeDetails’ and ‘EmployeeSalary.
Using subquery-

SELECT EmpId FROM 
EmployeeDetails 
where EmpId IN 
(SELECT EmpId FROM EmployeeSalary);

15. Write an SQL query to fetch the EmpIds that are present in EmployeeDetails but not in EmployeeSalary.
Using subquery-

SELECT EmpId FROM 
EmployeeDetails 
where EmpId Not IN 
(SELECT EmpId FROM EmployeeSalary);

16. Write an SQL query to fetch the employee’s full names and replace the space with ‘-’.
Using the ‘Replace’ function-

SELECT REPLACE(FullName, ' ', '-') 
FROM EmployeeDetails;

17. Write an SQL query to fetch the position of a given character(s) in a field.
Using the ‘Instr’ function-

SELECT INSTR(FullName, 'Snow')
FROM EmployeeDetails;

18. Write an SQL query to display both the EmpId and ManagerId together.
Here we can use the CONCAT command.

SELECT CONCAT(EmpId, ManagerId) as NewId
FROM EmployeeDetails;

19. Write a query to fetch only the first name(string before space) from the FullName column of the EmployeeDetails table.
In this question, we are required to first fetch the location of the space character in the FullName field and then extract the first name out of the FullName field.

For finding the location we will use the LOCATE method in MySQL and CHARINDEX in SQL SERVER and for fetching the string before space, we will use the SUBSTRING OR MID method.

MySQL – using MID

SELECT MID(FullName, 1, LOCATE(' ',FullName)) 
FROM EmployeeDetails;

SQL Server – using SUBSTRING

SELECT SUBSTRING(FullName, 1, CHARINDEX(' ',FullName)) 
FROM EmployeeDetails;

20. Write an SQL query to uppercase the name of the employee and lowercase the city values.
We can use SQL Upper and Lower functions to achieve the intended results.

SELECT UPPER(FullName), LOWER(City) 
FROM EmployeeDetails;

21. Write an SQL query to find the count of the total occurrences of a particular character – ‘n’ in the FullName field.
Here, we can use the ‘Length’ function. We can subtract the total length of the FullName field from the length of the FullName after replacing the character – ‘n’.

SELECT FullName, 
LENGTH(FullName) - LENGTH(REPLACE(FullName, 'n', ''))
FROM EmployeeDetails;

22. Write an SQL query to update the employee names by removing leading and trailing spaces.
Using the ‘Update’ command with the ‘LTRIM’ and ‘RTRIM’ functions.

UPDATE EmployeeDetails 
SET FullName = LTRIM(RTRIM(FullName));

23. Fetch all the employees who are not working on any project.
This is one of the very basic interview questions in which the interviewer wants to see if the person knows about the commonly used – Is NULL operator.

SELECT EmpId 
FROM EmployeeSalary 
WHERE Project IS NULL;

24. Write an SQL query to fetch employee names having a salary greater than or equal to 5000 and less than or equal to 10000.
Here, we will use BETWEEN in the ‘where’ clause to return the EmpId of the employees with salary satisfying the required criteria and then use it as a subquery to find the fullName of the employee from the EmployeeDetails table.

SELECT FullName 
FROM EmployeeDetails 
WHERE EmpId IN 
(SELECT EmpId FROM EmployeeSalary 
WHERE Salary BETWEEN 5000 AND 10000);

25. Write an SQL query to find the current date-time.
MySQL-

SELECT NOW();

SQL Server-

SELECT getdate();

Oracle-

SELECT SYSDATE FROM DUAL;

26. Write an SQL query to fetch all the Employee details from the EmployeeDetails table who joined in the Year 2020.
Using BETWEEN for the date range ’01-01-2020′ AND ’31-12-2020′-

SELECT * FROM EmployeeDetails
WHERE DateOfJoining BETWEEN '2020/01/01'
AND '2020/12/31';

Also, we can extract the year part from the joining date (using YEAR in MySQL)-

SELECT * FROM EmployeeDetails 
WHERE YEAR(DateOfJoining) = '2020';

27. Write an SQL query to fetch all employee records from the EmployeeDetails table who have a salary record in the EmployeeSalary table.
Using ‘Exists’-

SELECT * FROM EmployeeDetails E
WHERE EXISTS
(SELECT * FROM EmployeeSalary S 
WHERE  E.EmpId = S.EmpId);

28. Write an SQL query to fetch the project-wise count of employees sorted by project’s count in descending order.
The query has two requirements – first to fetch the project-wise count and then to sort the result by that count.

For project-wise count, we will be using the GROUP BY clause and for sorting, we will use the ORDER BY clause on the alias of the project count.

SELECT Project, count(EmpId) EmpProjectCount
FROM EmployeeSalary
GROUP BY Project
ORDER BY EmpProjectCount DESC;

29. Write a query to fetch employee names and salary records. Display the employee details even if the salary record is not present for the employee.
This is again one of the very common interview questions in which the interviewer just wants to check the basic knowledge of SQL JOINS.

Here, we can use the left join with the EmployeeDetail table on the left side of the EmployeeSalary table.

SELECT E.FullName, S.Salary 
FROM EmployeeDetails E 
LEFT JOIN 
EmployeeSalary S
ON E.EmpId = S.EmpId;

30. Write an SQL query to join 3 tables.
Considering 3 tables TableA, TableB, and TableC, we can use 2 joins clauses like below-

SELECT column1, column2
FROM TableA
JOIN TableB ON TableA.Column3 = TableB.Column3
JOIN TableC ON TableA.Column4 = TableC.Column4;

For more questions on SQL Joins, you can also check our top SQL Joins Interview Questions.

SQL Query Interview Questions for Experienced

Here is a list of some of the most frequently asked SQL query interview questions for experienced professionals. These questions cover SQL queries on advanced SQL JOIN concepts, fetching duplicate rows, odd and even rows, nth highest salary, etc.

31. Write an SQL query to fetch all the Employees who are also managers from the EmployeeDetails table.
Here, we have to use Self-Join as the requirement wants us to analyze the EmployeeDetails table as two tables. We will use different aliases ‘E’ and ‘M’ for the same EmployeeDetails table.

SELECT DISTINCT E.FullName
FROM EmployeeDetails E
INNER JOIN EmployeeDetails M
ON E.EmpID = M.ManagerID;

32. Write an SQL query to fetch duplicate records from EmployeeDetails (without considering the primary key – EmpId).
In order to find duplicate records from the table, we can use GROUP BY on all the fields and then use the HAVING clause to return only those fields whose count is greater than 1 i.e. the rows having duplicate records.

SELECT FullName, ManagerId, DateOfJoining, City, COUNT(*)
FROM EmployeeDetails
GROUP BY FullName, ManagerId, DateOfJoining, City
HAVING COUNT(*) > 1;

33. Write an SQL query to remove duplicates from a table without using a temporary table.
Here, we can use delete with alias and inner join. We will check for the equality of all the matching records and then remove the row with a higher EmpId.

DELETE E1 FROM EmployeeDetails E1
INNER JOIN EmployeeDetails E2 
WHERE E1.EmpId > E2.EmpId 
AND E1.FullName = E2.FullName 
AND E1.ManagerId = E2.ManagerId
AND E1.DateOfJoining = E2.DateOfJoining
AND E1.City = E2.City;

34. Write an SQL query to fetch only odd rows from the table.
In case we have an auto-increment field e.g. EmpId then we can simply use the below query-

SELECT * FROM EmployeeDetails 
WHERE MOD (EmpId, 2) <> 0;

In case we don’t have such a field then we can use the below queries.

Using Row_number in SQL server and checking that the remainder when divided by 2 is 1-

SELECT E.EmpId, E.Project, E.Salary
FROM (
    SELECT *, Row_Number() OVER(ORDER BY EmpId) AS RowNumber
    FROM EmployeeSalary
) E
WHERE E.RowNumber % 2 = 1;

Using a user-defined variable in MySQL-

SELECT *
FROM (
      SELECT *, @rowNumber := @rowNumber+ 1 rn
      FROM EmployeeSalary
      JOIN (SELECT @rowNumber:= 0) r
     ) t 
WHERE rn % 2 = 1;

35. Write an SQL query to fetch only even rows from the table.
In case we have an auto-increment field e.g. EmpId then we can simply use the below query-

SELECT * FROM EmployeeDetails 
WHERE MOD (EmpId, 2) = 0;

In case we don’t have such a field then we can use the below queries.

Using Row_number in SQL server and checking that the remainder, when divided by 2, is 1-

SELECT E.EmpId, E.Project, E.Salary
FROM (
    SELECT *, Row_Number() OVER(ORDER BY EmpId) AS RowNumber
    FROM EmployeeSalary
) E
WHERE E.RowNumber % 2 = 0;

Using a user-defined variable in MySQL-

SELECT *
FROM (
      SELECT *, @rowNumber := @rowNumber+ 1 rn
      FROM EmployeeSalary
      JOIN (SELECT @rowNumber:= 0) r
     ) t 
WHERE rn % 2 = 0;

36. Write an SQL query to create a new table with data and structure copied from another table.

CREATE TABLE NewTable 
SELECT * FROM EmployeeSalary;

37. Write an SQL query to create an empty table with the same structure as some other table.
Here, we can use the same query as above with the False ‘WHERE’ condition-

CREATE TABLE NewTable 
SELECT * FROM EmployeeSalary where 1=0;

38. Write an SQL query to fetch top n records.
In MySQL using LIMIT-

SELECT *
FROM EmployeeSalary
ORDER BY Salary DESC LIMIT N;

In SQL server using TOP command-

SELECT TOP N *
FROM EmployeeSalary
ORDER BY Salary DESC;

39. Write an SQL query to find the nth highest salary from a table.
Using Top keyword (SQL Server)-

SELECT TOP 1 Salary
FROM (
      SELECT DISTINCT TOP N Salary
      FROM Employee
      ORDER BY Salary DESC
      )
ORDER BY Salary ASC;

Using limit clause(MySQL)-

SELECT Salary
FROM Employee
ORDER BY Salary DESC LIMIT N-1,1;

40. Write SQL query to find the 3rd highest salary from a table without using the TOP/limit keyword.
This is one of the most commonly asked interview questions. For this, we will use a correlated subquery.

In order to find the 3rd highest salary, we will find the salary value until the inner query returns a count of 2 rows having a salary greater than other distinct salaries.

SELECT Salary
FROM EmployeeSalary Emp1
WHERE 2 = (
                SELECT COUNT( DISTINCT ( Emp2.Salary ) )
                FROM EmployeeSalary Emp2
                WHERE Emp2.Salary > Emp1.Salary
            )

For the nth highest salary-

SELECT Salary
FROM EmployeeSalary Emp1
WHERE N-1 = (
                SELECT COUNT( DISTINCT ( Emp2.Salary ) )
                FROM EmployeeSalary Emp2
                WHERE Emp2.Salary > Emp1.Salary
            )

Scenario-based SQL Query Interview Questions

Let’s see some interview questions based on different scenarios. The questions are of varying difficulty levels and the goal is to prepare you for different real-time scenario-based questions.

41. Consider a SalesData with columns SaleID, ProductID, RegionID, SaleAmount. Write a query to find the total sales amount for each product in each region.
The below query sums up SaleAmount for each combination of ProductID and RegionID, giving an insight into the total sales per product per region.

SELECT ProductID, RegionID, SUM(SaleAmount) AS TotalSales 
FROM SalesData 
GROUP BY ProductID, RegionID;

42. Write a query to find employees who earn more than their managers.
Here, we will write a query that joins the EmployeeDetails table with itself to compare the salaries of employees with their respective managers.

SELECT E.Name AS EmployeeName, 
M.Name AS ManagerName, 
E.Salary AS EmployeeSalary, 
M.Salary AS ManagerSalary 
FROM EmployeeDetails E JOIN EmployeeDetails M 
ON E.ManagerID = M.EmployeeID 
WHERE E.Salary > M.Salary;

43. Consider a BookCheckout table with columns – CheckoutID, MemberID, BookID, CheckoutDate, ReturnDate. Write an SQL query to find the number of books checked out by each member.

SELECT MemberID, COUNT(*) AS NumberOfBooksCheckedOut 
FROM BookCheckout 
GROUP BY MemberID;

44. Consider a StudentGrades table with columns – StudentID, CourseID, Grade. Write a query to find students who have scored an ‘A’ in more than three courses.
Here we will write an SQL query that filters students who have received an ‘A’ grade and groups them by StudentID, counting the number of ‘A’ grades per student.

SELECT StudentID FROM StudentGrades 
WHERE Grade = 'A' 
GROUP BY StudentID 
HAVING COUNT(*) > 3;

45. Consider a table OrderDetails with columns – OrderID, CustomerID, ProductID, OrderDate, Quantity, Price. Write a query to find the average order value for each customer.
The below query calculates the average order value (quantity multiplied by price) for each customer.

SELECT CustomerID, AVG(Quantity * Price) AS AvgOrderValue 
FROM OrderDetails 
GROUP BY CustomerID;

46. Consider a table PatientVisits with Columns VisitID, PatientID, DoctorID, VisitDate, Diagnosis. Write a query to find the latest visit date for each patient.

SELECT PatientID, MAX(VisitDate) AS LatestVisitDate 
FROM PatientVisits 
GROUP BY PatientID;

47. For a table FlightBookings with columns – BookingID, FlightID, PassengerID, BookingDate, TravelDate, Class, write a query to count the number of bookings for each flight class.
Here, we will write an SQL query that groups the bookings by Class and counts the number of bookings in each class.

SELECT Class, COUNT(*) AS NumberOfBookings 
FROM FlightBookings 
GROUP BY Class;

48. Consider a table FoodOrders with columns – OrderID, TableID, MenuItemID, OrderTime, Quantity. Write a query to find the most ordered menu item.
For the desired output, we will group the orders by MenuItemID and then sort the results by the count in descending order, fetching the top result.

SELECT MenuItemID 
FROM FoodOrders 
GROUP BY MenuItemID 
ORDER BY COUNT(*) DESC 
LIMIT 1;

49. Consider a table Transactions with columns – TransactionID, CustomerID, ProductID, TransactionDate, Amount. Write a query to find the total transaction amount for each month.
The below query sums the Amount for each month, giving a monthly total transaction amount.

SELECT MONTH(TransactionDate) AS Month, 
SUM(Amount) AS TotalAmount 
FROM Transactions 
GROUP BY MONTH(TransactionDate);

50. Consider a table EmployeeAttendance with columns – AttendanceID, EmployeeID, Date, Status. Write a query to find employees with more than 5 absences in a month.
This query filters the records for absent status, groups them by EmployeeID and month, and counts absences, filtering for more than 5 absences.

SELECT EmployeeID, 
MONTH(Date) AS Month, 
COUNT(*) AS Absences 
FROM EmployeeAttendance 
WHERE Status = 'Absent' 
GROUP BY EmployeeID, MONTH(Date) 
HAVING COUNT(*) > 5;

This concludes our post on frequently asked SQL query interview questions and answers. I hope you practice these questions and ace your database interviews.

If you feel, we have missed any of the common interview questions on SQL then do let us know in the comments and we will add those questions to our list.

Do check our article on – RDBMS Interview Questions, focussing on the theoretical interview questions based on the DBMS and SQL concepts.