Subqueries in FROM clause (derived tables) in MySQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of a simple derived table query

What is the output of this query?

SELECT dt.category, dt.total_sales FROM (SELECT category, SUM(sales) AS total_sales FROM products GROUP BY category) AS dt ORDER BY dt.total_sales DESC;

MySQL

CREATE TABLE products (id INT, category VARCHAR(20), sales INT);
INSERT INTO products VALUES (1, 'Books', 100), (2, 'Books', 150), (3, 'Electronics', 200), (4, 'Clothing', 50);

A [{"category": "Clothing", "total_sales": 50}, {"category": "Books", "total_sales": 250}, {"category": "Electronics", "total_sales": 200}]

B [{"category": "Books", "total_sales": 100}, {"category": "Books", "total_sales": 150}, {"category": "Electronics", "total_sales": 200}, {"category": "Clothing", "total_sales": 50}]

C [{"category": "Books", "total_sales": 250}, {"category": "Electronics", "total_sales": 200}, {"category": "Clothing", "total_sales": 50}]

D [{"category": "Electronics", "total_sales": 200}, {"category": "Books", "total_sales": 250}, {"category": "Clothing", "total_sales": 50}]

💡 Hint

Think about how the subquery groups sales by category and sums them, then the outer query orders by total_sales descending.

The subquery groups products by category and sums their sales. The outer query selects these results and orders them by total sales from highest to lowest. So Books with 250 comes first, then Electronics with 200, then Clothing with 50.

📝 Syntax

intermediate

Identify the syntax error in derived table usage

Which option contains a syntax error when using a derived table in the FROM clause?

MySQL

SELECT dt.category, dt.total FROM (SELECT category, SUM(amount) AS total FROM sales GROUP BY category) dt;

A SELECT dt.category, dt.total FROM (SELECT category, SUM(amount) AS total FROM sales GROUP BY category);

B SELECT dt.category, dt.total FROM (SELECT category, SUM(amount) AS total FROM sales GROUP BY category) AS dt;

C SELECT dt.category, dt.total FROM (SELECT category, SUM(amount) AS total FROM sales GROUP BY category) dt;

D SELECT dt.category, dt.total FROM (SELECT category, SUM(amount) AS total FROM sales GROUP BY category) AS dt WHERE dt.total > 100;

💡 Hint

Remember that every derived table must have an alias.

Option A is missing the alias for the derived table, which is required in SQL syntax. Options A and B correctly alias the derived table, and D adds a WHERE clause but still aliases correctly.

❓ optimization

advanced

Optimizing a query with multiple derived tables

Given two derived tables joined on a common key, which option is the most efficient way to write the query to avoid unnecessary computation?

MySQL

SELECT a.category, a.total_sales, b.total_returns FROM
(SELECT category, SUM(sales) AS total_sales FROM products GROUP BY category) AS a
JOIN
(SELECT category, SUM(returns) AS total_returns FROM products GROUP BY category) AS b
ON a.category = b.category;

A Keep the two derived tables and join them as is, since it is clearer and easier to read.

B Use a single derived table that calculates both sums in one pass: SELECT category, SUM(sales) AS total_sales, SUM(returns) AS total_returns FROM products GROUP BY category;

C Use two separate queries and combine results in application code to reduce database load.

D Replace derived tables with temporary tables to speed up the join.

💡 Hint

Think about how many times the table is scanned in each approach.

Option B scans the products table once and calculates both sums together, which is more efficient. Option B scans the table twice. Option B shifts load to application and may increase latency. Option B adds overhead of managing temp tables.

🔧 Debug

advanced

Debugging a query with nested derived tables

This query returns an error. What is the cause?

SELECT dt1.category, dt2.avg_price FROM
(SELECT category, COUNT(*) AS cnt FROM products GROUP BY category) dt1
JOIN
(SELECT category, AVG(price) AS avg_price FROM products GROUP BY category) dt2
ON dt1.category = dt2.category
WHERE dt1.cnt > 10;

A The query is valid and should run without errors.

B The WHERE clause cannot reference columns from derived tables; it should be in HAVING.

C The JOIN condition is invalid because it uses columns from different derived tables.

D The alias dt1 is missing in the WHERE clause.

💡 Hint

Check if referencing derived table columns in WHERE is allowed after the JOIN.

The query is valid. The WHERE clause filters after the JOIN and can reference columns from derived tables. The JOIN condition is valid. No alias is missing in WHERE.

🧠 Conceptual

expert

Understanding scope and visibility of derived tables

Consider this query:

SELECT category, total_sales FROM
(SELECT category, SUM(sales) AS total_sales FROM products GROUP BY category) AS dt
WHERE total_sales > (SELECT AVG(total_sales) FROM (SELECT category, SUM(sales) AS total_sales FROM products GROUP BY category) AS dt2);

What is the role of the derived table dt2 in the subquery?

A dt2 filters categories with total_sales above average before the outer query runs.

B dt2 is redundant and can be removed without affecting the query result.

C dt2 is the same as dt and causes a naming conflict leading to an error.

D dt2 provides the average total sales across all categories to compare each category's total_sales against.

💡 Hint

Think about how the subquery calculates the average and how it relates to the outer query.

dt2 is a derived table inside the subquery that calculates total sales per category. The subquery then computes the average of these totals. The outer query compares each category's total_sales to this average to filter results.

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