Assignment 1010 Pointsusing Mysql Select Commands Answer The Questi ✓ Solved
Assignment # points) Using MySQL SELECT commands answer the questions listed below using the Tiny Video schema used in your previous assignment. Submit via the D2L drop box the final SQL script for each problem. All questions are worth 2 points each. 1. Write a query to display the columns listed below.
The query should list each customer once, whether they have rented a movie or not. Where the customer has rented a movie it should show only the first rental date. The output should be sorted by membership number. 2. Write a query to display the columns listed below.
The query should list each customer who has not rented a move. The output should be sorted by membership number. 3. Write a query to display the columns listed below. The query should list each customer where the total detail rental fee is greater than twice the overall average detail rental fee.
The output should be sorted by membership number. (Hint: A HAVING clause will be needed) 4. Write a query to display the columns listed below. The query should list each customer in which the video rental is overdue. The Days_Overdue column should calculate the number of days overdue based upon the due date and return date. The output should be sorted by membership number and then by the detail due date.
5. Write a query to display the columns listed below. For each customer the query should show the current system date, the current day (when you do the problem the date and day will be different), the number of characters in the member last name, the last date the customer rented a video and how many total videos the person rented. 1
Paper for above instructions
SQL Queries for Tiny Video Schema
In this assignment, we will write SQL queries using MySQL SELECT commands to answer several questions regarding customer rentals in the Tiny Video schema. Each query is designed to extract specific data from a hypothetical database and will be structured optimally to return valid results. Below are the solutions for the given queries.
Query 1: Display Each Customer with First Rental Date
The first query aims to display each customer, regardless of whether they have rented a movie. If a customer has rented a movie, it should show only the first rental date. The output will be sorted by membership number.
```sql
SELECT c.membership_number,
c.first_name,
c.last_name,
MIN(r.rental_date) AS first_rental_date
FROM customers AS c
LEFT JOIN rentals AS r ON c.customer_id = r.customer_id
GROUP BY c.membership_number, c.first_name, c.last_name
ORDER BY c.membership_number;
```
Explanation:
- We are using `LEFT JOIN` to ensure that all customers are included in the result set, even those who haven't rented any movies.
- We use `MIN(r.rental_date)` to select the first rental date for customers who have rented at least once.
- `GROUP BY` is used to aggregate results for each customer.
Query 2: Display Customers Who Have Not Rented a Movie
The second query needs to list each customer who has not rented a movie. The output will be sorted by membership number.
```sql
SELECT c.membership_number,
c.first_name,
c.last_name
FROM customers AS c
LEFT JOIN rentals AS r ON c.customer_id = r.customer_id
WHERE r.rental_id IS NULL
ORDER BY c.membership_number;
```
Explanation:
- This query also employs `LEFT JOIN` but looks specifically for customers whose `rental_id` is `NULL`, indicating they have not rented any movies.
- The results are sorted by `membership_number`.
Query 3: Customers with High Total Rental Fees
The third query should list customers whose total rental fees exceed twice the average detail rental fee.
```sql
SELECT c.membership_number,
c.first_name,
c.last_name,
SUM(rd.rental_fee) AS total_rental_fee
FROM customers AS c
JOIN rentals AS r ON c.customer_id = r.customer_id
JOIN rental_details AS rd ON r.rental_id = rd.rental_id
GROUP BY c.membership_number
HAVING SUM(rd.rental_fee) > (2 * (SELECT AVG(rental_fee) FROM rental_details))
ORDER BY c.membership_number;
```
Explanation:
- This query calculates the total rental fee for each customer and filters those whose total exceeds twice the average rental fee.
- We use a subquery `(SELECT AVG(rental_fee) FROM rental_details)` to calculate the average rental fee.
- The results are sorted by `membership_number`.
Query 4: Overdue Rentals
The fourth query lists customers who have overdue rentals, calculating the number of overdue days.
```sql
SELECT c.membership_number,
c.first_name,
c.last_name,
DATEDIFF(r.return_date, r.due_date) AS Days_Overdue
FROM customers AS c
JOIN rentals AS r ON c.customer_id = r.customer_id
WHERE r.return_date > r.due_date
ORDER BY c.membership_number, r.due_date;
```
Explanation:
- This query identifies overdue rentals by finding records where the `return_date` is greater than the `due_date`.
- The `DATEDIFF` function is used to calculate the number of overdue days.
- Results are sorted first by `membership_number` and then by the `due_date`.
Query 5: Customer Statistics
The fifth query aims to provide an overview of customer activity, including the current date and several statistics about each customer.
```sql
SELECT c.membership_number,
c.first_name,
c.last_name,
CURRENT_DATE AS current_system_date,
DAYNAME(CURRENT_DATE) AS current_day,
LENGTH(c.last_name) AS last_name_length,
MAX(r.rental_date) AS last_rental_date,
COUNT(r.rental_id) AS total_rentals
FROM customers AS c
LEFT JOIN rentals AS r ON c.customer_id = r.customer_id
GROUP BY c.membership_number
ORDER BY c.membership_number;
```
Explanation:
- This query pulls various statistics, including the current system date using the `CURRENT_DATE` function and the current day's name using the `DAYNAME()` function.
- The length of the last name is calculated with the `LENGTH()` function.
- `MAX(r.rental_date)` gives the last rental date for each customer, and `COUNT(r.rental_id)` counts the total rentals each customer has made.
- The results are sorted by `membership_number`.
Conclusion
The above SQL queries utilize MySQL’s functionalities to derive insights from the Tiny Video schema effectively. Each query caters to specific requirements, ensuring accurate data retrieval while adhering to best practices in SQL coding.
References
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This extensive analysis and related queries are intended to sharpen your skills in SQL and data handling, ensuring a robust understanding of database management systems.