sharkawy98 / sql-case-studies

Contains solutions for #8WeekSQLChallenge case studies https://8weeksqlchallenge.com/

8 Weeks SQL Challenge

This repository contains solutions for #8WeekSQLChallenge, they are interesting real-world case studies that will allow you to apply and enhance your SQL skills in many use cases.
I used Microsoft SQL Server in writing SQL queries to solve these case studies.

Table of Contents

• SQL skills gained
• Case study 1
• Case study 2
• Case study 3
• Some interesting queries from my solutions

SQL skills gained

• Data cleaning & transformation
• Aggregations
• Joins
• CTEs
• Variables
• Window functions
  • Ranking (ROW_NUMBER, DENSE_RANK)
  • Analytics (LEAD, LAG)
• CASE WHEN statements
• Subqueries
• UNION & INTERSECT
• DATETIME functions
• Data type conversion
• TEXT functions, text and string manipulation

Case Study #1 : Danny's Diner

My solutions

ERD

Case Study #2 : Pizza Runner

My solutions

ERD

Case Study #3 : Foodie-Fi

My solutions

ERD

Case Study #4 : Data Bank

My solutions

ERD

Some interesting queries

-- Case study 1, question 10
-- Question: 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 dates_cte AS(
	SELECT *, 
		DATEADD(DAY, 6, join_date) AS valid_date, 
		EOMONTH('2021-01-1') AS last_date
	FROM members
)

SELECT
	s.customer_id,
	sum(CASE
		WHEN s.product_id = 1 THEN price*20
		WHEN s.order_date between d.join_date and d.valid_date THEN price*20
		ELSE price*10 
	END) as total_points
FROM
	dates_cte d,
	sales s,
	menu m
WHERE
	d.customer_id = s.customer_id
	AND
	m.product_id = s.product_id
	AND
	s.order_date <= d.last_date
GROUP BY s.customer_id;

---

-- Case study 2, data cleaning
SELECT 
    order_id,
    runner_id,
    cast(CASE 
        WHEN pickup_time = 'null' THEN null
        ELSE pickup_time
    END as datetime) as pickup_time,
    cast(CASE 
        WHEN distance = 'null' THEN null
        ELSE TRIM('km' from distance)
    END as float) as distance,
    cast(CASE
        WHEN duration = 'null' THEN null
        ELSE SUBSTRING(duration, 1, 2)
    END as int)as duration,
    CASE
        WHEN cancellation in ('null', '') THEN null
        ELSE cancellation
    END as cancellation
INTO #cleaned_runner_orders
FROM runner_orders;

---

-- Case study 2, c) ingredient optimisation data cleaning
-- Question: The ingredients of each pizza a stored at pizza_recipes table as a comma separated string including ids of all its toppings, change this string to multiple rows

 SELECT		
    p.pizza_id,
    TRIM(t.value) AS topping_id,
    pt.topping_name
 INTO #cleaned_toppings
 FROM 
    pizza_recipes as p
    CROSS APPLY string_split(p.toppings, ',') as t
    JOIN pizza_toppings as pt
    ON TRIM(t.value) = pt.topping_id 
 ;

---

-- Case study 2, c) ingredient optimisation question 3
-- Question: Generate an order item for each record in the customers_orders table in the format of one of the following:
-- Meat Lovers
-- Meat Lovers - Exclude Beef
-- Meat Lovers - Extra Bacon
-- Meat Lovers - Exclude Cheese, Bacon - Extra Mushroom, Peppers

WITH extras_cte AS
(
	SELECT 
		record_id,
		'Extra ' + STRING_AGG(t.topping_name, ', ') as record_options
	FROM
		#extras e,
		pizza_toppings t
	WHERE e.topping_id = t.topping_id
	GROUP BY record_id
),
exclusions_cte AS
(
	SELECT 
		record_id,
		'Exclude ' + STRING_AGG(t.topping_name, ', ') as record_options
	FROM
		#exclusions e,
		pizza_toppings t
	WHERE e.topping_id = t.topping_id
	GROUP BY record_id
),
union_cte AS
(
	SELECT * FROM extras_cte
	UNION
	SELECT * FROM exclusions_cte
)

SELECT 
	c.record_id,
	CONCAT_WS(' - ', p.pizza_name, STRING_AGG(cte.record_options, ' - '))
FROM 
	#cleaned_customer_orders c
	JOIN pizza_names p
	ON c.pizza_id = p.pizza_id
	LEFT JOIN union_cte cte
	ON c.record_id = cte.record_id
GROUP BY
	c.record_id,
	p.pizza_name
ORDER BY 1;

---

-- Case study 3, b) data analysis question 5
-- Question: How many customers have churned straight after their initial free trial - what percentage is this rounded to the nearest whole number?

WITH churned_after_trial AS(
  SELECT
    customer_id,
    CASE 
      WHEN 
        plan_id = 4  --churn plan
        AND
        LAG(plan_id) OVER (PARTITION BY customer_id ORDER BY start_date) = 0 --trial plan
      THEN 1
      ELSE 0
    END as is_churned
  FROM subscriptions
)

SELECT 
  SUM(is_churned) as churned_customers,
  FLOOR(SUM(is_churned) / CAST(COUNT(DISTINCT customer_id) AS float) * 100) as churn_perct
FROM churned_after_trial;

-- Case study 3, b) data analysis question 10
-- Question: Can you further breakdown this average value into 30 day periods (i.e. 0-30 days, 31-60 days etc)

WITH trial_plan AS(
  SELECT 
    customer_id,
    start_date AS join_date
  FROM subscriptions
  WHERE plan_id = 0
),
annual_plan AS(
  SELECT 
    customer_id,
    start_date AS annual_start_date
  FROM subscriptions
  WHERE plan_id = 3
),
buckets AS(
  SELECT 
    tp.customer_id,
    join_date,
    annual_start_date,
    -- create buckets of 30 days period from 1 to 12 (i.e monthly buckets)
    DATEDIFF(DAY, join_date, annual_start_date)/30 + 1 AS bucket
  FROM 
    trial_plan tp
    JOIN annual_plan ap
    ON tp.customer_id = ap.customer_id
)

SELECT 
  CASE 
    WHEN bucket = 1 THEN CONCAT(bucket-1, ' - ', bucket*30, ' days')
    ELSE CONCAT((bucket-1)*30 + 1, ' - ', bucket*30, ' days')
  END AS period,
  COUNT(customer_id) AS total_customers,
  CAST(AVG(DATEDIFF(DAY, join_date, annual_start_date)*1.0) AS decimal(5, 2)) AS average_days
FROM buckets
GROUP BY bucket;