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  • What SQL is, and why professionals learn it
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  • SELECT and FROM: your first real query
  • WHERE: keep only the rows you want
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  • GROUP BY: one number per group
  • HAVING: filter the groups
  • Why tables relate: keys and joins
  • INNER JOIN: matching rows from two tables
  • LEFT JOIN: keep everything, find what is missing
  • Subqueries: a query inside a query
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Combine tables with joins

LEFT JOIN: keep everything, find what is missing

Last updated 14 June 2026

An inner join drops rows with no match. Often the unmatched rows are the answer. Which customers have never ordered? Which products never sold? A LEFT JOIN keeps every row from the left table, and fills the right-side columns with NULL where there was no match.

Every customer and their orders. Watch the NULLs.

CREATE TABLE customers (
  id INTEGER PRIMARY KEY,
  name TEXT,
  city TEXT,
  signup_date TEXT
);
INSERT INTO customers VALUES
  (1, 'Aisyah Rahman', 'Kuala Lumpur', '2023-01-12'),
  (2, 'Lim Wei Jie', 'Penang', '2023-02-03'),
  (3, 'Arjun Pillai', 'Johor Bahru', '2023-02-20'),
  (4, 'Nurul Huda', 'Kuala Lumpur', '2023-03-15'),
  (5, 'Tan Mei Ling', 'Ipoh', '2023-05-01'),
  (6, 'Faiz Hassan', 'Penang', '2023-06-10');

CREATE TABLE products (
  id INTEGER PRIMARY KEY,
  name TEXT,
  category TEXT,
  price REAL
);
INSERT INTO products VALUES
  (1, 'Standard License', 'Software', 199.0),
  (2, 'Pro License', 'Software', 499.0),
  (3, 'Onboarding Workshop', 'Service', 1200.0),
  (4, 'Support Plan', 'Service', 300.0),
  (5, 'Data Pack', 'Add-on', 89.0);

CREATE TABLE orders (
  id INTEGER PRIMARY KEY,
  customer_id INTEGER,
  order_date TEXT,
  status TEXT
);
INSERT INTO orders VALUES
  (1001, 1, '2023-03-01', 'paid'),
  (1002, 1, '2023-04-12', 'paid'),
  (1003, 2, '2023-04-15', 'paid'),
  (1004, 3, '2023-05-02', 'refunded'),
  (1005, 4, '2023-05-20', 'paid'),
  (1006, 2, '2023-06-01', 'pending'),
  (1007, 5, '2023-06-18', 'paid');

CREATE TABLE order_items (
  order_id INTEGER,
  product_id INTEGER,
  quantity INTEGER
);
INSERT INTO order_items VALUES
  (1001, 1, 2), (1001, 5, 1),
  (1002, 2, 1),
  (1003, 1, 1), (1003, 4, 1),
  (1004, 3, 1),
  (1005, 2, 2), (1005, 5, 3),
  (1006, 1, 1),
  (1007, 3, 1), (1007, 4, 2);
SELECT c.name, o.id AS order_id, o.status
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id;

One customer comes back with NULL for the order columns. They have never placed an order. To isolate those customers, keep only the rows where the right side is NULL.

CREATE TABLE customers (
  id INTEGER PRIMARY KEY,
  name TEXT,
  city TEXT,
  signup_date TEXT
);
INSERT INTO customers VALUES
  (1, 'Aisyah Rahman', 'Kuala Lumpur', '2023-01-12'),
  (2, 'Lim Wei Jie', 'Penang', '2023-02-03'),
  (3, 'Arjun Pillai', 'Johor Bahru', '2023-02-20'),
  (4, 'Nurul Huda', 'Kuala Lumpur', '2023-03-15'),
  (5, 'Tan Mei Ling', 'Ipoh', '2023-05-01'),
  (6, 'Faiz Hassan', 'Penang', '2023-06-10');

CREATE TABLE products (
  id INTEGER PRIMARY KEY,
  name TEXT,
  category TEXT,
  price REAL
);
INSERT INTO products VALUES
  (1, 'Standard License', 'Software', 199.0),
  (2, 'Pro License', 'Software', 499.0),
  (3, 'Onboarding Workshop', 'Service', 1200.0),
  (4, 'Support Plan', 'Service', 300.0),
  (5, 'Data Pack', 'Add-on', 89.0);

CREATE TABLE orders (
  id INTEGER PRIMARY KEY,
  customer_id INTEGER,
  order_date TEXT,
  status TEXT
);
INSERT INTO orders VALUES
  (1001, 1, '2023-03-01', 'paid'),
  (1002, 1, '2023-04-12', 'paid'),
  (1003, 2, '2023-04-15', 'paid'),
  (1004, 3, '2023-05-02', 'refunded'),
  (1005, 4, '2023-05-20', 'paid'),
  (1006, 2, '2023-06-01', 'pending'),
  (1007, 5, '2023-06-18', 'paid');

CREATE TABLE order_items (
  order_id INTEGER,
  product_id INTEGER,
  quantity INTEGER
);
INSERT INTO order_items VALUES
  (1001, 1, 2), (1001, 5, 1),
  (1002, 2, 1),
  (1003, 1, 1), (1003, 4, 1),
  (1004, 3, 1),
  (1005, 2, 2), (1005, 5, 3),
  (1006, 1, 1),
  (1007, 3, 1), (1007, 4, 2);
SELECT c.name, c.city
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id
WHERE o.id IS NULL;

This "left join then filter for NULL" pattern answers every "which X has no Y" question, from customers who never bought to products that never sold.

A plain JOIN drops rows with no match without warning, so you can lose data and never notice. When the rows with no match are part of the answer, reach for LEFT JOIN.

Your turn

Find every product that has never appeared in an order. Return the product name. Hint: left join products to order_items and keep the unmatched ones.

Products that have never been ordered.

CREATE TABLE customers (
  id INTEGER PRIMARY KEY,
  name TEXT,
  city TEXT,
  signup_date TEXT
);
INSERT INTO customers VALUES
  (1, 'Aisyah Rahman', 'Kuala Lumpur', '2023-01-12'),
  (2, 'Lim Wei Jie', 'Penang', '2023-02-03'),
  (3, 'Arjun Pillai', 'Johor Bahru', '2023-02-20'),
  (4, 'Nurul Huda', 'Kuala Lumpur', '2023-03-15'),
  (5, 'Tan Mei Ling', 'Ipoh', '2023-05-01'),
  (6, 'Faiz Hassan', 'Penang', '2023-06-10');

CREATE TABLE products (
  id INTEGER PRIMARY KEY,
  name TEXT,
  category TEXT,
  price REAL
);
INSERT INTO products VALUES
  (1, 'Standard License', 'Software', 199.0),
  (2, 'Pro License', 'Software', 499.0),
  (3, 'Onboarding Workshop', 'Service', 1200.0),
  (4, 'Support Plan', 'Service', 300.0),
  (5, 'Data Pack', 'Add-on', 89.0);

CREATE TABLE orders (
  id INTEGER PRIMARY KEY,
  customer_id INTEGER,
  order_date TEXT,
  status TEXT
);
INSERT INTO orders VALUES
  (1001, 1, '2023-03-01', 'paid'),
  (1002, 1, '2023-04-12', 'paid'),
  (1003, 2, '2023-04-15', 'paid'),
  (1004, 3, '2023-05-02', 'refunded'),
  (1005, 4, '2023-05-20', 'paid'),
  (1006, 2, '2023-06-01', 'pending'),
  (1007, 5, '2023-06-18', 'paid');

CREATE TABLE order_items (
  order_id INTEGER,
  product_id INTEGER,
  quantity INTEGER
);
INSERT INTO order_items VALUES
  (1001, 1, 2), (1001, 5, 1),
  (1002, 2, 1),
  (1003, 1, 1), (1003, 4, 1),
  (1004, 3, 1),
  (1005, 2, 2), (1005, 5, 3),
  (1006, 1, 1),
  (1007, 3, 1), (1007, 4, 2);
SELECT p.name
FROM products AS p
LEFT JOIN order_items AS oi ON oi.product_id = p.id
WHERE ;
SELECT p.name FROM products AS p LEFT JOIN order_items AS oi ON oi.product_id = p.id WHERE oi.product_id IS NULL;

In a LEFT JOIN, what appears in the right-side columns for a left row that has no match?

  • NULL
  • Zero
  • The row is dropped from the result
A LEFT JOIN keeps every left row. Where there is no matching right row, the right-side columns come back as NULL, which is exactly what you filter on to find the gaps.
← PreviousINNER JOIN: matching rows from two tablesNext →Subqueries: a query inside a query