Subqueries in FROM (derived tables) in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

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

PostgreSQL

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

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

A*

Bemployees

Cname

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original table name instead of the alias.

Selecting a column not present in the derived table.

Using * selects all columns from the derived table sub.

2fill in blank

medium

Complete the code to calculate the average salary from a derived table named 'dept_salaries'.

PostgreSQL

SELECT AVG(salary) FROM (SELECT salary FROM employees WHERE department = [1]) AS dept_salaries;

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

A"Sales"

BSales

C'Sales'

Dsales

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting quotes around string literals.

Using double quotes instead of single quotes.

String literals in SQL must be enclosed in single quotes, so 'Sales' is correct.

3fill in blank

hard

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

PostgreSQL

SELECT sales_summary.id, sales_summary.total FROM (SELECT id, SUM(amount) AS total FROM sales GROUP BY id) [1];

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

AAS sales_summary

Bsummary

CAS

Dsales_summary

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the alias entirely.

Using AS without an alias name.

In PostgreSQL, the alias for a derived table can be given without the AS keyword, so sales_summary alone is correct.

4fill in blank

hard

Fill both blanks to select employee names and their department names using a derived table.

PostgreSQL

SELECT e.name, d.[1] FROM employees e JOIN (SELECT id, [2] FROM departments) d ON e.department_id = d.id;

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

Adepartment_name

Bname

Cdept_name

Ddepartment

Attempts:

3 left

💡 Hint

Common Mistakes

Using a column name that does not exist in the departments table.

Mixing up alias names.

The derived table selects id and name from departments, aliased as d. The main query selects d.name as the department name.

5fill in blank

hard

Fill all three blanks to create a derived table that counts employees per department and select departments with more than 5 employees.

PostgreSQL

SELECT dept_name, emp_count FROM (SELECT d.name AS [1], COUNT(e.id) AS [2] FROM departments d LEFT JOIN employees e ON d.id = e.department_id GROUP BY d.name) AS dept_counts WHERE [3] > 5;

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

Adept_name

Bemp_count

Ddepartment

Attempts:

3 left

💡 Hint

Common Mistakes

Using column names not aliased in the derived table.

Using the original column names in the WHERE clause instead of aliases.

The derived table aliases d.name as dept_name and the count as emp_count. The WHERE clause filters rows where emp_count is greater than 5.

Practice

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

easy

A. To create a permanent table in the database

B. To update data in multiple tables at once

C. To delete rows from a table based on a condition

D. To create a temporary table that can be used in the main query

Solution

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

   Subqueries in the FROM clause act like temporary tables that exist only during the query execution.

2. Step 2: Compare with other options

   Updating or deleting data and creating permanent tables are not done by subqueries in FROM.

3. Final Answer:

   To create a temporary table that can be used in the main query -> Option D

4. Quick Check:

   Subquery in FROM = temporary table [OK]

Hint: Subquery in FROM = think temporary table [OK]

Common Mistakes:

• Thinking subqueries in FROM create permanent tables
• Confusing subqueries with UPDATE or DELETE statements
• Forgetting subqueries need an alias

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

easy

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

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

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

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

Solution

1. Step 1: Identify correct subquery syntax in FROM

   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 either misuse WHERE, JOIN without ON, or lack alias.

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 subqueries in FROM with AS [OK]

Common Mistakes:

• Omitting alias after subquery
• Using subquery in WHERE without proper syntax
• Trying to JOIN without ON condition

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

SELECT sub.name, sub.total FROM (SELECT u.name, SUM(o.amount) AS total FROM users u LEFT JOIN orders o ON u.id = o.user_id GROUP BY u.name) AS sub WHERE sub.total > 100;

medium

A. Syntax error due to missing GROUP BY

B. All users with their total order amount including those with zero

C. Names of users with total order amount greater than 100

D. Empty result because no user has orders

Solution

1. Step 1: Analyze the subquery

   The subquery calculates total order amount per user using LEFT JOIN and GROUP BY user name.

2. Step 2: Apply the outer WHERE filter

   The outer query filters users whose total order amount is greater than 100.

3. Final Answer:

   Names of users with total order amount greater than 100 -> Option C

4. Quick Check:

   Subquery sums orders, outer filters total > 100 [OK]

Hint: Subquery computes totals; outer filters results [OK]

Common Mistakes:

• Ignoring GROUP BY causing syntax error
• Confusing LEFT JOIN with INNER JOIN results
• Assuming all users appear without filtering

4. Identify the error in this query:

SELECT sub.name, sub.order_count FROM (SELECT name, COUNT(*) FROM users GROUP BY name) sub;

medium

A. No error, query is correct

B. Missing alias for COUNT(*) column

C. Incorrect table name 'users'

D. Subquery missing parentheses

Solution

1. Step 1: Check subquery SELECT list

   COUNT(*) needs an alias to be referenced outside the subquery.

2. Step 2: Understand outer query references

   Outer query uses sub.order_count, but COUNT(*) column has no alias, causing error.

3. Final Answer:

   Missing alias for COUNT(*) column -> Option B

4. Quick Check:

   Aggregate columns need aliases in subqueries [OK]

Hint: Always alias aggregate columns in subqueries [OK]

Common Mistakes:

• Forgetting to alias aggregate functions
• Assuming subquery columns auto-named
• Ignoring parentheses around subquery

5. You want to find the average order amount per user but only for users who have placed more than 3 orders. Which query correctly uses a subquery in the FROM clause to achieve this?

hard

A. SELECT sub.user_id, sub.avg_amount FROM (SELECT user_id, AVG(amount) AS avg_amount, COUNT(*) AS order_count FROM orders GROUP BY user_id) AS sub WHERE sub.order_count > 3;

B. SELECT user_id, AVG(amount) FROM orders WHERE COUNT(*) > 3 GROUP BY user_id;

C. SELECT user_id, AVG(amount) FROM (SELECT * FROM orders WHERE COUNT(*) > 3) AS sub GROUP BY user_id;

D. SELECT user_id, AVG(amount) FROM orders GROUP BY user_id HAVING AVG(amount) > 3;

Solution

1. Step 1: Understand the requirement

   We need average order amount per user but only for users with more than 3 orders.

2. Step 2: Analyze each option

   SELECT sub.user_id, sub.avg_amount FROM (SELECT user_id, AVG(amount) AS avg_amount, COUNT(*) AS order_count FROM orders GROUP BY user_id) AS sub WHERE sub.order_count > 3; uses a subquery to calculate average and count per user, then filters users with order_count > 3 outside. Others misuse COUNT in WHERE or HAVING incorrectly.

3. Final Answer:

   SELECT sub.user_id, sub.avg_amount FROM (SELECT user_id, AVG(amount) AS avg_amount, COUNT(*) AS order_count FROM orders GROUP BY user_id) AS sub WHERE sub.order_count > 3; -> Option A

4. Quick Check:

   Subquery calculates aggregates; outer filters by count [OK]

Hint: Filter aggregates in outer query after subquery [OK]

Common Mistakes:

• Using COUNT(*) in WHERE clause directly
• Misplacing HAVING clause conditions
• Not aliasing subquery columns properly