LATERAL join for correlated subqueries in PostgreSQL

Introduction

LATERAL join lets you run a subquery for each row of a table, using values from that row. It helps when you want to get related data that depends on each row.

You want to find the top 3 recent orders for each customer.

You need to get the highest scoring review for each product.

You want to list employees along with their latest project details.

You want to join a table with a function that uses columns from that table.

Syntax

PostgreSQL

SELECT columns
FROM table1
JOIN LATERAL (
  SELECT columns
  FROM table2
  WHERE table2.column = table1.column
  LIMIT n
) alias ON true;

LATERAL allows the subquery to use columns from the main query's current row.

The ON condition is often ON true because the subquery is already filtered inside.

Examples

Get the most recent order for each customer.

PostgreSQL

SELECT c.customer_id, o.order_id, o.order_date
FROM customers c
JOIN LATERAL (
  SELECT order_id, order_date
  FROM orders
  WHERE orders.customer_id = c.customer_id
  ORDER BY order_date DESC
  LIMIT 1
) o ON true;

Get the highest rated review for each product.

PostgreSQL

SELECT p.product_id, r.review_text
FROM products p
JOIN LATERAL (
  SELECT review_text
  FROM reviews
  WHERE reviews.product_id = p.product_id
  ORDER BY rating DESC
  LIMIT 1
) r ON true;

Sample Program

This query finds each customer's latest order by using a LATERAL join to run a subquery for each customer.

PostgreSQL

CREATE TABLE customers (customer_id INT, name TEXT);
CREATE TABLE orders (order_id INT, customer_id INT, order_date DATE);

INSERT INTO customers VALUES (1, 'Alice'), (2, 'Bob');
INSERT INTO orders VALUES
  (101, 1, '2024-01-10'),
  (102, 1, '2024-02-15'),
  (103, 2, '2024-01-20');

SELECT c.customer_id, c.name, o.order_id, o.order_date
FROM customers c
JOIN LATERAL (
  SELECT order_id, order_date
  FROM orders
  WHERE orders.customer_id = c.customer_id
  ORDER BY order_date DESC
  LIMIT 1
) o ON true
ORDER BY c.customer_id;

Important Notes

LATERAL is useful when the subquery needs to know about the current row from the main table.

Without LATERAL, the subquery cannot refer to columns from the outer query.

Use LIMIT inside the lateral subquery to get top results per row.

Summary

LATERAL join runs a subquery for each row of the main table using that row's data.

It helps get related data that depends on each row, like top orders or best reviews.

The subquery can refer to outer query columns, making it very flexible.

Practice

1. What is the main purpose of using a LATERAL join in PostgreSQL?

easy

A. To run a subquery for each row of the main table using that row's data

B. To join two tables without any condition

C. To create a temporary table for all rows at once

D. To speed up queries by caching results

Solution

1. Step 1: Understand what LATERAL join does

   A LATERAL join allows a subquery to use columns from the current row of the main query.

2. Step 2: Identify the main purpose

   This means the subquery runs once per row, using that row's data to produce related results.

3. Final Answer:

   To run a subquery for each row of the main table using that row's data -> Option A

4. Quick Check:

   LATERAL join = per-row subquery execution [OK]

Hint: LATERAL runs subqueries per row using that row's data [OK]

Common Mistakes:

• Thinking LATERAL joins all rows at once
• Confusing LATERAL with regular JOIN
• Assuming LATERAL speeds up queries by caching

2. Which of the following is the correct syntax to use a LATERAL join in PostgreSQL?

easy

A. SELECT a.id, b.value FROM table_a a JOIN (SELECT value FROM table_b WHERE b.a_id = a.id) b ON true;

B. SELECT a.id, b.value FROM table_a a JOIN LATERAL (SELECT value FROM table_b WHERE b.a_id = a.id) b ON true;

C. SELECT a.id, b.value FROM table_a a LEFT JOIN LATERAL table_b b ON b.a_id = a.id;

D. SELECT a.id, b.value FROM table_a a CROSS JOIN table_b b WHERE b.a_id = a.id;

Solution

1. Step 1: Identify correct LATERAL syntax

   LATERAL must be explicitly stated before the subquery or table function to allow referencing outer query columns.

2. Step 2: Check each option

   SELECT a.id, b.value FROM table_a a JOIN LATERAL (SELECT value FROM table_b WHERE b.a_id = a.id) b ON true; correctly uses JOIN LATERAL with a subquery referencing outer alias 'a'. SELECT a.id, b.value FROM table_a a JOIN (SELECT value FROM table_b WHERE b.a_id = a.id) b ON true; misses LATERAL keyword. SELECT a.id, b.value FROM table_a a LEFT JOIN LATERAL table_b b ON b.a_id = a.id; uses LATERAL but tries to join a table directly without subquery syntax. SELECT a.id, b.value FROM table_a a CROSS JOIN table_b b WHERE b.a_id = a.id; uses CROSS JOIN without LATERAL, so no correlation.

3. Final Answer:

   SELECT a.id, b.value FROM table_a a JOIN LATERAL (SELECT value FROM table_b WHERE b.a_id = a.id) b ON true; -> Option B

4. Quick Check:

   Correct LATERAL syntax includes keyword before subquery [OK]

Hint: LATERAL keyword must precede subquery to access outer columns [OK]

Common Mistakes:

• Omitting LATERAL keyword before subquery
• Using regular JOIN without correlation
• Trying to join tables directly without subquery

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

SELECT u.name, o.amount FROM users u JOIN LATERAL (SELECT amount FROM orders WHERE user_id = u.id ORDER BY amount DESC LIMIT 1) o ON true;

medium

A. Users with all their orders listed

B. All orders with user names repeated

C. Syntax error due to missing ON condition

D. Each user with their highest order amount

Solution

1. Step 1: Understand the LATERAL subquery

   The subquery selects the highest order amount per user by ordering descending and limiting to 1.

2. Step 2: Analyze the JOIN LATERAL effect

   For each user row, the subquery runs and returns that user's top order amount, joining on true to include all users with orders.

3. Final Answer:

   Each user with their highest order amount -> Option D

4. Quick Check:

   LATERAL subquery returns top order per user [OK]

Hint: LATERAL runs subquery per user to get top order [OK]

Common Mistakes:

• Thinking it returns all orders per user
• Assuming syntax error due to ON true
• Confusing join with CROSS JOIN

4. Identify the error in this query:

SELECT u.name, o.amount FROM users u JOIN LATERAL (SELECT amount FROM orders WHERE user_id = u.id ORDER BY amount DESC LIMIT 1) o;

medium

A. Subquery cannot reference outer query columns

B. Incorrect alias usage for orders table

C. Missing ON clause after LATERAL join

D. ORDER BY not allowed in subqueries

Solution

1. Step 1: Check JOIN syntax

   Every JOIN requires an ON clause or USING clause to specify join condition.

2. Step 2: Verify the query

   The query uses JOIN LATERAL but omits ON clause, causing syntax error.

3. Final Answer:

   Missing ON clause after LATERAL join -> Option C

4. Quick Check:

   JOIN must have ON or USING clause [OK]

Hint: Always add ON clause after JOIN LATERAL [OK]

Common Mistakes:

• Forgetting ON clause after JOIN LATERAL
• Thinking subquery can't use outer columns
• Believing ORDER BY is disallowed in subqueries

5. You want to list each customer with their two most recent orders using LATERAL join. Which query correctly achieves this?

hard

A. SELECT c.name, o.order_id FROM customers c JOIN LATERAL (SELECT order_id FROM orders WHERE customer_id = c.id ORDER BY order_date DESC LIMIT 2) o ON true;

B. SELECT c.name, o.order_id FROM customers c JOIN orders o ON o.customer_id = c.id ORDER BY o.order_date DESC LIMIT 2;

C. SELECT c.name, o.order_id FROM customers c CROSS JOIN LATERAL orders o WHERE o.customer_id = c.id LIMIT 2;

D. SELECT c.name, o.order_id FROM customers c LEFT JOIN LATERAL (SELECT order_id FROM orders WHERE customer_id = c.id ORDER BY order_date ASC LIMIT 2) o ON true;

Solution

1. Step 1: Understand the requirement

   We want two most recent orders per customer, so order by order_date DESC and limit 2 in subquery.

2. Step 2: Check each option

   SELECT c.name, o.order_id FROM customers c JOIN LATERAL (SELECT order_id FROM orders WHERE customer_id = c.id ORDER BY order_date DESC LIMIT 2) o ON true; uses JOIN LATERAL with correct ordering and limit, joining on true to include all customers with orders. SELECT c.name, o.order_id FROM customers c JOIN orders o ON o.customer_id = c.id ORDER BY o.order_date DESC LIMIT 2; limits total rows, not per customer. SELECT c.name, o.order_id FROM customers c CROSS JOIN LATERAL orders o WHERE o.customer_id = c.id LIMIT 2; uses CROSS JOIN LATERAL incorrectly without subquery. SELECT c.name, o.order_id FROM customers c LEFT JOIN LATERAL (SELECT order_id FROM orders WHERE customer_id = c.id ORDER BY order_date ASC LIMIT 2) o ON true; orders ASC, which gives oldest orders, not recent.

3. Final Answer:

   SELECT c.name, o.order_id FROM customers c JOIN LATERAL (SELECT order_id FROM orders WHERE customer_id = c.id ORDER BY order_date DESC LIMIT 2) o ON true; -> Option A

4. Quick Check:

   LATERAL with ORDER BY DESC and LIMIT 2 per customer [OK]

Hint: Use LATERAL with ORDER BY DESC and LIMIT for top N per row [OK]

Common Mistakes:

• Limiting total rows instead of per customer
• Ordering ascending instead of descending
• Using CROSS JOIN LATERAL without subquery