Scalar subqueries in PostgreSQL

Introduction

Scalar subqueries let you use a small query inside another query to get a single value. This helps you get extra information without writing many separate queries.

When you want to find a value related to each row in a table, like the highest score for each player.

When you need to calculate a value on the fly, like the average price of products to compare with each product's price.

When you want to add extra details from another table without joining all rows, like showing the latest order date for each customer.

When you want to filter rows based on a single value from another query, like selecting employees whose salary is above the company average.

Syntax

PostgreSQL

SELECT column1, (SELECT single_value FROM table2 WHERE condition) AS alias FROM table1;

The subquery inside parentheses must return exactly one value (one row, one column).

If the subquery returns no rows, the result is NULL for that row.

Examples

Get each player's name and their highest game score using a scalar subquery.

PostgreSQL

SELECT name, (SELECT MAX(score) FROM games WHERE games.player_id = players.id) AS max_score FROM players;

Show each product's price and the average price of all products.

PostgreSQL

SELECT product_name, price, (SELECT AVG(price) FROM products) AS avg_price FROM products;

Select employees who earn more than the average salary.

PostgreSQL

SELECT employee_name, salary FROM employees WHERE salary > (SELECT AVG(salary) FROM employees);

Sample Program

This example creates two tables: employees and departments. Then it selects each employee's name, salary, and their department's name using a scalar subquery.

PostgreSQL

CREATE TABLE employees (id SERIAL PRIMARY KEY, name TEXT, department_id INT, salary INT);
INSERT INTO employees (name, department_id, salary) VALUES
('Alice', 1, 5000),
('Bob', 1, 6000),
('Charlie', 2, 5500);

CREATE TABLE departments (id SERIAL PRIMARY KEY, name TEXT);
INSERT INTO departments (name) VALUES ('HR'), ('IT');

SELECT name, salary, (SELECT name FROM departments WHERE departments.id = employees.department_id) AS department_name FROM employees;

Important Notes

Scalar subqueries must return only one value; otherwise, you get an error.

If the subquery returns no rows, the result is NULL for that row.

Scalar subqueries are useful for adding calculated or related values without complex joins.

Summary

Scalar subqueries return a single value used inside another query.

They help get related or calculated data for each row simply.

Make sure the subquery returns exactly one value to avoid errors.

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