Danny_ma-SQL-Challenge-week-2-soln

Introduction

Did you know that over 115 million kilograms of pizza is consumed daily worldwide??? (Well according to Wikipedia anyway…)

Danny was scrolling through his Instagram feed when something really caught his eye - “80s Retro Styling and Pizza Is The Future!”

Danny was sold on the idea, but he knew that pizza alone was not going to help him get seed funding to expand his new Pizza Empire - so he had one more genius idea to combine with it - he was going to Uberize it - and so Pizza Runner was launched!

Danny started by recruiting “runners” to deliver fresh pizza from Pizza Runner Headquarters (otherwise known as Danny’s house) and also maxed out his credit card to pay freelance developers to build a mobile app to accept orders from customers.

Available Data

Because Danny had a few years of experience as a data scientist - he was very aware that data collection was going to be critical for his business’ growth.

He has prepared for us an entity relationship diagram of his database design but requires further assistance to clean his data and apply some basic calculations so he can better direct his runners and optimise Pizza Runner’s operations.

All datasets exist within the pizza_runner database schema - be sure to include this reference within your SQL scripts as you start exploring the data and answering the case study questions.

Entity Relation Diagram

Case Study Questions

This case study has LOTS of questions - they are broken up by area of focus including:

A. Pizza Metrics

1. How many pizzas were ordered?

SELECT COUNT(pizza_id) as "Total number of pizzas ordered" 
FROM   pizza_runner.customer_orders;

Total number of pizzas ordered
14

2. How many unique customer orders were made?

SELECT COUNT(Distinct order_id) as " Number of unique orders" 
FROM   pizza_runner.customer_orders;

Number of unique orders
10

3. How many successful orders were delivered by each runner?

SELECT runner_id,COUNT(order_id) "number of orders completed "
FROM pizza_runner.runner_orders
WHERE pickup_time !='null'
GROUP BY 1 
ORDER BY 1;

runner_id	number of orders completed
1	4
2	3
3	1

4. How many of each type of pizza was delivered?

SELECT c.pizza_id,p.pizza_name,count(c.order_id) "number of times delivered "
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
GROUP BY 1,2
ORDER BY 1;

pizza_id	pizza_name	number of times delivered
1	Meatlovers	9
2	Vegetarian	3

5. How many Vegetarian and Meatlovers were ordered by each customer?

WITH meatveg as (
SELECT c.customer_id, CASE WHEN lower(p.pizza_name) ='meatlovers' THEN 1 
										ELSE 0 END AS meatlovers,
                                   CASE WHEN lower(p.pizza_name)='vegetarian' THEN 1 
                                   Else 0 END AS vegetarian      
FROM pizza_runner.customer_orders c



JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
ORDER BY 1)

SELECT customer_id ,SUM(meatlovers) meatlovers ,SUM(vegetarian) vegetarian
FROM meatveg 
GROUP BY 1
ORDER BY 1;

customer_id	meatlovers	vegetarian
101	2	1
102	2	1
103	3	1
104	3	0
105	0	1

6. What was the maximum number of pizzas delivered in a single order?

WITH max_num_pizza as (
SELECT c.order_id ,COUNT(p.pizza_id) pizza_count
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1 )

SELECT  pizza_count "Max number of pizza per order "
FROM max_num_pizza ;

Max number of pizza per order
3

7. For each customer, how many delivered pizzas had at least 1 change and how many had no changes?

WITH pizza_change as (
SELECT c.customer_id,r.order_id,p.pizza_name ,
	   CASE WHEN c.extras ='' THEN '0'
            WHEN c.extras IS NULL THEN '0' 
            WHEN c.extras='null' THEN '0'
            ELSE c.extras END as extras,
	   CASE WHEN exclusions ='' THEN '0'
		WHEN c.exclusions IS NULL THEN '0' 
            WHEN c.exclusions='null' THEN '0'
            ELSE c.exclusions END as exclusions
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
) ,

altered_pizza as (
SELECT customer_id,order_id ,pizza_name , 
       CASE WHEN exclusions != '0' or extras != '0' THEN 1 
		ELSE 0	END AS altered_pizza,
	   CASE WHEN exclusions ='0' AND extras = '0'  THEN 1 
            ELSE 0 END AS unaltered_pizza
FROM pizza_change 
ORDER BY 1 )

SELECT customer_id,SUM(altered_pizza) pizza_with_at_least_a_change,SUM(unaltered_pizza) pizza_with_no_change
FROM altered_pizza 
GROUP BY 1
ORDER BY 1;

customer_id	pizza_with_at_least_a_change	pizza_with_no_change
101	0	2
102	0	3
103	3	0
104	2	1
105	1	0

8. How many pizzas were delivered that had both exclusions and extras?

WITH pizza_change as (
SELECT c.customer_id,r.order_id,p.pizza_name ,
       CASE WHEN c.extras ='' THEN '0'
	        WHEN c.extras IS NULL THEN '0' 
            WHEN c.extras='null' THEN '0'
            ELSE c.extras END as extras,
       CASE WHEN exclusions ='' THEN '0'
		WHEN c.exclusions IS NULL THEN '0' 
            WHEN c.exclusions='null' THEN '0'
            ELSE c.exclusions END as exclusions
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
) ,

extra_and_exclusion as (
SELECT customer_id, order_id, pizza_name, 
       CASE WHEN exclusions != '0' AND extras != '0' THEN 1 
	        ELSE 0	END AS extra_and_exclusion
FROM pizza_change 
ORDER BY 1 )

SELECT COUNT(*) number_of_pizza_with_exclusion_and_extra
FROM extra_and_exclusion 
WHERE extra_and_exclusion != 0;

number_of_pizza_with_exclusion_and_extra
1

9. What was the total volume of pizzas ordered for each hour of the day?

SELECT DATE_PART('hour',order_time) hour_of_the_day, COUNT(order_id) volume_of_pizzas_ordered_by_hour
FROM pizza_runner.customer_orders
GROUP BY 1 
ORDER BY 2 DESC;

hour_of_the_day	volume_of_pizzas_ordered_by_hour
18	3
23	3
21	3
13	3
11	1
19	1

10. What was the volume of orders for each day of the week?

SELECT to_char(order_time,'Day') weekday ,COUNT(order_id) volume_of_pizzas_ordered
FROM pizza_runner.customer_orders
GROUP BY 1 
ORDER BY 2 DESC;

weekday	volume_of_pizzas_ordered
Saturday	5
Wednesday	5
Thursday	3
Friday	1

B. Runner and Customer Experience

1. How many runners signed up for each 1 week period? (i.e. week starts 2021-01-01)

SELECT runner_id, 
          CASE WHEN registration_date BETWEEN '2021-01-01' AND '2021-01-08' THEN '1'
            WHEN registration_date BETWEEN '2021-01-08' AND '2021-01-15' THEN '2'
            WHEN registration_date BETWEEN '2021-01-15' AND '2021-01-22' THEN '3'
			    END AS "signup week", registration_date
FROM pizza_runner.runners

runner_id	signup week	registration_date
1	1	2021-01-01T00:00:00.000Z
2	1	2021-01-03T00:00:00.000Z
3	1	2021-01-08T00:00:00.000Z
4	2	2021-01-15T00:00:00.000Z

2. What was the average time in minutes it took for each runner to arrive at the Pizza Runner HQ to pickup the order?

WITH mins_taken_to_pickup as (
SELECT Distinct r.runner_id ,r.order_id,Date_part('minutes',(TO_TIMESTAMP(r.pickup_time, 'YYYY/MM/DD/HH24:MI:ss')- c.order_time)) time_diff
FROM pizza_runner.runner_orders r
JOIN  pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null'
ORDER BY 2)

SELECT runner_id , ROUND(AVG(time_diff)) "avg_time_of_getting_to_HQ(mins)"
FROM mins_taken_to_pickup
GROUP BY 1
ORDER BY 1;

runner_id	avg_time_of_getting_to_HQ(mins)
1	14
2	20
3	10

3. Is there any relationship between the number of pizzas and how long the order takes to prepare?

WITH pizza_per_order as (SELECT c.order_id ,COUNT(p.pizza_id) pizza_count
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
GROUP BY 1 ) ,

mins_taken_to_pickup as (
SELECT Distinct r.runner_id ,r.order_id,Date_part('minutes',(TO_TIMESTAMP(r.pickup_time, 'YYYY/MM/DD/HH24:MI:ss')- c.order_time)) time_diff
FROM pizza_runner.runner_orders r
JOIN  pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null'
ORDER BY 2)

SELECT pp.order_id ,pp.pizza_count, ROUND(mp.time_diff) order_prep_time
FROM pizza_per_order pp
JOIN mins_taken_to_pickup mp
ON pp.order_id=mp.order_id ;

order_id	pizza_count	order_prep_time
1	1	10
2	1	10
3	2	21
4	3	29
5	1	10
7	1	10
8	1	20
10	2	15

Answer: Yes there is. Orders with more than 1 pizza have a longer prep time compared to order with just 1 pizza.

4. What was the average distance travelled for each customer?

WITH distance_travelled  AS (SELECT   Distinct c.customer_id ,r.order_id,
                             CASE  WHEN POSITION('k' IN distance) =0 THEN COALESCE(distance,'0') :: float
			                                ELSE SUBSTRING(distance,1,position('k' IN distance)-1) :: float END AS distance_km
FROM pizza_runner.runner_orders r 
JOIN pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null')

SELECT customer_id ,ROUND(AVG(distance_km)) avg_distance_travelled_for_km
FROM distance_travelled 
GROUP By 1 
ORDER BY 1 ;

customer_id	avg_distance_travelled_for_km
101	20
102	18
103	23
104	10
105	25

5. What was the difference between the longest and shortest delivery times for all orders?

WITH delivery_duration  AS (SELECT  Distinct c.customer_id ,r.order_id,LEFT(duration,2):: int duration_in_mins
FROM pizza_runner.runner_orders r 
JOIN pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null')

SELECT MAX(duration_in_mins)-MIN(duration_in_mins)  difference_between_the_longest_and_shortest_delivery_duration
FROM delivery_duration ;

difference_between_the_longest_and_shortest_delivery_duration
30

6. What was the average speed for each runner for each delivery and do you notice any trend for these values?

WITH distance_time as (SELECT   Distinct  r.runner_id ,r.order_id,LEFT(duration,2):: float duration,LEFT(distance,2):: float distance 
FROM pizza_runner.runner_orders r 
JOIN pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null'
)
SELECT *, ROUND((distance*60/(duration))::numeric,1) "avg_speed_km/hr"
FROM distance_time
ORDER BY 1 ,2;

runner_id	order_id	duration	distance	avg_speed_km/hr
1	1	32	20	37.5
1	2	27	20	44.4
1	3	20	13	39.0
1	10	10	10	60.0
2	4	40	23	34.5
2	7	25	25	60.0
2	8	15	23	92.0
3	5	15	10	40.0

7. What is the successful delivery percentage for each runner?

WITH successful_delivery as (
SELECT order_id,runner_id ,
       CASE WHEN pickup_time=Null or pickup_time='null' THEN 0
	   ELSE 1 END AS successful_delivery
FROM pizza_runner.runner_orders )

SELECT runner_id, SUM(successful_delivery)*100/COUNT(successful_delivery) successful_delivery_percentage
FROM successful_delivery
GROUP BY 1
ORDER BY 1;

runner_id	successful_delivery_percentage
1	100
2	75
3	50

C. Ingredient Optimisation

1. What are the standard ingredients for each pizza?

WITH meat_pizza as (SELECT topping_name ,row_number() over() as id 
FROM pizza_runner.pizza_toppings
WHERE topping_id  IN (1, 2, 3, 4, 5, 6, 8, 10)) ,

veg_pizza as (
SELECT topping_name,row_number() over() as id
FROM pizza_runner.pizza_toppings
WHERE topping_id IN  (4, 6, 7, 9, 11, 12))

SELECT mp.topping_name standard_meatlovers_ingredient,Coalesce(vp.topping_name,'') standard_vegetarian_ingredient
FROM meat_pizza mp 
FULL OUTER JOIN veg_pizza vp 
ON mp.id=vp.id ;

standard_meatlovers_ingredient	standard_vegetarian_ingredient
Bacon	Cheese
BBQ Sauce	Mushrooms
Beef	Onions
Cheese	Peppers
Chicken	Tomatoes
Mushrooms	Tomato Sauce
Pepperoni
Salami

3. What was the most commonly added extra?

WITH cleaned_extra AS (
    SELECT c.customer_id,r.order_id,p.pizza_name ,
           CASE 
                WHEN c.extras = '' THEN '0'
                WHEN c.extras IS NULL THEN '0' 
                WHEN c.extras = 'null' THEN '0'
                ELSE c.extras 
           END AS extras    FROM pizza_runner.runner_orders r
    JOIN pizza_runner.customer_orders c
        ON r.order_id=c.order_id 
    JOIN pizza_runner.pizza_names p
        ON p.pizza_id= c.pizza_id
), 
extra_split AS (
    SELECT SPLIT_PART(extras, ',', 1) AS extra1, SPLIT_PART(extras, ',', 2) AS extra2
    FROM cleaned_extra
    WHERE extras != '0'
)
SELECT pt.topping_name AS most_added_extra,
       extra AS extra_id,
       COUNT(extra) AS number_of_times_added 
FROM (
    SELECT extra1 AS extra FROM extra_split WHERE extra1 != ''
    UNION ALL
    SELECT extra2 AS extra FROM extra_split WHERE extra2 != ''
) sub
JOIN pizza_runner.pizza_toppings pt ON sub.extra::int = pt.topping_id
GROUP BY 1,2
ORDER BY 3 DESC 
LIMIT 1;

most_added_extra	extra_id	number_of_times_added
Bacon	1	4

3. What was the most common exclusion?

WITH cleaned_exclusions AS (
    SELECT c.customer_id, r.order_id, p.pizza_name,
           CASE 
                WHEN c.exclusions = '' THEN '0'
                WHEN c.exclusions IS NULL THEN '0' 
                WHEN c.exclusions = 'null' THEN '0'
                ELSE c.exclusions 
           END AS exclusions
    FROM pizza_runner.runner_orders r
    JOIN pizza_runner.customer_orders c 
    ON r.order_id = c.order_id 
    JOIN pizza_runner.pizza_names p 
    ON p.pizza_id = c.pizza_id
), 
exclusion_split AS (
    SELECT SPLIT_PART(exclusions, ',', 1) AS exclusion1, SPLIT_PART(exclusions, ',', 2) AS exclusion2
    FROM cleaned_exclusions
    WHERE exclusions != '0'
)
SELECT pt.topping_name AS most_common_exclusion, 
       exclusion AS exclusion_id, 
       COUNT(exclusion) AS number_of_times_excluded
FROM (
    SELECT exclusion1 AS exclusion FROM exclusion_split WHERE exclusion1 != ''
    UNION ALL
    SELECT exclusion2 AS exclusion FROM exclusion_split WHERE exclusion2 != ''
) sub
JOIN pizza_runner.pizza_toppings pt 
ON sub.exclusion::int = pt.topping_id
GROUP BY 1, 2
ORDER BY 3 DESC
LIMIT 1;

most_common_exclusion	exclusion_id	number_of_times_excluded
Cheese	4	4

D. Pricing and Ratings

1. If a Meat Lovers pizza costs $12 and Vegetarian costs $10 and there were no charges for changes - how much money has Pizza Runner made so far if there are no delivery fees?

WITH cost_per_order_with_no_charges_for_changes as ( 
SELECT c.customer_id, r.order_id,c.pizza_id, CASE WHEN c.pizza_id = 1 THEN 12 ELSE 10 END AS "price"
FROM pizza_runner.customer_orders c
JOIN pizza_runner.runner_orders r
ON c.order_id=r.order_id
WHERE r.pickup_time !='null'
ORDER BY 1,2
)
SELECT CONCAT('$',SUM(price)) "total_money_made"
FROM cost_per_order_with_no_charges_for_changes;

total_money_made
$138

1. If a Meat Lovers pizza costs $12 and Vegetarian costs $10 and there were no charges for changes - how much money has Pizza Runner made so far if there are no delivery fees?

WITH cost_per_order AS (
  SELECT
    c.customer_id,
    r.order_id,
    c.pizza_id,
    CASE 
      WHEN c.extras = '' THEN '0'
      WHEN c.extras IS NULL THEN '0' 
      WHEN c.extras = 'null' THEN '0'
      ELSE c.extras 
    END AS extras,
    CASE WHEN c.pizza_id = 1 THEN 12 ELSE 10 END AS price
  FROM pizza_runner.customer_orders c
  JOIN pizza_runner.runner_orders r
    ON c.order_id = r.order_id
  WHERE r.pickup_time!='null'
  ORDER BY 1, 2
),
cost_of_order_with_charges_for_extra AS (
  SELECT
    *,
    CASE  
      WHEN extras != '0' AND LENGTH(TRIM(extras)) = 1 THEN price + 1
      WHEN extras != '0' AND LENGTH(TRIM(extras)) > 1 THEN price + 2
      ELSE price
    END AS new_price
  FROM cost_per_order
)
SELECT CONCAT('$',SUM(new_price)) AS total_amt_made_including_charges_for_extras
FROM cost_of_order_with_charges_for_extra;

total_amt_made_including_charges_for_extras
$142

3. The Pizza Runner team now wants to add an additional ratings system that allows customers to rate their runner, how would you design an additional table for this new dataset - generate a schema for this new table and insert your own data for ratings for each successful customer order between 1 to 5.

CREATE TABLE IF NOT EXISTS runner_ratings (
  order_id INTEGER,
  runner_id INTEGER,
  rating INTEGER
);



INSERT INTO runner_ratings (order_id, runner_id, rating)
SELECT order_id, runner_id, FLOOR(1 + RANDOM() * 5) AS rating
FROM pizza_runner.runner_orders
WHERE pickup_time!= Null or pickup_time != 'null';


SELECT * 
FROM runner_ratings;

order_id	runner_id	rating
1	1	1
2	1	3
3	1	4
4	2	5
5	3	4
7	2	5
8	2	4
10	1	1

4. Using your newly generated table - can you join all of the information together to form a table which has the following information for successful deliveries? customer_id order_id runner_id rating order_time pickup_time Time between order and pickup Delivery duration Average speed Total number of pizzas

WITH pizza_per_order_table AS (
SELECT c.customer_id, c.order_id ,COUNT(p.pizza_id) pizza_count
FROM pizza_runner.runner_orders r
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id AND r.pickup_time != 'null'
JOIN pizza_runner.pizza_names p
ON p.pizza_id= c.pizza_id
GROUP BY 1,2
ORDER BY 1,2),

average_speed_table AS (
WITH distance_time as (SELECT   Distinct  r.runner_id ,r.order_id,LEFT(duration,2):: float duration,LEFT(distance,2):: float distance
FROM pizza_runner.runner_orders r 
JOIN pizza_runner.customer_orders c
ON c.order_id=r.order_id
WHERE r.pickup_time!= Null or r.pickup_time != 'null'
)
SELECT *, ROUND((distance*60/(duration))::numeric,1) "avg_speed"
FROM distance_time
ORDER BY 1,2
)
SELECT 
	DISTINCT pt.customer_id, pt.order_id, sp.runner_id, rr.rating, c.order_time, r.pickup_time, 
	ROUND(EXTRACT(EPOCH FROM age(to_timestamp(r.pickup_time, 'YYYY-MM-DD HH24:MI:SS'), c.order_time))/60) AS time_diff_minutes,
    sp.duration, sp.avg_speed, pt.pizza_count
FROM pizza_per_order_table pt
JOIN average_speed_table sp
ON sp.order_id=pt.order_id
JOIN  pizza_runner.runner_orders r
ON sp.order_id=r.order_id
JOIN pizza_runner.customer_orders c
ON r.order_id=c.order_id 
JOIN runner_ratings rr
ON rr.runner_id=r.runner_id AND rr.order_id=r.order_id
ORDER BY 2;

customer_id	order_id	runner_id	rating	order_time	pickup_time	time_diff_minutes	duration	avg_speed	pizza_count
101	1	1	1	2020-01-01T18:05:02.000Z	2020-01-01 18:15:34	11	32	37.5	1
101	2	1	3	2020-01-01T19:00:52.000Z	2020-01-01 19:10:54	10	27	44.4	1
102	3	1	4	2020-01-02T23:51:23.000Z	2020-01-03 00:12:37	21	20	39.0	2
103	4	2	5	2020-01-04T13:23:46.000Z	2020-01-04 13:53:03	29	40	34.5	3
104	5	3	4	2020-01-08T21:00:29.000Z	2020-01-08 21:10:57	10	15	40.0	1
105	7	2	5	2020-01-08T21:20:29.000Z	2020-01-08 21:30:45	10	25	60.0	1
102	8	2	4	2020-01-09T23:54:33.000Z	2020-01-10 00:15:02	20	15	92.0	1
104	10	1	1	2020-01-11T18:34:49.000Z	2020-01-11 18:50:20	16	10	60.0	2

5. IIf a Meat Lovers pizza was $12 and Vegetarian $10 fixed prices with no cost for extras and each runner is paid $0.30 per kilometre traveled - how much money does Pizza Runner have left over after these deliveries?

WITH cost_per_order_with_no_charges_for_changes AS (
SELECT 
    c.order_id,
    CASE WHEN c.pizza_id = 1 THEN 12 ELSE 10 END AS price
FROM pizza_runner.customer_orders c
JOIN pizza_runner.runner_orders r
ON c.order_id = r.order_id
WHERE r.pickup_time != 'null'
ORDER BY 1
),
distance_per_order AS (
SELECT r.order_id,
ROUND(SUM(CASE
WHEN POSITION('k' IN distance) = 0
THEN COALESCE(distance, '0')::float
ELSE SUBSTRING(distance, 1, POSITION('k' IN distance) - 1)::float
END)) AS distance_km
FROM pizza_runner.runner_orders r
WHERE r.pickup_time != 'null'
GROUP BY 1
)
SELECT CONCAT('$',SUM(c.price) - (SUM(d.distance_km) * 0.3)) AS money_left_over
FROM cost_per_order_with_no_charges_for_changes c
JOIN distance_per_order d
ON c.order_id = d.order_id;

money_left_over
$74.1

C. Bonus Questions

1. If Danny wants to expand his range of pizzas - how would this impact the existing data design?

Answer:

If Danny decides to expand his range of pizzas, This will impact the existing data design by requiring updates to the "pizza_names" and "pizza_recipes" tables. In order to add a new pizza to the menu, a new row would need to be inserted into the "pizza_names" table with a unique pizza_id and the name of the new pizza. A new row would also need to be inserted into the "pizza_recipes" table with the same pizza_id and a comma-separated list of topping IDs for the new pizza.

2. Write an INSERT statement to demonstrate what would happen if a new Supreme pizza with all the toppings was added to the Pizza Runner menu?

INSERT INTO pizza_runner.pizza_names (pizza_id, pizza_name)
VALUES (3, 'Supreme');



SELECT * 
FROM pizza_runner.pizza_names;

pizza_id	pizza_name
1	Meatlovers
2	Vegetarian
3	Supreme

INSERT INTO pizza_runner.pizza_recipes (pizza_id, toppings)
VALUES (3, '1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ');


SELECT * 
FROM pizza_runner.pizza_recipes;

pizza_id	toppings
1	1, 2, 3, 4, 5, 6, 8, 10
2	4, 6, 7, 9, 11, 12
3	1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Name		Name	Last commit message	Last commit date
Latest commit History 15 Commits
Case_study2_pizza_runner.sql		Case_study2_pizza_runner.sql
ERD1.jpg		ERD1.jpg
LICENSE		LICENSE
README.md		README.md
Report.jpg		Report.jpg

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Danny_ma-SQL-Challenge-week-2-soln

Introduction

Available Data

Entity Relation Diagram

Case Study Questions

A. Pizza Metrics

B. Runner and Customer Experience

C. Ingredient Optimisation

D. Pricing and Ratings

C. Bonus Questions

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Danny_ma-SQL-Challenge-week-2-soln

Introduction

Available Data

Entity Relation Diagram

Case Study Questions

A. Pizza Metrics

B. Runner and Customer Experience

C. Ingredient Optimisation

D. Pricing and Ratings

C. Bonus Questions

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