One-to-many relationship design in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a table named 'orders' with a primary key column 'order_id'.

SQL

CREATE TABLE orders (order_id [1]);

Drag options to blanks, or click blank then click option'

AINT PRIMARY KEY

BVARCHAR(255)

CFLOAT

DDATE

Attempts:

3 left

💡 Hint

Common Mistakes

Using VARCHAR instead of INT for the primary key.

Not specifying PRIMARY KEY.

The 'order_id' column should be an integer and set as the primary key to uniquely identify each order.

2fill in blank

medium

Complete the code to add a foreign key column 'customer_id' to the 'orders' table.

SQL

ALTER TABLE orders ADD COLUMN customer_id [1];

Drag options to blanks, or click blank then click option'

AINT

BDATE

CTEXT

DBOOLEAN

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different data type than the referenced primary key.

Using a non-numeric type for IDs.

The 'customer_id' should be an integer to match the primary key type of the 'customers' table.

3fill in blank

hard

Fix the error in the foreign key constraint syntax to link 'customer_id' in 'orders' to 'id' in 'customers'.

SQL

ALTER TABLE orders ADD CONSTRAINT fk_customer FOREIGN KEY (customer_id) REFERENCES customers([1]);

Drag options to blanks, or click blank then click option'

Acustomer_id

Border_id

Cid

Dorder_date

Attempts:

3 left

💡 Hint

Common Mistakes

Referencing a non-primary key column.

Using the foreign key column name instead of the referenced column.

The foreign key must reference the primary key column 'id' in the 'customers' table.

4fill in blank

hard

Fill both blanks to create a 'customers' table with a primary key 'id' and a 'name' column.

SQL

CREATE TABLE customers (id [1], name [2]);

Drag options to blanks, or click blank then click option'

AINT PRIMARY KEY

BVARCHAR(100)

CTEXT

DDATE

Attempts:

3 left

💡 Hint

Common Mistakes

Using DATE for the name column.

Not setting the primary key on 'id'.

The 'id' column is an integer primary key, and 'name' is a variable character string with max length 100.

5fill in blank

hard

Fill all three blanks to insert a new order with order_id 101, customer_id 5, and order_date '2024-06-01'.

SQL

INSERT INTO orders (order_id, customer_id, order_date) VALUES ([1], [2], '[3]');

Drag options to blanks, or click blank then click option'

A100

B5

C2024-06-01

D101

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping order_id and customer_id values.

Not quoting the date string.

The order_id is 101, customer_id is 5, and the order_date is '2024-06-01'.

Practice

1. What does a one-to-many relationship in a database mean?

easy

A. One record in a table relates to many records in another table

B. Many records in a table relate to one record in the same table

C. One record relates to exactly one record in another table

D. Many records relate to many records in another table

Solution

1. Step 1: Understand relationship types

   A one-to-many relationship means one record in a table connects to multiple records in another table.

2. Step 2: Match definition to options

   One record in a table relates to many records in another table correctly describes this as one record relating to many records in another table.

3. Final Answer:

   One record in a table relates to many records in another table -> Option A

4. Quick Check:

   One-to-many = one record to many records [OK]

Hint: One-to-many means one record links to many records [OK]

Common Mistakes:

• Confusing one-to-many with many-to-many
• Thinking one-to-many means one record links to one record
• Mixing up the direction of the relationship

2. Which SQL statement correctly creates a foreign key for a one-to-many relationship from Orders to Customers?

easy

A. ALTER TABLE Customers ADD FOREIGN KEY (OrderID) REFERENCES Orders(OrderID);

B. ALTER TABLE Customers ADD FOREIGN KEY (CustomerID) REFERENCES Orders(OrderID);

C. ALTER TABLE Orders ADD FOREIGN KEY (CustomerID) REFERENCES Customers(CustomerID);

D. ALTER TABLE Orders ADD FOREIGN KEY (OrderID) REFERENCES Customers(CustomerID);

Solution

1. Step 1: Identify the 'many' and 'one' tables

   Orders is the 'many' side, Customers is the 'one' side in a one-to-many relationship.

2. Step 2: Add foreign key in 'many' table referencing 'one' table

   The foreign key should be in Orders referencing Customers, so ALTER TABLE Orders ADD FOREIGN KEY (CustomerID) REFERENCES Customers(CustomerID); is correct.

3. Final Answer:

   ALTER TABLE Orders ADD FOREIGN KEY (CustomerID) REFERENCES Customers(CustomerID); -> Option C

4. Quick Check:

   Foreign key in 'many' table references 'one' table [OK]

Hint: Foreign key goes in 'many' table pointing to 'one' table [OK]

Common Mistakes:

• Placing foreign key in the 'one' table instead of 'many'
• Referencing wrong columns between tables
• Mixing table names in foreign key definition

3. Given these tables:
Customers(CustomerID, Name)
Orders(OrderID, CustomerID, Amount)
What will this query return?

SELECT Customers.Name, COUNT(Orders.OrderID) AS OrderCount FROM Customers LEFT JOIN Orders ON Customers.CustomerID = Orders.CustomerID GROUP BY Customers.Name;

medium

A. List of customers who have placed at least one order

B. List of customers with total amount spent on orders

C. List of orders with customer names repeated for each order

D. List of customers with the number of orders each placed, including customers with zero orders

Solution

1. Step 1: Understand the LEFT JOIN usage

   LEFT JOIN keeps all customers, even those without matching orders.

2. Step 2: COUNT(Orders.OrderID) counts orders per customer

   Grouping by customer name counts how many orders each customer has, zero if none.

3. Final Answer:

   List of customers with the number of orders each placed, including customers with zero orders -> Option D

4. Quick Check:

   LEFT JOIN + COUNT = all customers with order counts [OK]

Hint: LEFT JOIN + COUNT counts all, including zero matches [OK]

Common Mistakes:

• Thinking COUNT counts total amount spent
• Assuming only customers with orders appear
• Confusing JOIN types and their effects

4. You wrote this SQL to create a one-to-many relationship:

CREATE TABLE Orders (OrderID INT PRIMARY KEY, CustomerID INT, FOREIGN KEY (CustomerID) REFERENCES Customers(CustomerID));

But you get an error. What is the most likely cause?

medium

A. OrderID should not be primary key in Orders

B. Customers table does not exist or CustomerID is not a primary key

C. Foreign key should be in Customers table, not Orders

D. CustomerID column type must be VARCHAR, not INT

Solution

1. Step 1: Check foreign key reference validity

   Foreign key must reference an existing table and a primary or unique key column.

2. Step 2: Verify Customers table and CustomerID key

   If Customers table or CustomerID primary key is missing, error occurs.

3. Final Answer:

   Customers table does not exist or CustomerID is not a primary key -> Option B

4. Quick Check:

   Foreign key references must exist and be keys [OK]

Hint: Foreign key target must exist and be primary/unique key [OK]

Common Mistakes:

• Assuming foreign key can reference non-key columns
• Placing foreign key in wrong table
• Mismatching data types between foreign key and referenced key

5. You have two tables:
Authors(AuthorID, Name)
Books(BookID, Title, AuthorID)
You want to find authors who have written more than 3 books. Which query is correct?

hard

A. SELECT Name FROM Authors JOIN Books ON Authors.AuthorID = Books.AuthorID GROUP BY Name HAVING COUNT(BookID) > 3;

B. SELECT Name FROM Authors LEFT JOIN Books ON Authors.AuthorID = Books.AuthorID WHERE COUNT(BookID) > 3;

C. SELECT Name FROM Books GROUP BY AuthorID HAVING COUNT(BookID) > 3;

D. SELECT Name FROM Authors WHERE AuthorID IN (SELECT AuthorID FROM Books WHERE COUNT(BookID) > 3);

Solution

1. Step 1: Join Authors and Books on AuthorID

   We join to connect authors with their books.

2. Step 2: Group by author name and filter by book count

   Use GROUP BY Name and HAVING COUNT(BookID) > 3 to find authors with more than 3 books.

3. Final Answer:

   SELECT Name FROM Authors JOIN Books ON Authors.AuthorID = Books.AuthorID GROUP BY Name HAVING COUNT(BookID) > 3; -> Option A

4. Quick Check:

   JOIN + GROUP BY + HAVING filters authors by book count [OK]

Hint: Use GROUP BY and HAVING to filter by count in one-to-many [OK]

Common Mistakes:

• Using WHERE with aggregate functions instead of HAVING
• Missing GROUP BY clause
• Using LEFT JOIN but filtering with WHERE on aggregate