Independent practice · SQLite

Attach names to every order

Practice dataset

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Your task

Return order_id and name for all orders.
Keep unmatched orders; use NULL for their name.

Pass every output check to complete this task. Suggested active time: 10 minutes. You can keep working after that.

Required output

Return exactly these columns in this order.

order_idnumber
namestring or NULL

Expected output: 8 rows. Row order does not affect the check.

Values, data types and the number of copies of each row must match. A missing value is distinct from zero and from an empty string.

Preserve SQL NULL unless the task explicitly gives a replacement rule. Check JOINs for extra copies of rows.

Reconciliation checks

Return exactly the named columns in the stated order.
Preserve row multiplicity, numeric types and the declared missing-value policy.

Need a next step?

Hint 1: where to start

Use LEFT JOIN and preserve SQL NULL.

Hint 2: build the method

Check the declared business key, eligibility rules and missing-value policy before editing.

Hint 3: check your work

Compare the exact output columns, row count and each value with the contract; matching a total alone is insufficient.

How your work is checked

Every output check must pass. The score shows how much matched and helps you find mistakes; completion comes from passing all checks. Active time is recorded separately.

Passing records a correct practice output on this device. It is not a professional certification.

Source materials

29 original source records.

Data dictionary

orders table: Orders with paid, pending and cancelled statuses, integer whole-dollar amounts and nullable customer/region keys.
customers table: Customer dimension keyed by customer_id; emails may be shared, empty or NULL.
returns table: Partial refunds in whole dollars, with an orphan order reference.
inventory table: Physical counts and reorder points; on_hand can be NULL.
events table: Customer events; repeated visits are separate events.

View database setup and records

CREATE TABLE orders(order_id INTEGER, customer_id INTEGER, status TEXT, amount INTEGER, region TEXT, ordered_at TEXT);
INSERT INTO orders VALUES (1,1,'paid',120,'East','2026-09-01'),(2,2,'pending',140,'West','2026-09-02'),(3,1,'paid',100,'East','2026-09-03'),(4,3,'paid',180,'West','2026-09-03'),(5,99,'paid',80,'East','2026-09-04'),(6,NULL,'cancelled',0,NULL,'2026-09-04'),(7,2,'paid',120,'West','2026-09-05'),(8,4,'paid',30,'North','2026-09-06');
CREATE TABLE customers(customer_id INTEGER PRIMARY KEY,name TEXT,email TEXT);
INSERT INTO customers VALUES(1,'Avery','team@example.test'),(2,'Morgan','team@example.test'),(3,'Riley',NULL),(4,'Casey',''),(5,'Sky','sky@example.test');
CREATE TABLE returns(return_id INTEGER,order_id INTEGER,refund INTEGER);
INSERT INTO returns VALUES(1,1,20),(2,1,10),(3,4,180),(4,99,25);
CREATE TABLE inventory(sku TEXT,on_hand INTEGER,reorder_at INTEGER);
INSERT INTO inventory VALUES('PEN',5,10),('BOOK',20,10),('MUG',0,5),('BAG',NULL,4),('CARD',4,4);
CREATE TABLE events(event_id INTEGER,customer_id INTEGER,stage TEXT);
INSERT INTO events VALUES(1,1,'visit'),(2,1,'visit'),(3,1,'checkout'),(4,2,'visit'),(5,3,'visit'),(6,3,'checkout'),(7,4,'visit');

Your work

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SQL query

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Show the reference answer

SELECT o.order_id, c.name FROM orders o LEFT JOIN customers c ON o.customer_id=c.customer_id

An inner join drops order 5 and order 6. Join on the ID.