Kamran Ahmed
2 weeks ago
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--- |
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title: Readers Like You |
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description: Practice using joins to find customers who share reading interests with a specific customer |
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order: 160 |
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type: challenge |
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setup: | |
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```sql |
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CREATE TABLE customer ( |
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id INT PRIMARY KEY, |
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name VARCHAR(255), |
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email VARCHAR(255) |
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); |
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|
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CREATE TABLE sale ( |
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id INT PRIMARY KEY, |
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customer_id INT, |
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book_id INT, |
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order_date DATE, |
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total_amount DECIMAL(10,2) |
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); |
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|
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CREATE TABLE book ( |
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id INT PRIMARY KEY, |
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title VARCHAR(255), |
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author VARCHAR(255) |
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); |
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|
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INSERT INTO book (id, title, author) |
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VALUES |
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(1, 'The Great Gatsby', 'F. Scott Fitzgerald'), |
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(2, 'To Kill a Mockingbird', 'Harper Lee'), |
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(3, '1984', 'George Orwell'), |
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(4, 'Pride and Prejudice', 'Jane Austen'), |
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(5, 'The Catcher in the Rye', 'J.D. Salinger'); |
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|
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INSERT INTO customer (id, name, email) |
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VALUES |
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(1, 'Alice Smith', 'alice@example.com'), |
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(2, 'Bob Johnson', 'bob@example.com'), |
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(3, 'Carol White', 'carol@example.com'), |
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(4, 'David Brown', 'david@example.com'), |
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(5, 'Emily Davis', 'emily@example.com'), |
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(6, 'Frank Wilson', 'frank@example.com'), |
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(7, 'Grace Taylor', 'grace@example.com'); |
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INSERT INTO sale (id, customer_id, book_id, order_date, total_amount) |
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VALUES |
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(1, 1, 1, '2024-01-01', 10.00), -- Alice: Great Gatsby |
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(2, 1, 2, '2024-01-15', 15.00), -- Alice: Mockingbird |
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(3, 1, 3, '2024-02-01', 20.00), -- Alice: 1984 |
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(4, 2, 1, '2024-02-15', 10.00), -- Bob: Great Gatsby |
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(5, 2, 3, '2024-03-01', 20.00), -- Bob: 1984 |
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(6, 3, 2, '2024-03-15', 15.00), -- Carol: Mockingbird |
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(7, 3, 3, '2024-04-01', 20.00), -- Carol: 1984 |
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(8, 4, 1, '2024-04-15', 10.00), -- David: Great Gatsby |
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(9, 5, 4, '2024-05-01', 25.00), -- Emily: Pride and Prejudice |
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(10, 6, 3, '2024-05-15', 20.00); -- Frank: 1984 |
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``` |
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--- |
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|
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The bookstore wants to implement a "Readers Like You" feature that shows customers who have similar reading interests. Given a specific customer ID, we want to find all other customers who have purchased at least one of the same books. |
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Given the following data in table `customer` |
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|
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| id | name | email | |
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| --- | ------------ | ----------------- | |
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| 1 | Alice Smith | alice@example.com | |
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| 2 | Bob Johnson | bob@example.com | |
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| 3 | Carol White | carol@example.com | |
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| 4 | David Brown | david@example.com | |
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| 5 | Emily Davis | emily@example.com | |
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| 6 | Frank Wilson | frank@example.com | |
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| 7 | Grace Taylor | grace@example.com | |
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|
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The following data in table `sale` |
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|
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| id | customer_id | book_id | order_date | total_amount | |
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| --- | ----------- | ------- | ---------- | ------------ | |
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| 1 | 1 | 1 | 2024-01-01 | 10.00 | |
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| 2 | 1 | 2 | 2024-01-15 | 15.00 | |
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| 3 | 1 | 3 | 2024-02-01 | 20.00 | |
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| 4 | 2 | 1 | 2024-02-15 | 10.00 | |
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| 5 | 2 | 3 | 2024-03-01 | 20.00 | |
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| 6 | 3 | 2 | 2024-03-15 | 15.00 | |
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| 7 | 3 | 3 | 2024-04-01 | 20.00 | |
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| 8 | 4 | 1 | 2024-04-15 | 10.00 | |
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| 9 | 5 | 4 | 2024-05-01 | 25.00 | |
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| 10 | 6 | 3 | 2024-05-15 | 20.00 | |
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And the following data in table `book` |
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| id | title | author | |
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| --- | ---------------------- | ------------------- | |
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| 1 | The Great Gatsby | F. Scott Fitzgerald | |
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| 2 | To Kill a Mockingbird | Harper Lee | |
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| 3 | 1984 | George Orwell | |
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| 4 | Pride and Prejudice | Jane Austen | |
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| 5 | The Catcher in the Rye | J.D. Salinger | |
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Write a query that takes customer ID `1` (Alice Smith) and finds all other customers who have purchased any of the same books as Alice, along with the books they have in common. |
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Your output should contain the following columns: |
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- `customer_name` - name of the other customer |
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- `common_book` - title of the book they both purchased |
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- `purchase_date` - when the other customer purchased the book |
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## Expected Output |
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| customer_name | common_book | purchase_date | |
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| ------------- | --------------------- | ------------------------ | |
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| Bob Johnson | 1984 | 2024-03-01T00:00:00.000Z | |
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| Bob Johnson | The Great Gatsby | 2024-02-15T00:00:00.000Z | |
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| Carol White | 1984 | 2024-04-01T00:00:00.000Z | |
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| Carol White | To Kill a Mockingbird | 2024-03-15T00:00:00.000Z | |
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| David Brown | The Great Gatsby | 2024-04-15T00:00:00.000Z | |
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| Frank Wilson | 1984 | 2024-05-15T00:00:00.000Z | |
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## Solution |
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```sql |
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SELECT |
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c2.name AS customer_name, |
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b.title AS common_book, |
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s2.order_date AS purchase_date |
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FROM sale s1 |
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INNER JOIN sale s2 |
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ON s1.book_id = s2.book_id |
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AND s1.customer_id = 1 |
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AND s2.customer_id != 1 |
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INNER JOIN customer c2 |
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ON s2.customer_id = c2.id |
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INNER JOIN book b |
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ON s1.book_id = b.id |
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ORDER BY c2.name, b.title; |
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``` |
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|
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## Explanation |
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Let's break down how this query works: |
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We first start with all the purchases of Alice (customer_id = 1) to find the books she has purchased i.e. |
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```sql |
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FROM sale s1 |
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``` |
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Now, because we want to find all the other sales that have the same book_id as Alice's sales, we need to join the `sale` table again (aliased as `s2`) to compare purchases: |
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```sql |
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INNER JOIN sale s2 ON s1.book_id = s2.book_id |
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``` |
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This will give us all the sales that have the same book_id as Alice's sales. But we are only interested in the sales to other customers, so we need to add a condition to exclude Alice's sales: |
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```sql |
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AND s2.customer_id != 1 |
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``` |
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Now we need to join the `customer` table to get the name of the other customer: |
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```sql |
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INNER JOIN customer c2 ON s2.customer_id = c2.id |
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``` |
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Finally, we need to join the `book` table to get the title of the book: |
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```sql |
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INNER JOIN book b ON s1.book_id = b.id |
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``` |
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Finally, we order the results by customer name and book title for easy reading: |
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```sql |
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ORDER BY c2.name, b.title; |
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``` |
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This query efficiently finds all customers who share at least one book purchase with Alice, showing which books they have in common and when they made their purchase. |
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