SQLTasks Tier 2.sql

/* Welcome to the SQL mini project. You will carry out this project partly in
the PHPMyAdmin interface, and partly in Jupyter via a Python connection.

This is Tier 2 of the case study, which means that there'll be less guidance for you about how to setup
your local SQLite connection in PART 2 of the case study. This will make the case study more challenging for you: 
you might need to do some digging, aand revise the Working with Relational Databases in Python chapter in the previous resource.

Otherwise, the questions in the case study are exactly the same as with Tier 1. 

PART 1: PHPMyAdmin
You will complete questions 1-9 below in the PHPMyAdmin interface. 
Log in by pasting the following URL into your browser, and
using the following Username and Password:

URL: https://sql.springboard.com/
Username: student
Password: learn_sql@springboard

The data you need is in the "country_club" database. This database
contains 3 tables:
    i) the "Bookings" table,
    ii) the "Facilities" table, and
    iii) the "Members" table.

In this case study, you'll be asked a series of questions. You can solve them using the platform, but for the final deliverable, paste the code for each solution into this script, and upload it to your GitHub.

Before starting with the questions, feel free to take your time,
exploring the data, and getting acquainted with the 3 tables. */


/* QUESTIONS 
/* Q1: Some of the facilities charge a fee to members, but some do not.
Write a SQL query to produce a list of the names of the facilities that do. */

SELECT name FROM Facilities WHERE membercost > 0;


/* Q2: How many facilities do not charge a fee to members? */

SELECT COUNT(name) FROM Facilities WHERE membercost = 0;


/* Q3: Write an SQL query to show a list of facilities that charge a fee to members, ­
where the fee is less than 20% of the facility's monthly maintenance cost.
Return the facid, facility name, member cost, and monthly maintenance of the
facilities in question. */

SELECT facid, name, membercost, monthlymaintenance  FROM Facilities
WHERE membercost > 0 
AND membercost < monthlymaintenance/.2;

/* Q4: Write an SQL query to retrieve the details of facilities with ID 1 and 5.
Try writing the query without using the OR operator. */

SELECT * FROM Facilities
WHERE facid IN (1,5);

/* Q5: Produce a list of facilities, with each labelled as
'cheap' or 'expensive', depending on if their monthly maintenance cost is
more than $100. Return the name and monthly maintenance of the facilities
in question. */

SELECT name, monthlymaintenance, 
CASE WHEN monthlymaintenance > 100 THEN 'Expensive'
	 WHEN monthlymaintenance < 100 THEN	'Cheap'
	 END AS Cost
FROM Facilities;

/* Q6: You'd like to get the first and last name of the last member(s)
who signed up. Try not to use the LIMIT clause for your solution. */

SELECT firstname, surname, joindate FROM Members
WHERE joindate IN (SELECT max(joindate) FROM Members);

/* Q7: Produce a list of all members who have used a tennis court.
Include in your output the name of the court, and the name of the member
formatted as a single column. Ensure no duplicate data, and order by
the member name. */

SELECT CONCAT(m.firstname,' ', m.surname) AS Name, f.name
FROM Members AS m
JOIN Facilities AS f
WHERE f.name LIKE 'Tennis%'
GROUP BY m.surname,f.name
ORDER BY Name;



/* Q8: Produce a list of bookings on the day of 2012-09-14 which
will cost the member (or guest) more than $30. Remember that guests have
different costs to members (the listed costs are per half-hour 'slot'), and
the guest user's ID is always 0. Include in your output the name of the
facility, the name of the member formatted as a single column, and the cost.
Order by descending cost, and do not use any subqueries. */

SELECT f.name as Facility, 
	   CONCAT(m.firstname,' ', m.surname) AS Member,
	   CASE WHEN b.memid = 0 THEN f.guestcost * b.slots
			ELSE f.membercost * b.slots END AS Cost
FROM Bookings as b
INNER JOIN Facilities AS f ON b.facid = f.facid
INNER JOIN Members AS m ON b.memid = m.memid
WHERE b.starttime LIKE '%2012-09-14%'
AND (b.memid = 0 AND f.guestcost * b.slots > 30)
OR (b.memid != 0 and f.membercost * b.slots > 30)
ORDER BY Cost DESC;

/* Q9: This time, produce the same result as in Q8, but using a subquery. */


/* PART 2: SQLite

Export the country club data from PHPMyAdmin, and connect to a local SQLite instance from Jupyter notebook 
for the following questions.  

QUESTIONS:
/* Q10: Produce a list of facilities with a total revenue less than 1000.
The output of facility name and total revenue, sorted by revenue. Remember
that there's a different cost for guests and members! */

SELECT name, revenue FROM
	(SELECT f.name, sum(CASE WHEN b.memid = 0 THEN b.slots * f.guestcost
                        	 ELSE b.slots * f.membercost
                        END) AS revenue
     		FROM Bookings as b
     		INNER JOIN Facilities AS f ON b.facid = f.facid
     		GROUP BY f.name)
			AS agg WHERE revenue < 1000
ORDER BY revenue;


/* Q11: Produce a report of members and who recommended them in alphabetic surname,firstname order */

SELECT a.surname||', '||a.firstname AS Member, b.firstname||' '||b.surname AS 'Rec By'
FROM Members AS a
JOIN Members AS b
ON (a.recommendedby = b.memid AND b.memid != 0 AND a.recommendedby IS NOT NULL)
ORDER BY a.surname

/* Q12: Find the facilities with their usage by member, but not guests */

SELECT f.name AS Name, CONCAT(m.firstname,' ', m.surname) AS Member, count(f.name) AS Used
FROM Facilities AS f
JOIN Bookings AS b ON b.facid = f.facid
JOIN Members AS m ON b.memid = m.memid
WHERE b.memid != 0
GROUP BY m.surname, f.name
ORDER BY f.name, m.surname;

/* Q13: Find the facilities usage by month, but not guests */


SELECT f.name as Name, strftime('%m', b.starttime) as Month, COUNT(f.name) AS Used
FROM Bookings AS b
JOIN Facilities AS f ON b.facid = f.facid
GROUP BY f.name, strftime('%m', b.starttime);