Scalar subqueries in MySQL - Mini Project: Build & Apply

Using Scalar Subqueries in MySQL

📖 Scenario: You are managing a small bookstore database. You want to find out the price of the most expensive book in each category and display it alongside the category name.

🎯 Goal: Build a MySQL query using scalar subqueries to show each book category with the highest book price in that category.

📋 What You'll Learn

Create a table called books with columns id (integer), title (varchar), category (varchar), and price (decimal).

Insert at least 5 rows with different categories and prices.

Write a scalar subquery to find the maximum price for each category.

Select the category and the maximum price using the scalar subquery.

💡 Why This Matters

🌍 Real World

Scalar subqueries help you get single values from related data, like finding max prices per category in a bookstore.

💼 Career

Database developers and analysts often use scalar subqueries to write efficient queries that summarize or compare data within groups.

1

Create the books table and insert data

Create a table called books with columns id as integer, title as varchar(100), category as varchar(50), and price as decimal(5,2). Then insert these exact rows: (1, 'Learn SQL', 'Programming', 29.99), (2, 'Cooking 101', 'Cooking', 15.50), (3, 'Advanced SQL', 'Programming', 39.99), (4, 'Baking Basics', 'Cooking', 12.00), (5, 'Gardening Tips', 'Gardening', 22.00).

MySQL

-- Create the books table and insert data
-- Your code here

Hint

Use CREATE TABLE to define the table and INSERT INTO to add rows.

2

Write a scalar subquery to find the maximum price per category

Create a scalar subquery that selects the maximum price from books where the category matches the outer query's category. Assign this subquery to a column alias called max_price.

MySQL

CREATE TABLE books (
  id INT,
  title VARCHAR(100),
  category VARCHAR(50),
  price DECIMAL(5,2)
);

INSERT INTO books (id, title, category, price) VALUES
(1, 'Learn SQL', 'Programming', 29.99),
(2, 'Cooking 101', 'Cooking', 15.50),
(3, 'Advanced SQL', 'Programming', 39.99),
(4, 'Baking Basics', 'Cooking', 12.00),
(5, 'Gardening Tips', 'Gardening', 22.00);

-- Write a scalar subquery to find max price per category
-- Your code here

Hint

Use a subquery inside the SELECT clause that compares the outer query's category with the inner query's category.

3

Select distinct categories with their maximum price

Modify the query to select distinct category values from books and use the scalar subquery to show the maximum price for each category. Use DISTINCT in the outer query.

MySQL

CREATE TABLE books (
  id INT,
  title VARCHAR(100),
  category VARCHAR(50),
  price DECIMAL(5,2)
);

INSERT INTO books (id, title, category, price) VALUES
(1, 'Learn SQL', 'Programming', 29.99),
(2, 'Cooking 101', 'Cooking', 15.50),
(3, 'Advanced SQL', 'Programming', 39.99),
(4, 'Baking Basics', 'Cooking', 12.00),
(5, 'Gardening Tips', 'Gardening', 22.00);

-- Select distinct categories with max price using scalar subquery
-- Your code here

Hint

Use SELECT DISTINCT to get unique categories and the scalar subquery for max price.

4

Complete the query with ordering by category

Add an ORDER BY category clause at the end of the query to sort the results alphabetically by category.

MySQL

CREATE TABLE books (
  id INT,
  title VARCHAR(100),
  category VARCHAR(50),
  price DECIMAL(5,2)
);

INSERT INTO books (id, title, category, price) VALUES
(1, 'Learn SQL', 'Programming', 29.99),
(2, 'Cooking 101', 'Cooking', 15.50),
(3, 'Advanced SQL', 'Programming', 39.99),
(4, 'Baking Basics', 'Cooking', 12.00),
(5, 'Gardening Tips', 'Gardening', 22.00);

-- Add ORDER BY category to sort results
-- Your code here

Hint

Use ORDER BY category at the end of the query to sort alphabetically.

Practice

1. What does a scalar subquery return in MySQL?

easy

A. A list of tables

B. A single value that can be used in another query

C. Only column names

D. Multiple rows and columns

Solution

1. Step 1: Understand scalar subquery definition

   A scalar subquery returns exactly one value, not multiple rows or columns.

2. Step 2: Identify usage in queries

   This single value can be used inside another query, like in WHERE or SELECT clauses.

3. Final Answer:

   A single value that can be used in another query -> Option B

4. Quick Check:

   Scalar subquery = single value [OK]

Hint: Scalar subqueries always return one value only [OK]

Common Mistakes:

• Thinking scalar subqueries return multiple rows
• Confusing scalar subqueries with table subqueries
• Assuming scalar subqueries return column names only

2. Which of the following is the correct syntax for a scalar subquery in a SELECT statement?

easy

A. SELECT name, MAX(score) FROM scores FROM students;

B. SELECT name, MAX(score) AS max_score WHERE score > 50;

C. SELECT name, (SELECT MAX(score) FROM scores) AS max_score FROM students;

D. SELECT name, (MAX(score) FROM scores) AS max_score FROM students;

Solution

1. Step 1: Analyze correct scalar subquery syntax

   The scalar subquery must be enclosed in parentheses and return a single value, as in (SELECT MAX(score) FROM scores).

2. Step 2: Check placement in SELECT clause

   SELECT name, (SELECT MAX(score) FROM scores) AS max_score FROM students; correctly places the scalar subquery inside the SELECT list with an alias.

3. Final Answer:

   SELECT name, (SELECT MAX(score) FROM scores) AS max_score FROM students; -> Option C

4. Quick Check:

   Scalar subquery syntax = SELECT name, (SELECT MAX(score) FROM scores) AS max_score FROM students; [OK]

Hint: Scalar subqueries must be in parentheses inside SELECT or WHERE [OK]

Common Mistakes:

• Missing parentheses around subquery
• Incorrect FROM clause placement
• Using aggregate functions without subquery

3. Given the tables:
students(id, name)
scores(student_id, score)
What will this query return?

SELECT name, (SELECT AVG(score) FROM scores WHERE student_id = students.id) AS avg_score FROM students;

medium

A. Each student's name with their average score

B. The overall average score for all students repeated for each name

C. Syntax error due to subquery

D. List of student names only without scores

Solution

1. Step 1: Understand the subquery correlation

   The subquery calculates the average score for each student by matching student_id with students.id.

2. Step 2: Result of the main query

   The main query returns each student's name along with their individual average score from the subquery.

3. Final Answer:

   Each student's name with their average score -> Option A

4. Quick Check:

   Scalar subquery returns per-student average [OK]

Hint: Correlated scalar subqueries return values per row [OK]

Common Mistakes:

• Assuming subquery returns overall average for all rows
• Confusing correlated and uncorrelated subqueries
• Expecting syntax error when none exists

4. Identify the error in this query using a scalar subquery:

SELECT name, (SELECT score FROM scores) AS score FROM students;

medium

A. No error, query runs fine

B. Missing alias for subquery

C. Incorrect table name in subquery

D. Subquery returns multiple rows causing an error

Solution

1. Step 1: Analyze the subquery result

   The subquery SELECT score FROM scores returns multiple rows because no WHERE clause limits it.

2. Step 2: Understand scalar subquery requirement

   Scalar subqueries must return exactly one value; multiple rows cause an error.

3. Final Answer:

   Subquery returns multiple rows causing an error -> Option D

4. Quick Check:

   Scalar subquery must return one value [OK]

Hint: Scalar subqueries must return one value, not many [OK]

Common Mistakes:

• Ignoring multiple rows returned by subquery
• Assuming alias is mandatory for subquery
• Thinking query runs without error

5. You want to find students whose score is above the average score of all students. Which query correctly uses a scalar subquery to achieve this?

hard

A. SELECT name FROM students WHERE score > (SELECT AVG(score) FROM scores);

B. SELECT name FROM students WHERE score > SELECT AVG(score) FROM scores;

C. SELECT name FROM students WHERE (SELECT AVG(score) FROM scores) < (SELECT score FROM scores WHERE student_id = students.id);

D. SELECT name FROM students WHERE score > (SELECT score FROM scores);

Solution

1. Step 1: Understand the goal

   We want students whose individual score is greater than the average score of all students.

2. Step 2: Analyze each option

   SELECT name FROM students WHERE score > (SELECT AVG(score) FROM scores); correctly compares each student's score with the average score using a scalar subquery.

3. Final Answer:

   SELECT name FROM students WHERE score > (SELECT AVG(score) FROM scores); -> Option A

4. Quick Check:

   Scalar subquery compares average with individual score [OK]

Hint: Use scalar subquery for average, correlate for individual score [OK]

Common Mistakes:

• Missing parentheses around subqueries
• Using subquery without correlation for individual scores
• Syntax errors from missing SELECT in WHERE