Scalar subqueries in PostgreSQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the average salary from the employees table.

PostgreSQL

SELECT (SELECT [1](salary) FROM employees) AS avg_salary;

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

AAVG

BCOUNT

CSUM

DMAX

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM instead of AVG returns total salary, not average.

Using COUNT returns number of rows, not average salary.

The AVG function calculates the average value of a column. Here, it returns the average salary.

2fill in blank

medium

Complete the code to find employees whose salary is greater than the average salary.

PostgreSQL

SELECT name FROM employees WHERE salary > (SELECT [1](salary) FROM employees);

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

ACOUNT

BSUM

CMIN

DAVG

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM returns total salary, causing incorrect comparison.

Using MIN returns lowest salary, not average.

The subquery calculates the average salary using AVG. The main query selects employees earning more than this average.

3fill in blank

hard

Fix the error in the scalar subquery to correctly get the maximum salary.

PostgreSQL

SELECT name, salary FROM employees WHERE salary = (SELECT [1](salary) FROM employees);

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

AMAX()

BMAX(

CMAX

DMAX)

Attempts:

3 left

💡 Hint

Common Mistakes

Writing MAX without parentheses causes syntax error.

Adding parentheses incorrectly breaks the query.

Use MAX( to form the correct aggregate function syntax MAX(salary) in the scalar subquery.

4fill in blank

hard

Fill both blanks to select employees whose salary is between the minimum and maximum salaries.

PostgreSQL

SELECT name FROM employees WHERE salary BETWEEN (SELECT [1](salary) FROM employees) AND (SELECT [2](salary) FROM employees);

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

AMIN

BAVG

CMAX

DSUM

Attempts:

3 left

💡 Hint

Common Mistakes

Using AVG instead of MIN or MAX changes the range incorrectly.

Using SUM does not make sense for range boundaries.

The MIN function gets the lowest salary, and MAX gets the highest. Using BETWEEN with these selects salaries in that range.

5fill in blank

hard

Fill all three blanks to select employees whose salary is greater than the average salary and less than the maximum salary.

PostgreSQL

SELECT name FROM employees WHERE salary > (SELECT [1](salary) FROM employees) AND salary < (SELECT [2](salary) FROM employees) ORDER BY [3];

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

AMIN

BAVG

CMAX

Dsalary

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up MIN and AVG in comparisons.

Forgetting to order by a valid column.

The query compares salary to the average and maximum using AVG and MAX. It orders results by the salary column.

Practice

1. What does a scalar subquery return in PostgreSQL?

easy

A. A single value used inside another query

B. Multiple rows and columns

C. Only table names

D. A list of table columns

Solution

1. Step 1: Understand scalar subquery definition

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

2. Step 2: Compare options with definition

   Only A single value used inside another query describes a single value used inside another query, matching the scalar subquery concept.

3. Final Answer:

   A single value used inside another query -> Option A

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 names
• Assuming scalar subqueries return lists

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

easy

A. SELECT (SELECT MAX(price) FROM products) AS max_price;

B. SELECT MAX(price) FROM products WHERE price = (SELECT);

C. SELECT MAX(price) FROM products SELECT price;

D. SELECT MAX(price) FROM (SELECT price FROM products);

Solution

1. Step 1: Identify correct scalar subquery syntax

   The scalar subquery must be enclosed in parentheses and return a single value, used inside the main SELECT.

2. Step 2: Check each option

   SELECT (SELECT MAX(price) FROM products) AS max_price; correctly uses (SELECT MAX(price) FROM products) as a scalar subquery. Options A, B, and D have syntax errors or misuse subqueries.

3. Final Answer:

   SELECT (SELECT MAX(price) FROM products) AS max_price; -> Option A

4. Quick Check:

   Scalar subquery syntax = SELECT (SELECT MAX(price) FROM products) AS max_price; [OK]

Hint: Scalar subqueries are enclosed in parentheses inside SELECT [OK]

Common Mistakes:

• Omitting parentheses around subquery
• Using incomplete or invalid subquery syntax
• Confusing subqueries with JOINs

3. Given the tables:
products(id, name, price)
and
sales(product_id, quantity)
What will this query return?

SELECT name, (SELECT SUM(quantity) FROM sales WHERE product_id = products.id) AS total_sold FROM products;

medium

A. Syntax error due to subquery

B. Only product names without sales data

C. Sum of all sales quantities for all products

D. Each product's name with total quantity sold

Solution

1. Step 1: Understand the subquery role

   The subquery sums quantity from sales where product_id matches the current product's id, returning total sold per product.

2. Step 2: Analyze the main query output

   The main query selects product name and the scalar subquery result as total_sold for each product row.

3. Final Answer:

   Each product's name with total quantity sold -> Option D

4. Quick Check:

   Scalar subquery sums sales per product = Each product's name with total quantity sold [OK]

Hint: Scalar subquery calculates per-row related value [OK]

Common Mistakes:

• Thinking subquery sums all sales once
• Ignoring WHERE clause linking tables
• Assuming syntax error without checking carefully

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

SELECT name, (SELECT price FROM products) AS product_price FROM products;

medium

A. Missing alias for subquery

B. Incorrect table name in subquery

C. Subquery returns multiple rows causing error

D. No error, query runs fine

Solution

1. Step 1: Check subquery result cardinality

   The subquery SELECT price FROM products returns multiple prices (rows), not a single value.

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 error -> Option C

4. Quick Check:

   Scalar subquery must return one value [OK]

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

Common Mistakes:

• Ignoring multiple rows returned by subquery
• Assuming alias fixes multiple rows error
• Confusing syntax error with runtime error

5. You want to list each product's name and its price difference from the average price of all products. Which query correctly uses a scalar subquery to achieve this?

hard

A. SELECT name, price - AVG(price) FROM products;

B. SELECT name, price - (SELECT AVG(price) FROM products) AS price_diff FROM products;

C. SELECT name, price - (SELECT price FROM products) AS price_diff FROM products;

D. SELECT name, price - (SELECT SUM(price) FROM products) AS price_diff FROM products;

Solution

1. Step 1: Understand the goal

   We want each product's price minus the average price of all products.

2. Step 2: Analyze each option

   SELECT name, price - (SELECT AVG(price) FROM products) AS price_diff FROM products; correctly uses a scalar subquery to get AVG(price) once and subtracts it from each product's price. SELECT name, price - AVG(price) FROM products; lacks GROUP BY and will error. SELECT name, price - (SELECT price FROM products) AS price_diff FROM products; returns multiple prices, causing error. SELECT name, price - (SELECT SUM(price) FROM products) AS price_diff FROM products; subtracts sum, not average, which is incorrect.

3. Final Answer:

   SELECT name, price - (SELECT AVG(price) FROM products) AS price_diff FROM products; -> Option B

4. Quick Check:

   Scalar subquery for AVG(price) = SELECT name, price - (SELECT AVG(price) FROM products) AS price_diff FROM products; [OK]

Hint: Use scalar subquery for single aggregate value [OK]

Common Mistakes:

• Using aggregate without GROUP BY causing error
• Subquery returning multiple rows instead of one
• Confusing SUM with AVG in calculation