Unit 8.3.3_SQL_Tasks_Country_Club.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. */

A1: SELECT name FROM `Facilities` WHERE membercost > 0;

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

A2: 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. */

A3: SELECT facid, name, membercost, monthlymaintenance
    FROM `Facilities` WHERE membercost > 0 AND membercost < 0.2*monthlymaintenance;

/* 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. */

A4: 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. */

A5: SELECT name, monthlymaintenance,
           CASE WHEN monthlymaintenance > 100 THEN 'expensive'
                ELSE 'cheap' END AS expense_level
    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. */

A6: SELECT firstname, surname
    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. */

A7: SELECT DISTINCT(CONCAT(m.surname,' ',m.firstname)) AS member_name,
           f.name AS facility
    FROM `Bookings` AS b 
    LEFT JOIN `Members` AS m
    ON b.memid = m.memid 
    LEFT JOIN `Facilities` AS f 
    ON b.facid = f.facid
    WHERE b.facid IN (0,1) AND b.memid != 0
    ORDER BY member_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. */

A8: SELECT b.bookid, f.name, 
           CASE WHEN b.memid =0 THEN b.slots * f.guestcost 
                ELSE b.slots * f.membercost END AS cost,
           CASE WHEN b.memid =0 THEN 'Guest'
                ELSE CONCAT(m.surname,' ',m.firstname) END AS member_name
    FROM `Bookings` AS b 
    LEFT JOIN `Members` AS m
    ON b.memid = m.memid 
    LEFT JOIN `Facilities` AS f 
    ON b.facid = f.facid
    WHERE b.starttime LIKE '2012-09-14%' 
    AND (CASE WHEN b.memid =0 THEN b.slots * f.guestcost ELSE b.slots * f.membercost END) > 30
    ORDER BY cost DESC;

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

A9: WITH gb AS (SELECT b.bookid, b.memid, f.name, b.slots * f.guestcost AS cost, 'Guest' AS member_name
                FROM `Bookings` AS b
                LEFT JOIN Facilities AS f
                ON b.facid = f.facid
                WHERE b.memid = 0 AND b.starttime LIKE '2012-09-14%' AND b.slots * f.guestcost > 30), 
                
         mb AS (SELECT b.bookid, b.memid, f.name, b.slots * f.membercost AS cost
                FROM (`Bookings` AS b
                LEFT JOIN Facilities AS f
                ON b.facid = f.facid), `Members`
                WHERE b.memid != 0 AND b.starttime LIKE '2012-09-14%' AND b.slots * f.membercost > 30)

    SELECT gb.bookid, gb.name, gb.cost, gb.member_name
    FROM gb
    UNION 
    SELECT mb.bookid, mb.name, mb.cost,  CONCAT(m.surname,' ',m.firstname)
    FROM mb INNER JOIN `Members` AS m USING (memid)
    ORDER BY cost DESC;     

/* 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! */

a10: WITH gr AS (SELECT Facilities.name,
                 SUM(Bookings.slots)*Facilities.guestcost AS revenue
                 FROM Bookings
                 LEFT JOIN Facilities
                 USING (facid)
                 WHERE Bookings.memid = 0
                 GROUP BY Facilities.name),
                       
           mr AS (SELECT Facilities.name, 
                  SUM(Bookings.slots)*Facilities.membercost AS revenue
                  FROM Bookings
                  LEFT JOIN Facilities
                  USING (facid)
                  WHERE Bookings.memid != 0
                  GROUP BY Facilities.name)
                       
     SELECT gr.name, gr.revenue + mr.revenue AS total_revenue
     FROM gr INNER JOIN mr USING (name)
     WHERE gr.revenue + mr.revenue < 1000
     ORDER BY total_revenue;

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

A11: SELECT m1.surname, m1.firstname, 
            m2.surname AS recommender_surname,
            m2.firstname AS recommender_firstname
     FROM Members AS m1
     INNER JOIN Members AS m2
     ON m1.recommendedby = m2.memid
     WHERE m1.recommendedby != 0
     ORDER BY m1.surname, m1.firstname;

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

A12: SELECT Bookings.memid, Facilities.name AS facility, SUM(Bookings.slots)*0.5 AS usage_hours
     FROM Facilities
     LEFT JOIN Bookings
     ON Facilities.facid = Bookings.facid
     WHERE Bookings.memid != 0
     GROUP BY Bookings.memid, facility;

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

A13: SELECT SUBSTRING(Bookings.starttime,6,2) AS month, 
            Facilities.name AS facility, SUM(Bookings.slots)*0.5 AS usage_hours
     FROM Facilities
     LEFT JOIN Bookings
     ON Facilities.facid = Bookings.facid
     WHERE Bookings.memid != 0
     GROUP BY month, facility;