Subqueries with IN operator in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all employees who work in departments listed in the subquery.

MySQL

SELECT employee_name FROM employees WHERE department_id [1] (SELECT department_id FROM departments WHERE location = 'New York');

Drag options to blanks, or click blank then click option'

A=

BNOT IN

CIN

DLIKE

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of 'IN' when comparing to a subquery result.

Using 'LIKE' which is for pattern matching, not list membership.

The IN operator is used to check if a value matches any value in a list or subquery result.

2fill in blank

medium

Complete the code to find all products whose category_id is in the list of categories with more than 100 products.

MySQL

SELECT product_name FROM products WHERE category_id [1] (SELECT category_id FROM products GROUP BY category_id HAVING COUNT(*) > 100);

Drag options to blanks, or click blank then click option'

AIN

BNOT IN

C=

DBETWEEN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which only compares to a single value, not a list.

Using 'BETWEEN' which is for ranges, not lists.

The IN operator allows filtering products whose category_id matches any category_id from the subquery.

3fill in blank

hard

Fix the error in the query to select customers who have orders in the USA.

MySQL

SELECT customer_name FROM customers WHERE customer_id [1] (SELECT customer_id FROM orders WHERE country = 'USA');

Drag options to blanks, or click blank then click option'

AIN

BLIKE

CNOT IN

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which causes an error if subquery returns multiple rows.

Using 'LIKE' which is for pattern matching, not list membership.

The IN operator is needed to compare a column to multiple values returned by the subquery.

4fill in blank

hard

Fill both blanks to select orders where the product_id is in the list of products priced above 50.

MySQL

SELECT order_id FROM orders WHERE product_id [1] (SELECT product_id FROM products WHERE price [2] 50);

Drag options to blanks, or click blank then click option'

AIN

B>

C<

DNOT IN

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'NOT IN' instead of 'IN' for the first blank.

Using '<' instead of '>' for the price comparison.

Use IN to check if product_id is in the subquery list, and > to filter products priced above 50.

5fill in blank

hard

Fill all three blanks to select employee names who work in departments located in 'London' and have a salary greater than 50000.

MySQL

SELECT employee_name FROM employees WHERE department_id [1] (SELECT department_id FROM departments WHERE location = [2]) AND salary [3] 50000;

Drag options to blanks, or click blank then click option'

AIN

B'London'

C>

DNOT IN

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'NOT IN' instead of 'IN' for the first blank.

Forgetting quotes around 'London' in the second blank.

Using '<' instead of '>' for the salary comparison.

The query uses IN to filter department_id by those in London, compares location to the string 'London', and filters salary with > for values greater than 50000.