Question

Given the following schema:

Furniture f id category style price int char(20) char(15) dec(5,2) primary key not null values restricted to 'modern', 'rustic', traditional' not null [Example: sofa, bed,...] Per week rental cost] Customers c id name phone email int char(10) char(13) char(15) primary key not null Checkout c id f id ck out_date date ret date foreign key, references c id of customers foreign key, references f id of furniture not null int int date [c id and f id together form the primary key]

I would like to know the SQL queries to:

1. Total up the number of checkouts made on each month of the year. Then print the month (spelled out completely) and the total number of corresponding checkouts next to it. That is, print "January" and total number of checkouts made in January, and so on.

2. Show the furniture styles that are checked-out on Sundays only. That is, none of the furniture in that style must have been checked out on any other days.

3. Find names of people who have checked out any piece of furniture within the last 15 days (not counting the day the query is being run) but have not returned that yet. Print their names and the string "* To be contacted" next to each name.

4. For all the furniture that were checked out in October 2018 and returned in November 2018, print their F_ID, category and style. However, while printing style, print 'M' for 'modern', 'R' for 'rustic', and 'T' for 'traditional'.

5. Assume that the price attribute of Furniture table indicates weekly (7 days) rental cost. Write a view to calculate the total money earned from all "rustic" style furniture, which has been rented and returned (i.e., do not take into consideration the ones that are yet to be returned). Print the amount.

Answer 1

/*1. Total up the number of checkouts made on each month of the year.

Then print the month (spelled out completely) and the total number of corresponding checkouts next to it.

That is, print "January" and total number of checkouts made in January, and so on.*/

--sql serve

SELECT FORMAT(ck_out_date, 'MMMM') AS Month,COUNT(c_id) AS TotalNumberOfCheckOuts FROM Check_out GROUP BY FORMAT(ck_out_date, 'MMMM');

--Mysql

SELECT MONTHNAME(ck_out_date) AS Month,COUNT(c_id) AS TotalNumberOfCheckOuts FROM Check_out GROUP BY MONTHNAME(ck_out_date);

/*2. Show the furniture styles that are checked-out on Sundays only.

That is, none of the furniture in that style must have been checked out on any other days.*/

--Msql

SELECT f.style, FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE DAYNAME(c.ck_out_date)='SUNDAY';

--SQLSERVER

SELECT f.style, FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE datename(dw,c.ck_out_date)='SUNDAY';

/*3. Find names of people who have checked out any piece of furniture within the last 15 days

(not counting the day the query is being run) but have not returned that yet.

Print their names and the string "* To be contacted" next to each name.*/

--SQL SERVER

SELECT c.name FROM customers c INNER JOIN Check_out co ON c.c_id=co.c_id

WHERE convert(date,co.ck_out_date)>=convert(date,GETDATE()-16)

AND co.ck_out_date is not null

--Mysql

SELECT c.name FROM customers c INNER JOIN Check_out co ON c.c_id=co.c_id

WHERE convert(date,co.ck_out_date)>=convert(subdate(NOW(), 16),date)

AND co.ck_out_date is not null

/*4. For all the furniture that were checked out in October 2018 and returned in November 2018,

print their F_ID, category and style. However, while printing style,

print 'M' for 'modern', 'R' for 'rustic', and 'T' for 'traditional'.*/

--MYSQL

SELECT c.F_ID, (CASE WHEN style = 'modern' THEN 'M'

WHEN style = 'rustic' THEN 'R'

WHEN style = 'traditional' THEN 'T'

END )AS Style

FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE (MONTHNAME(c.ck_out_date) ='October' AND YEAR(c.ck_out_date)=2018)

AND (MONTHNAME(c.ret_date) ='November' AND YEAR(c.ret_date)=2018)

--SQL SERVER

SELECT c.F_ID, (CASE WHEN style = 'modern' THEN 'M'

WHEN style = 'rustic' THEN 'R'

WHEN style = 'traditional' THEN 'T'

END )AS Style

FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE (FORMAT(c.ck_out_date, 'MMMM') ='October' AND YEAR(c.ck_out_date)=2018)

AND (FORMAT(c.ret_date, 'MMMM') ='November' AND YEAR(ret_date)=2018)

/*5. Assume that the price attribute of Furniture table indicates weekly (7 days) rental cost.

Write a view to calculate the total money earned from all "rustic" style furniture,

which has been rented and returned

(i.e., do not take into consideration the ones that are yet to be returned). Print the amount.*/

-- Mysql

SELECT SUM(f.price * (timestampdiff(DAY,c.ck_out_date,c.ret_date)/7))

FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE f.style='rustic' and c.ret_date is not null

GROUP BY f.style

--Sql server

SELECT SUM(f.price * (DATEDIFF(DAY,c.ck_out_date,c.ret_date)/7))

FROM Furniture f INNER JOIN Check_out c on f.f_id=c.f_id

WHERE f.style='rustic' and c.ret_date is not null

GROUP BY f.style