Danny_ma_SQL_challenge_week1_soln

My solution to Danny_ma's 8 weeks SQL challenge,Case Study 1. Solution contains Aggregate functions ,Window Functions ,JOINS and Common Table Expressions with easy readability.

Introduction

Danny seriously loves Japanese food so in the beginning of 2021, he decides to embark upon a risky venture and opens up a cute little restaurant that sells his 3 favourite foods: sushi, curry and ramen.

Danny’s Diner is in need of your assistance to help the restaurant stay afloat - the restaurant has captured some very basic data from their few months of operation but have no idea how to use their data to help them run the business.

Problem Statement

Danny wants to use the data to answer a few simple questions about his customers, especially about their visiting patterns, how much money they’ve spent and also which menu items are their favourite. Having this deeper connection with his customers will help him deliver a better and more personalised experience for his loyal customers.
He plans on using these insights to help him decide whether he should expand the existing customer loyalty program - additionally he needs help to generate some basic datasets so his team can easily inspect the data without needing to use SQL.

Danny has provided you with a sample of his overall customer data due to privacy issues - but he hopes that these examples are enough for you to write fully functioning SQL queries to help him answer his questions!

Danny has shared with you 3 key datasets for this case study:

• sales
  
• menu
  
• members
  You can inspect the entity relationship diagram below.

Entity Relationship Diagram

Case Study Questions

---

1. What is the total amount each customer spent at the restaurant?

   SELECT s.customer_id,SUM(m.price)  "total_amt_spent($)"
   FROM dannys_diner.sales s
   JOIN dannys_diner.menu m
   ON m.product_id=s.product_id
   GROUP BY 1
   ORDER BY 2 DESC;

customer_id	total_amt_spent($)
A	76
B	74
C	36

---

2. How many days has each customer visited the restaurant?

WITH unique_days_count as (
SELECT DISTINCT s.customer_id ,s.order_date days
FROM dannys_diner.sales s
)

SELECT customer_id ,COUNT(days) visitation_count
FROM unique_days_count
GROUP BY 1
ORDER BY 1;

customer_id	visitation_count
A	4
B	6
C	2

---

3 What was the first item from the menu purchased by each customer?

WITH  first_order_date as (
SELECT s.Customer_id ,MIN(s.order_date) order_date
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON m.product_id = s.product_id
GROUP BY 1
ORDER BY 1
)
SELECT Distinct  s.Customer_id , m.product_name, s.order_date date_of_first_order
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON m.product_id = s.product_id
JOIN first_order_date f
ON f.customer_id=s.customer_id AND f.order_date=s.order_date 
ORDER BY 1,3;

customer_id	product_name	date_of_first_order
A	curry	2021-01-01T00:00:00.000Z
A	sushi	2021-01-01T00:00:00.000Z
B	curry	2021-01-01T00:00:00.000Z
C	ramen	2021-01-01T00:00:00.000Z

---

4. What is the most purchased item on the menu and how many times was it purchased by all customers?

WITH most_purchased_item as 
(
SELECT m.product_name, COUNT(m.product_name),SUM(PRICE)
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1 
)

SELECT s.customer_id, m.product_name , COUNT(m.product_name)
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
WHERE m.product_name= (SELECT product_name FROM most_purchased_item)
GROUP BY 1,2
ORDER BY 1;

customer_id	product_name	count
A	ramen	3
B	ramen	2
C	ramen	3

---

5. Which item was the most popular for each customer?

WITH customer_item_count as ( 
SELECT s.customer_id , m.product_name, COUNT(m.product_name) popularity_count
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
GROUP BY 1,2 
) ,
customer_most_popular_item as (
SELECT customer_id, MAX(popularity_count) popularity_count
FROM customer_item_count
GROUP BY 1
)

SELECT  c.customer_id ,c.product_name, c.popularity_count
FROM customer_item_count c
JOIN customer_most_popular_item p
ON c.customer_id = p.customer_id AND c.popularity_count = p.popularity_count
ORDER BY 3 DESC,1;

customer_id	product_name	popularity_count
A	ramen	3
C	ramen	3
B	sushi	2
B	curry	2
B	ramen	2

---

6. Which item was purchased first by the customer after they became a member?

WITH members_order as (
SELECT mb.Customer_id, m.product_name , s.order_date
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id and s.order_date >= mb.join_date  
), first_date_order_as_a_member as (
SELECT customer_id ,MIN(order_date) order_date
FROM members_order
GROUP BY 1 
)
SELECT f.customer_id ,m.product_name, f.order_date first_order_date_as_a_member
FROM dannys_diner.sales s
JOIN first_date_order_as_a_member f 
ON f.customer_id=s.customer_id AND f.order_date=s.order_date
JOIN dannys_diner.menu m
ON  s.product_id=m.product_id 
ORDER BY 1;

customer_id	product_name	first_order_date_as_a_member
A	curry	2021-01-07T00:00:00.000Z
B	sushi	2021-01-11T00:00:00.000Z

---

7. Which item was purchased just before the customer became a member?

WITH members_order_before_becoming_members as (
SELECT mb.Customer_id, m.product_name ,s.order_date
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id and s.order_date < mb.join_date  
), last_order_date_as_a_non_member as (
SELECT customer_id ,MAX(order_date) order_date
FROM members_order_before_becoming_members 
GROUP BY 1 
)
SELECT l.customer_id , m.product_name, l.order_date last_order_date_as_a_non_member
FROM dannys_diner.sales s
JOIN last_order_date_as_a_non_member l 
ON l.customer_id=s.customer_id AND l.order_date=s.order_date
JOIN dannys_diner.menu m
ON  s.product_id=m.product_id 
ORDER BY 1;

customer_id	product_name	last_order_date_as_a_non_member
A	sushi	2021-01-01T00:00:00.000Z
A	curry	2021-01-01T00:00:00.000Z
B	sushi	2021-01-04T00:00:00.000Z

---

8. What is the total items and amount spent for each member before they became a member?

SELECT s.Customer_id, SUM(m.price) "total_amt_Spent_before_becoming_a_member($)", COUNT(m.product_id) number_of_orders
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id and s.order_date < mb.join_date  
GROUP BY 1 
ORDER BY 2 DESC ,1;

customer_id	total_amt_Spent_before_becoming_a_member($)	number_of_orders
B	40	3
A	25	2

---

9. If each $1 spent equates to 10 points and sushi has a 2x points multiplier - how many points would each customer have?

WITH customer_points as(
SELECT s.customer_id, 
CASE WHEN lower(m.product_name)='sushi' Then (m.price)*20 
					ELSE  price*10 END AS points
from dannys_diner.sales s
JOIN dannys_diner.menu m 
ON s.product_id=m.product_id)

SELECT customer_id , SUM(points) total_points
FROM customer_points
GROUP BY 1 
ORDER BY  1 ;

customer_id	total_points
A	860
B	940
C	360

---

10. In the first week after a customer joins the program (including their join date) they earn 2x points on all items, not just sushi - how many points do customer A and B have at the end of January?

WITH members_jan_order as (
SELECT s.customer_id , m.product_name, m.price, s.order_date , mb.join_date, (mb.join_date + integer '6')  first_week_order_date_as_a_member
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id and (s.order_date >= mb.join_date AND  s.order_date < '2021-02-01')
 ), 
members_jan_points as (SELECT customer_id, product_name, price, order_date, join_date , 
   CASE WHEN order_date <= first_week_order_date_as_a_member  THEN 20 * price 
   WHEN order_date >first_week_order_date_as_a_member and lower(product_name)='sushi' THEN  20 * price 
   ELSE price *10 END as points 
   FROM members_jan_order )

SELECT customer_id, SUM(points) total_january_points
FROM members_jan_points
GROUP BY 1
ORDER BY 1;

customer_id	total_january_points
A	1020
B	320

---

Bonus Questions

Join All The Things

Bonus Question 1

SELECT s.customer_id, s.order_date, m.product_name , m.price ,
CASE WHEN s.customer_id=mb.customer_id AND s.order_date>= join_date THEN  'Y'
ELSE 'N' END as Members 
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
LEFT JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id
ORDER By 1 , 2 ;

customer_id	order_date	product_name	price	members
A	2021-01-01T00:00:00.000Z	sushi	10	N
A	2021-01-01T00:00:00.000Z	curry	15	N
A	2021-01-07T00:00:00.000Z	curry	15	Y
A	2021-01-10T00:00:00.000Z	ramen	12	Y
A	2021-01-11T00:00:00.000Z	ramen	12	Y
A	2021-01-11T00:00:00.000Z	ramen	12	Y
B	2021-01-01T00:00:00.000Z	curry	15	N
B	2021-01-02T00:00:00.000Z	curry	15	N
B	2021-01-04T00:00:00.000Z	sushi	10	N
B	2021-01-11T00:00:00.000Z	sushi	10	Y
B	2021-01-16T00:00:00.000Z	ramen	12	Y
B	2021-02-01T00:00:00.000Z	ramen	12	Y
C	2021-01-01T00:00:00.000Z	ramen	12	N
C	2021-01-01T00:00:00.000Z	ramen	12	N
C	2021-01-07T00:00:00.000Z	ramen	12	N

---

Bonus Question 2

WITH member_ranking as (
SELECT s.customer_id ,s.order_date, m.product_name ,m.price ,
CASE WHEN s.customer_id=mb.customer_id AND s.order_date>= join_date THEN  'Y'
ELSE 'N' END as Members 
FROM dannys_diner.sales s
JOIN dannys_diner.menu m
ON s.product_id=m.product_id
LEFT JOIN dannys_diner.members mb
ON  mb.customer_id=s.customer_id
ORDER By 1 , 2)

SELECT m.customer_id , m.order_date, m.product_name, m.Price , m.members,
CASE WHEN m.customer_id=mb.customer_id AND m.order_date>=mb.join_date 
THEN  DENSE_RANK() OVER (PARTITION BY m.customer_id,m.members ORDER BY m.order_date ) END as Ranking
FROM member_ranking m
LEFT JOIN dannys_diner.members mb
ON  mb.customer_id=m.customer_id 

customer_id	order_date	product_name	price	members	ranking
A	2021-01-01T00:00:00.000Z	sushi	10	N	null
A	2021-01-01T00:00:00.000Z	curry	15	N	null
A	2021-01-07T00:00:00.000Z	curry	15	Y	1
A	2021-01-10T00:00:00.000Z	ramen	12	Y	2
A	2021-01-11T00:00:00.000Z	ramen	12	Y	3
A	2021-01-11T00:00:00.000Z	ramen	12	Y	3
B	2021-01-01T00:00:00.000Z	curry	15	N	null
B	2021-01-02T00:00:00.000Z	curry	15	N	null
B	2021-01-04T00:00:00.000Z	sushi	10	N	null
B	2021-01-11T00:00:00.000Z	sushi	10	Y	1
B	2021-01-16T00:00:00.000Z	ramen	12	Y	2
B	2021-02-01T00:00:00.000Z	ramen	12	Y	3
C	2021-01-01T00:00:00.000Z	ramen	12	N	null
C	2021-01-01T00:00:00.000Z	ramen	12	N	null
C	2021-01-07T00:00:00.000Z	ramen	12	N	null

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Thanks for taking your time in going through my solution for danny_ma SQL challenge week 1 .