Correlated subqueries in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select employees whose salary is greater than the average salary in their department.

MySQL

SELECT employee_id, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.[1]);

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

Adepartment_id

Bemployee_id

Csalary

Dmanager_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using employee_id instead of department_id causes no correlation.

Using salary in the WHERE clause of subquery is incorrect here.

The subquery compares each employee's salary to the average salary of their own department. The correlation is done using department_id.

2fill in blank

medium

Complete the code to find customers who have placed more orders than the average number of orders per customer.

MySQL

SELECT customer_id FROM orders o GROUP BY customer_id HAVING COUNT(*) > (SELECT AVG([1]) FROM (SELECT customer_id, COUNT(*) AS order_count FROM orders GROUP BY customer_id) AS subquery);

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

Aorder_date

Border_id

Corder_count

Dcustomer_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using customer_id would average customer IDs instead of order counts.

Using order_date or order_id does not represent the number of orders.

The subquery calculates the average number of orders per customer using the order_count column from the derived table.

3fill in blank

hard

Fix the error in the correlated subquery to find products priced higher than the average price in their category.

MySQL

SELECT product_id, price FROM products p WHERE price > (SELECT AVG(price) FROM products WHERE category_id = [1].category_id);

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

Acategory

Bproducts

Cc

Dp

Attempts:

3 left

💡 Hint

Common Mistakes

Using the table name 'products' inside the subquery instead of the alias causes an error.

Using an undefined alias like 'c' or 'category' causes syntax errors.

The subquery must refer to the alias of the outer query table, which is p, to correlate by category_id.

4fill in blank

hard

Fill both blanks to select employees who earn more than the average salary of their department and have a job title starting with 'Senior'.

MySQL

SELECT employee_id, salary FROM employees e WHERE salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.[1]) AND job_title [2] 'Senior%';

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

Adepartment_id

Bemployee_id

CLIKE

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of LIKE for pattern matching causes no results.

Using employee_id instead of department_id breaks the correlation.

The first blank needs the department_id to correlate the subquery. The second blank uses LIKE to match job titles starting with 'Senior'.

5fill in blank

hard

Fill both blanks to select orders where the order amount is greater than the average order amount for the same customer and the order status is 'Completed'.

MySQL

SELECT order_id, order_amount FROM orders o WHERE order_amount > (SELECT AVG(order_amount) FROM orders WHERE customer_id = o.[1]) AND order_status = [2];

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

Acustomer_id

B'Completed'

D"Completed"

Attempts:

3 left

💡 Hint

Common Mistakes

Using double quotes for string literals causes syntax errors in MySQL.

Not correlating on customer_id causes wrong average calculation.

The subquery correlates on customer_id. The order status comparison requires the string 'Completed' enclosed in single quotes.

Practice

1. What is a correlated subquery in MySQL?

easy

A. A subquery that uses values from the outer query for each row.

B. A subquery that runs only once before the outer query.

C. A query that joins two tables without any subqueries.

D. A subquery that returns a fixed constant value.

Solution

1. Step 1: Understand subquery behavior

   A correlated subquery depends on the outer query's current row values to execute.

2. Step 2: Compare options

   The option describing a subquery that uses values from the outer query for each row correctly defines a correlated subquery.

3. Final Answer:

   A subquery that uses values from the outer query for each row. -> Option A

4. Quick Check:

   Correlated subquery = uses outer row values [OK]

Hint: Correlated subqueries depend on outer query row values [OK]

Common Mistakes:

• Thinking subqueries always run once
• Confusing joins with subqueries
• Assuming subqueries return constants only

2. Which of the following is the correct syntax for a correlated subquery in MySQL?

easy

A. SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id);

B. SELECT * FROM employees WHERE salary IN (SELECT salary FROM employees);

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

D. SELECT * FROM employees WHERE salary = 50000;

Solution

1. Step 1: Identify correlated subquery syntax

   A correlated subquery references a column from the outer query inside the subquery, like e2.department_id = employees.department_id.

2. Step 2: Check each option

   SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); uses the outer table's column inside the subquery condition, making it a correlated subquery.

3. Final Answer:

   SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); -> Option A

4. Quick Check:

   Correlated subquery syntax = SELECT * FROM employees WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = employees.department_id); [OK]

Hint: Look for subquery referencing outer query table alias [OK]

Common Mistakes:

• Missing outer table reference in subquery
• Using subquery without WHERE clause for correlation
• Confusing correlated with simple subqueries

3. Given the tables:
employees(id, name, department_id, salary)
and
departments(id, name)
What will this query return?

SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees WHERE department_id = e.department_id);

medium

A. Employees whose salary is above the average salary of all departments combined.

B. Employees whose salary is above the average salary of all employees.

C. Employees whose salary is above the average salary of their department.

D. Syntax error due to subquery referencing outer query.

Solution

1. Step 1: Understand the subquery condition

   The subquery calculates the average salary for the department of the current employee (using e.department_id).

2. Step 2: Compare employee salary with department average

   The outer query selects employees whose salary is greater than their department's average salary.

3. Final Answer:

   Employees whose salary is above the average salary of their department. -> Option C

4. Quick Check:

   Correlated subquery filters by department average salary [OK]

Hint: Check if subquery filters by outer row's department_id [OK]

Common Mistakes:

• Thinking subquery calculates overall average salary
• Assuming syntax error due to correlation
• Confusing department and employee averages

4. Identify the error in this correlated subquery:

SELECT e1.name FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department = e1.department_id);

medium

A. Subquery cannot reference outer query columns.

B. Alias 'e1' is not allowed in subquery.

C. Missing GROUP BY clause in subquery.

D. Column 'department' does not exist in employees table.

Solution

1. Step 1: Check column names in subquery condition

   The subquery compares e2.department to e1.department_id, but 'department' column does not exist in employees table.

2. Step 2: Confirm correct column usage

   The correct column should be e2.department_id to match e1.department_id for correlation.

3. Final Answer:

   Column 'department' does not exist in employees table. -> Option D

4. Quick Check:

   Invalid column name causes error [OK]

Hint: Verify column names match exactly in both queries [OK]

Common Mistakes:

• Using wrong column names for correlation
• Assuming alias usage causes error
• Thinking GROUP BY is required in subquery

5. You want to find all departments where every employee earns more than the average salary of all employees in the company. Which query correctly uses a correlated subquery to achieve this?

hard

A. SELECT d.name FROM departments d WHERE ALL (SELECT salary FROM employees WHERE department_id = d.id) > (SELECT AVG(salary) FROM employees);

B. SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees));

C. SELECT d.name FROM departments d WHERE EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary > (SELECT AVG(salary) FROM employees));

D. SELECT d.name FROM departments d WHERE EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary > (SELECT AVG(salary) FROM employees WHERE department_id = d.id));

Solution

1. Step 1: Understand the requirement

   We want departments where every employee earns more than the company-wide average salary.

2. Step 2: Analyze the NOT EXISTS query

   SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees)); uses NOT EXISTS to ensure no employee in the department earns less or equal to the company average, meaning all earn more.

3. Step 3: Check other options

   The options using EXISTS check for the existence of at least one qualifying employee, not all. The option attempting to use ALL with a subquery on the left side uses invalid syntax in MySQL.

4. Final Answer:

   SELECT d.name FROM departments d WHERE NOT EXISTS (SELECT 1 FROM employees e WHERE e.department_id = d.id AND e.salary <= (SELECT AVG(salary) FROM employees)); -> Option B

5. Quick Check:

   NOT EXISTS ensures all employees earn above average [OK]

Hint: Use NOT EXISTS to check all employees meet condition [OK]

Common Mistakes:

• Using EXISTS instead of NOT EXISTS for 'all' condition
• Misusing ALL keyword in MySQL
• Comparing with department average instead of company average