1520094343_sql_project.sql

/* Welcome to the SQL mini project. For this project, you will use
Springboard' online SQL platform, which you can log into through the
following link:

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.

Note that, if you need to, you can also download these tables locally.

In the mini project, 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. */



/* Q1: Some of the facilities charge a fee to members, but some do not.
Please list the names of the facilities that do. */

SELECT name
  FROM Facilities
WHERE membercost != 0.0


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

SELECT COUNT(name) AS count_free_facilities
  FROM Facilities
WHERE membercost = 0.0


/* Q3: How can you produce 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 != 0.0
    AND membercost / monthlymaintenance < 0.20


/* Q4: How can you retrieve the details of facilities with ID 1 and 5?
Write the query without using the OR operator. */

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


/* Q5: How can you 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 'cheap'
           ELSE 'expensive' END AS monthlymaintenance_level
  FROM Facilities


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

SELECT firstname, surname
FROM Members
WHERE memid =37


/* Q7: How can you 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( firstname, ' ', surname ) AS name,
           CASE WHEN facid =0 THEN 'Tennis Court 1'
           ELSE 'Tennis Court 2' END AS tennis_court_#
  FROM Members
    LEFT JOIN Bookings 
      ON Members.memid = Bookings.memid
 WHERE facid IN ( 0, 1 )
 GROUP BY name
 ORDER BY name


/* Q8: How can you 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 Facilities.name AS facility_name, 
           CONCAT( Members.firstname, ' ', Members.surname ) AS member_name,
           CASE WHEN Members.memid =0 THEN Facilities.guestcost * Bookings.slots
           ELSE Facilities.membercost * Bookings.slots END AS cost
  FROM Bookings
   LEFT JOIN Members 
     ON Bookings.memid = Members.memid
   LEFT JOIN Facilities 
     ON Bookings.facid = Facilities.facid
WHERE Bookings.starttime LIKE '2012-09-14%'
HAVING cost >30
ORDER BY cost DESC


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

SELECT (SELECT name AS facility_name
              FROM Facilities
            WHERE Bookings.facid = facid),
           (SELECT CONCAT( firstname, ' ', surname ) AS member_name
              FROM Members
            WHERE Bookings.memid = memid),
            CASE WHEN memid =0 THEN slots * (SELECT guestcost
                                                               FROM Facilities
                                                             WHERE Bookings.facid = facid )
            ELSE slots * (SELECT membercost
                              FROM Facilities
                              WHERE Bookings.facid = facid )END AS cost
 FROM Bookings
WHERE starttime LIKE '2012-09-14%'
HAVING cost >30
ORDER BY cost DESC


/* 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 Facilities.name AS facility_name, 
           SUM(CASE WHEN Members.memid =0 THEN Facilities.guestcost * Bookings.slots
                  ELSE Facilities.membercost * Bookings.slots END ) AS total_revenue
  FROM Bookings
   LEFT JOIN Members ON Bookings.memid = Members.memid
   LEFT JOIN Facilities ON Bookings.facid = Facilities.facid
GROUP BY facility_name
HAVING total_revenue <1000
ORDER BY total_revenue