JOIN with aggregate functions in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the total number of orders for each customer.

MySQL

SELECT customers.name, COUNT(orders.id) AS total_orders FROM customers LEFT JOIN orders ON customers.id = orders.customer_id GROUP BY [1];

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

Acustomers.name

Borders.id

Corders.customer_id

Dcustomers.id

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by order ID instead of customer name.

Not using GROUP BY at all.

The GROUP BY clause should group by the customer's unique ID to get the total orders per customer. Grouping by name can cause issues if names are not unique.

2fill in blank

medium

Complete the code to find the average order amount per customer.

MySQL

SELECT customers.name, AVG([1]) AS avg_order_amount FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name;

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

Acustomers.name

Borders.amount

Corders.id

Dcustomers.id

Attempts:

3 left

💡 Hint

Common Mistakes

Using customer ID or name inside AVG function.

Applying AVG to order ID instead of amount.

The AVG function should be applied to the order amount to find the average per customer.

3fill in blank

hard

Fix the error in the code to get the maximum order amount per customer.

MySQL

SELECT customers.name, MAX([1]) AS max_amount FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name;

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

Aorders.id

Bcustomers.name

Corders.amount

Dcustomers.id

Attempts:

3 left

💡 Hint

Common Mistakes

Using MAX on customer name or ID.

Using MAX on order ID instead of amount.

The MAX function should be applied to the order amount to find the highest order per customer.

4fill in blank

hard

Fill both blanks to calculate the total quantity ordered per product.

MySQL

SELECT products.name, SUM([1]) AS total_quantity FROM products JOIN order_items ON products.id = order_items.product_id GROUP BY [2];

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

Aorder_items.quantity

Bproducts.id

Cproducts.name

Dorder_items.product_id

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by product ID instead of name.

Summing product ID or name instead of quantity.

SUM should be applied to the quantity column to get total quantity. GROUP BY should be on product name to group results by product.

5fill in blank

hard

Fill all three blanks to find the average price per category.

MySQL

SELECT categories.name AS category_name, AVG([1]) AS avg_price FROM categories JOIN products ON categories.id = products.category_id GROUP BY [2] HAVING AVG([3]) > 50;

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

Aproducts.price

Bcategories.name

Dcategories.id

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by category ID instead of name.

Using HAVING on wrong column or without AVG.

AVG should be applied to products.price to calculate average price. GROUP BY should be on category name. HAVING filters categories with average price greater than 50.

Practice

1. What does the following SQL query do?
SELECT department, COUNT(employee_id) FROM employees JOIN departments ON employees.dept_id = departments.id GROUP BY department;

easy

A. Counts employees in each department by joining employees and departments tables.

B. Lists all employees without grouping by department.

C. Calculates the average salary per department.

D. Deletes employees who are not assigned to any department.

Solution

1. Step 1: Understand the JOIN operation

   The query joins the employees and departments tables using the department ID to combine related data.

2. Step 2: Analyze the GROUP BY and COUNT

   It groups the results by department and counts the number of employees in each group.

3. Final Answer:

   Counts employees in each department by joining employees and departments tables. -> Option A

4. Quick Check:

   JOIN + GROUP BY + COUNT = Employee count per department [OK]

Hint: JOIN tables, GROUP BY key, then apply COUNT [OK]

Common Mistakes:

• Ignoring GROUP BY causes wrong aggregation
• Using COUNT without JOIN returns incomplete data
• Confusing COUNT with SUM or AVG

2. Which of the following is the correct syntax to get the total sales per customer using JOIN and SUM?
SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name;

easy

A. SELECT customers.name, SUM(orders.amount) FROM customers LEFT JOIN orders GROUP BY customers.name;

B. SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders WHERE customers.id = orders.customer_id GROUP BY customers.name;

C. SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name;

D. SELECT customers.name, SUM(orders.amount) FROM customers, orders GROUP BY customers.name;

Solution

1. Step 1: Check JOIN syntax

   SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name; uses correct INNER JOIN syntax with ON clause to link customers and orders.

2. Step 2: Verify GROUP BY and SUM usage

   SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name; correctly groups by customer name and sums order amounts.

3. Final Answer:

   SELECT customers.name, SUM(orders.amount) FROM customers JOIN orders ON customers.id = orders.customer_id GROUP BY customers.name; -> Option C

4. Quick Check:

   JOIN with ON + GROUP BY + SUM = Correct syntax [OK]

Hint: Use JOIN ... ON ... with GROUP BY for aggregates [OK]

Common Mistakes:

• Using WHERE instead of ON in JOIN
• Missing ON clause causes syntax error
• Grouping by wrong column

3. Given tables:
employees(id, name, dept_id)
departments(id, department)
What is the output of this query?

SELECT d.department, COUNT(e.id) AS emp_count FROM employees e LEFT JOIN departments d ON e.dept_id = d.id GROUP BY d.department ORDER BY emp_count DESC;

medium

A. Syntax error due to missing GROUP BY column.

B. List of departments with employee counts, including departments with zero employees.

C. List of employees with their department names.

D. List of departments with employee counts, excluding departments with zero employees.

Solution

1. Step 1: Understand LEFT JOIN effect

   FROM employees e LEFT JOIN departments d ON e.dept_id = d.id includes all employees and matching departments if they exist. Departments with no employees do not appear because no employee row matches them.

2. Step 2: GROUP BY and COUNT behavior

   Grouping by d.department counts employees only for departments that have at least one employee (count >=1). Departments with zero employees are excluded.

3. Final Answer:

   List of departments with employee counts, excluding departments with zero employees. -> Option D

4. Quick Check:

   LEFT JOIN from employees + GROUP BY department excludes zero counts [OK]

Hint: LEFT JOIN keeps all left rows, COUNT counts matches [OK]

Common Mistakes:

• Assuming INNER JOIN behavior excludes zero counts
• Confusing COUNT(*) with COUNT(column)
• Ignoring NULL values in grouping

4. Identify the error in this query:

SELECT d.department, COUNT(e.id) FROM employees e JOIN departments d ON e.dept_id = d.id WHERE e.salary > 50000 GROUP BY e.dept_id;

medium

A. GROUP BY should use d.department instead of e.dept_id.

B. Missing HAVING clause for salary filter.

C. COUNT(e.id) cannot be used with JOIN.

D. ON clause is incorrect syntax.

Solution

1. Step 1: Check GROUP BY columns

   The SELECT lists d.department but GROUP BY uses e.dept_id, causing mismatch.

2. Step 2: Fix GROUP BY to match SELECT

   GROUP BY should use d.department or include it to avoid errors.

3. Final Answer:

   GROUP BY should use d.department instead of e.dept_id. -> Option A

4. Quick Check:

   GROUP BY columns must match SELECT non-aggregates [OK]

Hint: GROUP BY columns must match SELECT non-aggregates [OK]

Common Mistakes:

• Grouping by wrong or missing columns
• Using WHERE instead of HAVING for aggregates
• Incorrect JOIN ON syntax

5. You have tables:
orders(order_id, customer_id, amount)
customers(customer_id, name)
Write a query to find customers who have total orders amount greater than 1000, including customers with no orders showing zero total.
Which query is correct?

hard

A. SELECT c.name, SUM(o.amount) AS total FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id WHERE total > 1000 GROUP BY c.name;

B. SELECT c.name, COALESCE(SUM(o.amount), 0) AS total FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.name HAVING total > 1000;

C. SELECT c.name, SUM(o.amount) AS total FROM customers c JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.name HAVING total > 1000;

D. SELECT c.name, SUM(o.amount) AS total FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.name WHERE total > 1000;

Solution

1. Step 1: Use LEFT JOIN to include customers without orders

   LEFT JOIN ensures all customers appear, even if no matching orders exist.

2. Step 2: Use COALESCE to convert NULL sums to zero and HAVING to filter totals

   COALESCE replaces NULL with 0; HAVING filters groups by aggregate result.

3. Final Answer:

   SELECT c.name, COALESCE(SUM(o.amount), 0) AS total FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.name HAVING total > 1000; -> Option B

4. Quick Check:

   LEFT JOIN + COALESCE + HAVING filters totals correctly [OK]

Hint: Use LEFT JOIN + COALESCE + HAVING for conditional aggregates [OK]

Common Mistakes:

• Using WHERE instead of HAVING for aggregate filters
• Ignoring NULL sums for customers without orders
• Using INNER JOIN excludes customers with zero orders