lesson 19 Joins & sets · intermediate

Keep unmatched rows

Keep rows even when the other table has no match.

the idea

LEFT JOIN keeps every row from the left table. When no match exists, right-side columns are NULL. A right-table filter in WHERE may remove those unmatched rows.

tables in this lesson

The challenges use this data. Each column shows its type underneath; → marks a link to another table.

customers4 columns · 8 rows

idinteger · key	nametext	citytext	joined_attext
1	Alice	Chicago	2024-01-10
2	Ben	Austin	2024-02-12
3	Cara	Seattle	2024-03-05
4	Dev	Chicago	2024-04-20
5	Elena	NULL	2024-05-01

showing 5 of 8 rows

orders6 columns · 12 rows

idinteger · key	customer_idinteger	product_idinteger	quantityinteger	statustext	ordered_attext
101	1	1	2	paid	2025-01-01
102	2	3	1	paid	2025-01-02
103	1	2	1	paid	2025-01-05
104	3	5	2	pending	2025-01-08
105	4	4	5	paid	2025-02-01

showing 5 of 12 rows

example

The idea in one query. Press Run to see what it returns; running examples never affects your progress.

example.sql

SELECT c.name, COUNT(o.id) AS order_count
FROM customers c
LEFT JOIN orders o ON o.customer_id = c.id
GROUP BY c.id, c.name;

Watch for: Using INNER JOIN loses Hugo, who has no orders.

build it step by step

Each step changes one thing. Run them in order and watch the result change.

1
Identify the preserved side. Every customer must stay, even Hugo without an order.
step-1.sql
sql›
```
SELECT *
FROM customers;
```
2
Inspect the joined rows. Customers repeat for each order. Hugo gets a NULL right-side id.
step-2.sql
sql›
```
SELECT c.id,c.name,o.id AS order_id
FROM customers c
LEFT JOIN orders o ON o.customer_id=c.id;
```

Count matching right-side ids. COUNT(o.id) ignores the unmatched NULL and gives Hugo zero.

step-3.sql

SELECT c.name,COUNT(o.id) AS order_count
FROM customers c
LEFT JOIN orders o ON o.customer_id=c.id
GROUP BY c.id,c.name;