Work

Jessica_Williams_SQLTasks Tier 2.sql

@@ -0,0 +1,201 @@
+/* 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. */
+
+Answer:
+SELECT name, membercost AS is_memcost
+FROM Facilities
+WHERE membercost>0;
+
+/* Q2: How many facilities do not charge a fee to members? */
+
+Answer:
+SELECT COUNT(membercost) AS no_memcost
+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. */
+
+Answer:
+SELECT name, membercost AS less_memcost
+FROM Facilities
+WHERE membercost>0 AND membercost<(.20 * 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. */
+
+Answer:
+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. */
+
+Answer:
+SELECT name, monthlymaintenance, 
+CASE WHEN monthlymaintenance<= 100 THEN 'cheap' 
+     WHEN monthlymaintenance>100 THEN 'expensive' END 
+     AS price_category 
+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. */
+
+Answer:
+SELECT surname, firstname, joindate
+FROM Members
+WHERE joindate=(
+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. */
+
+Answer:
+SELECT DISTINCT f.name AS Facility,memid, CONCAT(firstName , ' ' , surname) as full_name
+FROM Bookings AS b
+INNER JOIN Facilities AS f USING(facid)
+INNER JOIN Members AS m USING (memid)
+WHERE facid IN
+	(SELECT facid
+     FROM Bookings 
+     WHERE facid=0 OR facid=1)
+ORDER BY full_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. */
+
+Answer:
+SELECT f.name AS Facility,
+    concat(m.firstname, ' ', m.surname) AS full_name,
+    IF(b.memid = 0, b.slots * f.guestcost, b.slots * f.membercost) AS Cost
+FROM Bookings as b
+INNER JOIN Facilities AS f USING(facid)
+INNER JOIN Members AS m USING (memid)
+WHERE starttime LIKE '2012-09-14%' AND IF(b.memid = 0, b.slots * f.guestcost, b.slots * f.membercost > 30)
+ORDER BY Cost DESC;
+
+/* Q9: This time, produce the same result as in Q8, but using a subquery. */
+
+Answer:
+SELECT f.name AS Facility,
+    concat(m.firstname, ' ', m.surname) AS full_name,
+    IF(b.memid = 0, b.slots * f.guestcost, b.slots * f.membercost) AS Cost
+FROM Bookings as b
+INNER JOIN Facilities AS f USING(facid)
+INNER JOIN Members AS m USING (memid)
+WHERE starttime LIKE '2012-09-14%'
+	AND bookid IN
+    (SELECT
+    	bookid
+        FROM Bookings
+        WHERE IF(b.memid = 0, b.slots * f.guestcost > 30, b.slots * f.membercost > 30))
+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! */
+
+Answer:
+SELECT * 
+FROM (
+SELECT lrev.facility, SUM( lrev.cost ) AS total_revenue
+FROM (
+SELECT Facilities.name AS facility, 
+CASE WHEN Bookings.memid =0
+THEN Facilities.guestcost * Bookings.slots
+ELSE Facilities.membercost * Bookings.slots
+END AS cost
+FROM Bookings
+INNER JOIN Facilities ON Bookings.facid = Facilities.facid
+INNER JOIN Members ON Bookings.memid = Members.memid
+)lrev
+GROUP BY lrev.facility
+)lrev2
+WHERE lrev2.total_revenue <1000;
+
+/* Q11: Produce a report of members and who recommended them in alphabetic surname,firstname order */
+
+Answer:
+SELECT m.surname, m.firstname, m.recommendedby AS recomender_id, r.surname AS recomender_surname, r.firstname AS recomender_firstname
+FROM Members AS m
+LEFT JOIN Members AS r ON m.recommendedby = r.memid
+WHERE m.recommendedby != 0
+ORDER BY r.surname, r.firstname;
+
+
+/* Q12: Find the facilities with their usage by member, but not guests */
+
+Answer:
+SELECT b.facid, COUNT( b.memid ) AS mem_usage, f.name
+FROM (
+SELECT facid, memid
+FROM Bookings
+WHERE memid !=0
+) AS b
+LEFT JOIN Facilities AS f ON b.facid = f.facid
+GROUP BY b.facid;
+
+/* Q13: Find the facilities usage by month, but not guests */
+
+Answer:
+SELECT b.months, COUNT( b.memid ) AS mem_usage
+FROM (
+SELECT MONTH( starttime ) AS months, memid
+FROM Bookings
+WHERE memid !=0
+) AS b
+GROUP BY b.months;