Ecosyste.ms: Awesome
An open API service indexing awesome lists of open source software.
https://github.com/ayonika2001/music_data_sql
https://github.com/ayonika2001/music_data_sql
csv dbms mysql mysql-database mysql-server sql
Last synced: 19 days ago
JSON representation
- Host: GitHub
- URL: https://github.com/ayonika2001/music_data_sql
- Owner: Ayonika2001
- Created: 2025-01-24T17:19:51.000Z (22 days ago)
- Default Branch: main
- Last Pushed: 2025-01-24T17:34:43.000Z (22 days ago)
- Last Synced: 2025-01-24T18:24:39.366Z (22 days ago)
- Topics: csv, dbms, mysql, mysql-database, mysql-server, sql
- Homepage:
- Size: 253 KB
- Stars: 0
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
Awesome Lists containing this project
README
# SQL Project: Music Store Data Analysis
This project involves analyzing the data of a music store using SQL. The analysis is divided into three question sets: **Easy**, **Moderate**, and **Advanced**, with each set addressing specific business questions. The dataset includes tables such as `employee`, `invoice`, `customer`, `track`, `album`, `artist`, `genre`, and `invoice_line`. The queries aim to extract actionable insights such as identifying top customers, best-performing artists, and popular genres.
---
## Database Setup
```sql
CREATE DATABASE music;
USE music;
```
---
## Question Set 1 - Easy
### 1. Who is the senior most employee based on job title?
```sql
SELECT first_name, last_name, title, MAX(levels) AS senior_most_employee
FROM employee
GROUP BY first_name, last_name, title
ORDER BY senior_most_employee DESC LIMIT 1;
```
### 2. Which countries have the most invoices?
```sql
SELECT billing_country, COUNT(invoice_id) AS most_invoice
FROM invoice
GROUP BY billing_country;
```
### 3. What are the top 3 values of total invoice?
```sql
SELECT total AS total_invoice
FROM invoice
ORDER BY total_invoice DESC LIMIT 3;
```
### 4. Which city has the best customers?
Return the city with the highest sum of invoice totals.
```sql
SELECT billing_city, SUM(total) AS invoice_totals
FROM invoice
GROUP BY billing_city
ORDER BY invoice_totals DESC LIMIT 1;
```
### 5. Who is the best customer?
The customer who has spent the most money will be declared the best customer.
```sql
SELECT customer.customer_id, customer.first_name, customer.last_name, SUM(invoice.total) AS invoice_totals
FROM customer
JOIN invoice ON customer.customer_id = invoice.customer_id
GROUP BY customer.customer_id, customer.first_name, customer.last_name
ORDER BY invoice_totals DESC LIMIT 1;
```
---
## Question Set 2 - Moderate
### 1. Rock Music Listeners
Return the email, first name, last name, and genre of all rock music listeners, ordered alphabetically by email.
```sql
SELECT c.first_name, c.last_name, c.email, g.name
FROM customer AS c
JOIN invoice AS i ON c.customer_id = i.customer_id
JOIN invoice_line AS il ON i.invoice_id = il.invoice_id
JOIN track AS t ON il.track_id = t.track_id
JOIN genre AS g ON t.genre_id = g.genre_id
WHERE g.name = "Rock"
ORDER BY c.email;
```
### 2. Top 10 Rock Bands
Return the artist name and total track count for the top 10 rock bands.
```sql
SELECT a.name, g.name, COUNT(t.track_id) AS total_track
FROM artist AS a
JOIN album AS al ON a.artist_id = al.artist_id
JOIN track AS t ON al.album_id = t.album_id
JOIN genre AS g ON t.genre_id = g.genre_id
WHERE g.name = "Rock"
GROUP BY a.name, g.name
ORDER BY total_track DESC LIMIT 10;
```
### 3. Tracks Longer than Average Length
Return all track names with a song length longer than the average, sorted by the longest song first.
```sql
SELECT name, milliseconds
FROM track
WHERE milliseconds > (SELECT AVG(milliseconds) FROM track)
ORDER BY milliseconds DESC;
```
---
## Question Set 3 - Advanced
### 1. Amount Spent by Each Customer on Artists
Return customer name, artist name, and the total amount spent.
```sql
WITH best_selling_artist AS (
SELECT a.artist_id, a.name, SUM(il.unit_price * il.quantity) AS total_spent
FROM artist AS a
JOIN album AS al ON a.artist_id = al.artist_id
JOIN track AS t ON al.album_id = t.album_id
JOIN invoice_line AS il ON t.track_id = il.track_id
GROUP BY a.artist_id
ORDER BY total_spent DESC LIMIT 1
)
SELECT c.customer_id, c.first_name, c.last_name, bsa.name AS artist_name, SUM(il.unit_price * il.quantity) AS amount_spent
FROM best_selling_artist AS bsa
JOIN album AS al ON bsa.artist_id = al.artist_id
JOIN track AS t ON al.album_id = t.album_id
JOIN invoice_line AS il ON t.track_id = il.track_id
JOIN invoice AS i ON il.invoice_id = i.invoice_id
JOIN customer AS c ON i.customer_id = c.customer_id
GROUP BY c.customer_id, c.first_name, c.last_name, artist_name
ORDER BY amount_spent DESC;
```
### 2. Most Popular Genre for Each Country
Determine the most popular genre for each country based on purchases.
```sql
WITH most_popular_genre AS (
SELECT g.name, i.billing_country, COUNT(il.quantity) AS purchase,
ROW_NUMBER() OVER(PARTITION BY i.billing_country ORDER BY COUNT(il.quantity) DESC) AS row_no
FROM genre AS g
JOIN track AS t ON g.genre_id = t.genre_id
JOIN invoice_line AS il ON t.track_id = il.track_id
JOIN invoice AS i ON il.invoice_id = i.invoice_id
GROUP BY g.name, i.billing_country
)
SELECT *
FROM most_popular_genre
WHERE row_no = 1;
```
### 3. Top Customer in Each Country
Return the country, top customer, and the amount spent.
```sql
WITH top_customer AS (
SELECT c.customer_id, c.first_name, c.last_name, i.billing_country, SUM(il.unit_price * il.quantity) AS amount_spent,
ROW_NUMBER() OVER(PARTITION BY i.billing_country ORDER BY SUM(il.unit_price * il.quantity) DESC) AS row_num
FROM customer AS c
JOIN invoice AS i ON c.customer_id = i.customer_id
JOIN invoice_line AS il ON i.invoice_id = il.invoice_id
GROUP BY c.customer_id, c.first_name, c.last_name, i.billing_country
)
SELECT *
FROM top_customer
WHERE row_num = 1;