Subqueries in FROM clause (derived tables) in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all columns from a derived table named 'sub'.

MySQL

SELECT [1] FROM (SELECT id, name FROM users) AS sub;

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

A*

Bid

Cname

Dsub

Attempts:

3 left

💡 Hint

Common Mistakes

Selecting the alias name instead of columns.

Forgetting to use the alias in the FROM clause.

Using * selects all columns from the derived table sub.

2fill in blank

medium

Complete the code to count the number of users in the derived table.

MySQL

SELECT COUNT([1]) FROM (SELECT id FROM users) AS user_count;

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

Auser_count

B*

Cname

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Counting a column not selected in the derived table.

Using the alias name inside COUNT().

Counting id counts the number of rows in the derived table.

3fill in blank

hard

Fix the error in the code by completing the alias for the derived table.

MySQL

SELECT sub.id FROM (SELECT id FROM users) [1];

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

AAS sub

BBY

CON

DIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect keywords like BY, ON, or IN for aliasing.

Omitting the alias entirely.

The AS keyword is used to alias the derived table.

4fill in blank

hard

Fill both blanks to select user names and their total orders from a derived table.

MySQL

SELECT [1], orders FROM (SELECT user_id, COUNT(*) [2] orders FROM orders GROUP BY user_id) AS order_summary JOIN users ON users.id = order_summary.user_id;

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

Ausers.name

BAS

Cuser_name

DON

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect aliasing keywords.

Selecting columns without table prefix causing ambiguity.

users.name selects the user name, and AS aliases the count as orders.

5fill in blank

hard

Fill all three blanks to create a derived table that calculates average scores and select students with average above 80.

MySQL

SELECT [1], avg_score FROM (SELECT student_id, AVG(score) [2] avg_score FROM scores GROUP BY student_id) [3] WHERE avg_score > 80;

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

Astudent_id

BAS

Chigh_achievers

Dscore

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to alias the derived table.

Not aliasing the average score column.

Selecting wrong columns in the outer query.

student_id is selected, AS aliases the average score, and high_achievers is the alias for the derived table.

Practice

1. What is the main purpose of using a subquery in the FROM clause in MySQL?

easy

A. To create a permanent table in the database

B. To update data in multiple tables at once

C. To delete rows based on a condition

D. To create a temporary table that the outer query can use

Solution

1. Step 1: Understand the role of subqueries in FROM clause

   Subqueries in the FROM clause act like temporary tables that the outer query can use to simplify complex queries.

2. Step 2: Differentiate from other SQL operations

   Updating or deleting data is not the purpose of subqueries in FROM; they are for organizing data selection.

3. Final Answer:

   To create a temporary table that the outer query can use -> Option D

4. Quick Check:

   Subquery in FROM = temporary table [OK]

Hint: Think of subquery in FROM as a temporary table [OK]

Common Mistakes:

• Confusing subqueries in FROM with UPDATE or DELETE statements
• Thinking subqueries create permanent tables
• Ignoring the need for an alias for the subquery

2. Which of the following is the correct syntax for using a subquery in the FROM clause in MySQL?

easy

A. SELECT * FROM (SELECT id FROM users) AS sub;

B. SELECT * FROM users WHERE (SELECT id FROM users);

C. SELECT * FROM users (SELECT id FROM users);

D. SELECT * FROM users AS (SELECT id FROM users);

Solution

1. Step 1: Identify correct subquery syntax in FROM clause

   The subquery must be enclosed in parentheses and given an alias using AS.

2. Step 2: Check each option

   SELECT * FROM (SELECT id FROM users) AS sub; correctly uses parentheses and alias. Others misuse syntax or place subquery incorrectly.

3. Final Answer:

   SELECT * FROM (SELECT id FROM users) AS sub; -> Option A

4. Quick Check:

   Subquery in FROM needs parentheses + alias [OK]

Hint: Always alias your subquery in FROM clause [OK]

Common Mistakes:

• Forgetting to alias the subquery
• Placing subquery outside parentheses
• Using subquery in WHERE or FROM incorrectly

3. Given the tables:
users(id, name)
orders(id, user_id, amount)
What will the following query return?

SELECT sub.user_id, sub.total_amount FROM (SELECT user_id, SUM(amount) AS total_amount FROM orders GROUP BY user_id) AS sub WHERE sub.total_amount > 100;

medium

A. List of all orders with amount greater than 100

B. List of user IDs with total order amount greater than 100

C. Syntax error due to missing alias

D. Empty result because no user has orders

Solution

1. Step 1: Understand the subquery

   The subquery calculates total order amount per user by grouping orders by user_id and summing amounts.

2. Step 2: Apply the outer query filter

   The outer query selects users whose total order amount is greater than 100 from the subquery result.

3. Final Answer:

   List of user IDs with total order amount greater than 100 -> Option B

4. Quick Check:

   Subquery sums orders; outer filters total_amount > 100 [OK]

Hint: Subquery groups and sums; outer filters results [OK]

Common Mistakes:

• Confusing total order amount with individual order amounts
• Forgetting to alias the subquery
• Assuming syntax error without checking alias

4. Identify the error in the following query:

SELECT sub.user_id, sub.total FROM (SELECT user_id, SUM(amount) FROM orders GROUP BY user_id);

medium

A. GROUP BY cannot be used in subqueries

B. SUM(amount) must be renamed with AS

C. Missing alias for the subquery

D. SELECT cannot use subqueries in FROM clause

Solution

1. Step 1: Check subquery alias requirement

   Subqueries in FROM clause must have an alias; this query lacks it.

2. Step 2: Verify other parts

   SUM(amount) without alias is allowed but referencing it as sub.total will fail; aliasing subquery is mandatory.

3. Final Answer:

   Missing alias for the subquery -> Option C

4. Quick Check:

   Subquery in FROM needs alias [OK]

Hint: Always give an alias to subqueries in FROM [OK]

Common Mistakes:

• Assuming SUM() must be aliased
• Thinking GROUP BY is disallowed in subqueries
• Ignoring alias requirement for subqueries

5. You want to find users who have placed orders totaling more than 500, but only include users who have more than 3 orders. Which query correctly uses a subquery in the FROM clause to achieve this?

Tables:
users(id, name)
orders(id, user_id, amount)

hard

A. SELECT u.name, stats.total_amount FROM users u JOIN (SELECT user_id, COUNT(*) AS order_count, SUM(amount) AS total_amount FROM orders GROUP BY user_id) AS stats ON u.id = stats.user_id WHERE stats.total_amount > 500 AND stats.order_count > 3;

B. SELECT u.name FROM users u WHERE (SELECT SUM(amount) FROM orders WHERE user_id = u.id) > 500 AND (SELECT COUNT(*) FROM orders WHERE user_id = u.id) > 3;

C. SELECT u.name, SUM(o.amount) FROM users u JOIN orders o ON u.id = o.user_id GROUP BY u.id HAVING SUM(o.amount) > 500 AND COUNT(o.id) > 3;

D. SELECT name FROM users WHERE id IN (SELECT user_id FROM orders GROUP BY user_id HAVING SUM(amount) > 500 AND COUNT(*) > 3);

Solution

1. Step 1: Understand the requirement

   We need users with total order amount > 500 and more than 3 orders.

2. Step 2: Analyze each option's approach

   SELECT u.name, stats.total_amount FROM users u JOIN (SELECT user_id, COUNT(*) AS order_count, SUM(amount) AS total_amount FROM orders GROUP BY user_id) AS stats ON u.id = stats.user_id WHERE stats.total_amount > 500 AND stats.order_count > 3; uses a subquery in FROM to calculate order count and total per user, then joins users to filter correctly. SELECT u.name FROM users u WHERE (SELECT SUM(amount) FROM orders WHERE user_id = u.id) > 500 AND (SELECT COUNT(*) FROM orders WHERE user_id = u.id) > 3; uses correlated subqueries but not FROM clause subquery. SELECT u.name, SUM(o.amount) FROM users u JOIN orders o ON u.id = o.user_id GROUP BY u.id HAVING SUM(o.amount) > 500 AND COUNT(o.id) > 3; uses HAVING but no subquery in FROM. SELECT name FROM users WHERE id IN (SELECT user_id FROM orders GROUP BY user_id HAVING SUM(amount) > 500 AND COUNT(*) > 3); uses IN with subquery but no join.

3. Step 3: Confirm correct use of subquery in FROM

   Only SELECT u.name, stats.total_amount FROM users u JOIN (SELECT user_id, COUNT(*) AS order_count, SUM(amount) AS total_amount FROM orders GROUP BY user_id) AS stats ON u.id = stats.user_id WHERE stats.total_amount > 500 AND stats.order_count > 3; uses a derived table (subquery in FROM) with alias and joins properly.

4. Final Answer:

   SELECT u.name, stats.total_amount FROM users u JOIN (SELECT user_id, COUNT(*) AS order_count, SUM(amount) AS total_amount FROM orders GROUP BY user_id) AS stats ON u.id = stats.user_id WHERE stats.total_amount > 500 AND stats.order_count > 3; -> Option A

5. Quick Check:

   Subquery in FROM calculates stats; outer query filters users [OK]

Hint: Use subquery in FROM to pre-aggregate data, then join [OK]

Common Mistakes:

• Not aliasing the subquery in FROM
• Using correlated subqueries instead of derived tables
• Missing join between users and subquery results