Add exercises

src/data/courses/sql-mastery/chapters/scalar-functions/lessons/book-sale-categories.md

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+---
+title: Book Categories
+description: Practice using CASE expressions to categorize books and inventory levels
+order: 220
+type: challenge
+setup: |
+  ```sql
+  CREATE TABLE book (
+      id INT PRIMARY KEY,
+      title VARCHAR(100),
+      page_count INT,
+      publish_year INT,
+      copies_in_stock INT,
+      monthly_rentals INT
+  );
+
+  INSERT INTO book (id, title, page_count, publish_year, copies_in_stock, monthly_rentals) VALUES
+      (1, 'SQL Basics', 250, 2020, 5, 12),
+      (2, 'Advanced Queries', 450, 2021, 3, 8),
+      (3, 'Database Design', 350, 2019, 10, 15),
+      (4, 'Query Optimization', 275, 2022, 2, 6),
+      (5, 'Data Modeling', 400, 2021, 1, 4),
+      (6, 'PostgreSQL Tips', 200, 2023, 8, 20);
+  ```
+---
+
+A library needs to analyze their SQL book collection. They want a report that categorizes books based on various criteria.
+
+Given the following data in table `book`:
+
+| id  | title              | page_count | publish_year | copies_in_stock | monthly_rentals |
+| --- | ------------------ | ---------- | ------------ | --------------- | --------------- |
+| 1   | SQL Basics         | 250        | 2020         | 5               | 12              |
+| 2   | Advanced Queries   | 450        | 2021         | 3               | 8               |
+| 3   | Database Design    | 350        | 2019         | 10              | 15              |
+| 4   | Query Optimization | 275        | 2022         | 2               | 6               |
+| 5   | Data Modeling      | 400        | 2021         | 1               | 4               |
+| 6   | PostgreSQL Tips    | 200        | 2023         | 8               | 20              |
+
+Write a query that shows:
+
+- Book title
+- Book length category:
+  - `Short` if page_count < 300
+  - `Medium` if page_count between 300 and 400
+  - `Long` if page_count > 400
+- Publication status:
+  - `Recent` if publish_year >= 2022
+  - `Current` if publish_year is 2020 or 2021
+  - `Older` if publish_year < 2020
+- Stock status:
+  - `Low` if copies_in_stock < 3
+  - `Medium` if copies_in_stock between 3 and 7
+  - `High` if copies_in_stock > 7
+- Popularity:
+  - `High Demand` if monthly_rentals > 12
+  - `Medium Demand` if monthly_rentals between 6 and 12
+  - `Low Demand` if monthly_rentals < 6
+
+## Expected Output
+
+| title              | length_category | publication_status | stock_status | popularity    |
+| ------------------ | --------------- | ------------------ | ------------ | ------------- |
+| PostgreSQL Tips    | Short           | Recent             | High         | High Demand   |
+| Database Design    | Medium          | Older              | High         | High Demand   |
+| SQL Basics         | Short           | Current            | Medium       | Medium Demand |
+| Advanced Queries   | Long            | Current            | Medium       | Medium Demand |
+| Query Optimization | Short           | Recent             | Low          | Medium Demand |
+
+Data Modeling Medium Current Low Low Demand
+
+## Solution
+
+```sql
+SELECT
+    title,
+    CASE
+        WHEN page_count < 300 THEN 'Short'
+        WHEN page_count <= 400 THEN 'Medium'
+        ELSE 'Long'
+    END as length_category,
+    CASE
+        WHEN publish_year >= 2022 THEN 'Recent'
+        WHEN publish_year >= 2020 THEN 'Current'
+        ELSE 'Older'
+    END as publication_status,
+    CASE
+        WHEN copies_in_stock < 3 THEN 'Low'
+        WHEN copies_in_stock <= 7 THEN 'Medium'
+        ELSE 'High'
+    END as stock_status,
+    CASE
+        WHEN monthly_rentals > 12 THEN 'High Demand'
+        WHEN monthly_rentals >= 6 THEN 'Medium Demand'
+        ELSE 'Low Demand'
+    END as popularity
+FROM books
+ORDER BY monthly_rentals DESC;
+```

src/data/courses/sql-mastery/chapters/scalar-functions/lessons/customer-contact-list.md

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+---
+title: Customer Contact List
+description: Practice using string and conversion functions to standardize customer contact information
+order: 210
+type: challenge
+setup: |
+  ```sql
+  CREATE TABLE customer (
+      id INT PRIMARY KEY,
+      first_name VARCHAR(50),
+      last_name VARCHAR(50),
+      email VARCHAR(255),
+      phone VARCHAR(20)
+  );
+
+  INSERT INTO customer (id, first_name, last_name, email, phone) VALUES
+      (1, 'john', 'DOE', 'JOHN.DOE@email.com', '  555-123-4567  '),
+      (2, 'JANE', 'smith', 'jane.smith@email.com', '555.123.4568'),
+      (3, 'Bob', 'WILSON', 'bob.wilson@email.com', '5551234569'),
+      (4, 'ALICE', 'brown', 'alice.b@email.com', NULL),
+      (5, 'charlie', 'DAVIS', 'charlie.d@email.com', '555-123-4570');
+  ```
+---
+
+The bookstore is updating their customer contact list and needs to standardize how customer information is displayed. They want all names properly capitalized, emails in lowercase, and phone numbers in a consistent format.
+
+Given the following data in table `customer`
+
+| id  | first_name | last_name | email                | phone        |
+| --- | ---------- | --------- | -------------------- | ------------ |
+| 1   | john       | DOE       | JOHN.DOE@email.com   | 555-123-4567 |
+| 2   | JANE       | smith     | jane.smith@email.com | 555.123.4568 |
+| 3   | Bob        | WILSON    | bob.wilson@email.com | 5551234569   |
+| 4   | ALICE      | brown     | alice.b@email.com    | NULL         |
+| 5   | charlie    | DAVIS     | charlie.d@email.com  | 555-123-4570 |
+
+Write a query that formats the customer information according to these requirements:
+
+- Full name (first name and last name properly capitalized)
+- Email address (in lowercase)
+- Phone number (Numbers only, or `No phone` if `NULL`)
+
+## Expected Output
+
+You output should look like this:
+
+| full_name     | email                | phone      |
+| ------------- | -------------------- | ---------- |
+| John Doe      | john.doe@email.com   | 5551234567 |
+| Jane Smith    | jane.smith@email.com | 5551234568 |
+| Bob Wilson    | bob.wilson@email.com | 5551234569 |
+| Alice Brown   | alice.b@email.com    | No phone   |
+| Charlie Davis | charlie.d@email.com  | 5551234570 |
+
+> **Hint:** Use the `REGEXP_REPLACE` function to remove non-numeric characters from the phone number. Use the `g` flag to replace all occurrences of the pattern.
+
+## Solution
+
+```sql
+SELECT
+    CONCAT(
+        INITCAP(first_name),
+        ' ',
+        INITCAP(last_name)
+    ) as full_name,
+    LOWER(email) as email,
+    COALESCE(
+        REGEXP_REPLACE(
+            TRIM(phone),
+            '[^0-9]',
+            '',
+            'g'
+        ),
+        'No phone'
+    ) as phone
+FROM customer
+ORDER BY id;
+```

src/data/courses/sql-mastery/chapters/scalar-functions/lessons/membership-duration.md

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+---
+title: Membership Duration
+description: Practice using date functions to calculate membership length and categories
+order: 210
+type: challenge
+setup: |
+  ```sql
+  CREATE TABLE member (
+      id INT PRIMARY KEY,
+      name VARCHAR(100),
+      join_date DATE
+  );
+
+  INSERT INTO member (id, name, join_date) VALUES
+      (1, 'John Smith', '2022-01-15'),
+      (2, 'Mary Johnson', '2023-06-20'),
+      (3, 'Bob Wilson', '2024-01-10'),
+      (4, 'Alice Brown', '2023-03-01'),
+      (5, 'Charlie Davis', '2023-12-25');
+  ```
+---
+
+The bookstore wants to categorize their members based on how long they've been part of the loyalty program. They need a report showing each member's duration and their corresponding membership tier.
+
+Given the following data in table `member`
+
+| id  | name          | join_date  |
+| --- | ------------- | ---------- |
+| 1   | John Smith    | 2022-01-15 |
+| 2   | Mary Johnson  | 2023-06-20 |
+| 3   | Bob Wilson    | 2024-01-10 |
+| 4   | Alice Brown   | 2023-03-01 |
+| 5   | Charlie Davis | 2023-12-25 |
+
+Write a query that shows:
+
+- Member's name
+- Number of months they've been a member
+- Membership tier:
+  - `Bronze` if less than 6 months
+  - `Silver` if between 6 and 12 months
+  - `Gold` if between 12 and 24 months
+  - `Platinum` if more than 24 months
+
+## Expected Output
+
+| name          | months | tier     |
+| ------------- | ------ | -------- |
+| John Smith    | 35     | Platinum |
+| Alice Brown   | 22     | Gold     |
+| Mary Johnson  | 18     | Gold     |
+| Charlie Davis | 12     | Gold     |
+| Bob Wilson    | 11     | Silver   |
+
+> **Note:** The months will vary based on current date. This example assumes CURRENT_DATE is '2025-01-06'.
+
+## Solution
+
+```sql
+SELECT
+    name,
+    EXTRACT(YEAR FROM AGE(CURRENT_DATE, join_date)) * 12 +
+    EXTRACT(MONTH FROM AGE(CURRENT_DATE, join_date)) AS months,
+    CASE
+        WHEN AGE(CURRENT_DATE, join_date) < INTERVAL '6 months' THEN 'Bronze'
+        WHEN AGE(CURRENT_DATE, join_date) < INTERVAL '12 months' THEN 'Silver'
+        WHEN AGE(CURRENT_DATE, join_date) < INTERVAL '24 months' THEN 'Gold'
+        ELSE 'Platinum'
+    END AS tier
+FROM member
+ORDER BY months DESC;
+```