1. CS 3630 Database Design and Implementation

2. Joins Retrieve data from two or more tables Join Conditions
PK and FK (Natural Join)
Other attributes (Theta Join)
Based on Cartesian Product
Implementation
Nested loops
Optimization

3. Tables Schemas
Assignment 7

4. Joins List all guests who have at least one booking.
(We keep all guests even they don’t have any bookings.)
-- Which table(s)?
-- Guest (Guest_no, Guest_Name, Address)
PK: Guest_no
-- Booking (HoteL_no, Guest_no, ...)
PK: Hotel_no, Guest_No, Date_From
FK: Guest_no references Guest
Select Guest.*
From Guest, Booking
Where Guest.Guest_no = Booking.Guest_no;

5. Joins: New Style Select Guest.* From Guest, Booking
Where Guest.Guest_no = Booking.Guest_no;
From Guest
Join Booking
on Guest.Guest_no = Booking.Guest_no;
We use the new style!
-- Run in Oracle
Duplicate records

6. Joins List all guests who have at least one booking. Select Guest.*
From Guest
Join Booking
on Guest.Guest_no = Booking.Guest_no;
Why duplicate records?
For every g in Guest
For every b in Booking
If g.Guest_no = b.Guest_no Then
Select g.*

7. Always remove duplicate records!
Joins
Always remove duplicate records!
Select Distinct Guest.*
From Guest
Join Booking
on Guest.Guest_no = Booking.Guest_no;
Select Unique Guest.*

8. Using Short Table Names
List all guests who have at least one booking.
Select Distinct G.*
From Guest G
Join Booking B
on G.Guest_no = B.Guest_no;
Pause
-- Cannot use the original table name any more
Select Distinct Guest.*
on Guest.Guest_no = B.Guest_no;
ERROR at line 1:
ORA-00904: "GUEST"."GUEST_NO": invalid identifier

9. Joins List all guests who have at least one booking
with the details of each booking.
-- No need for Distinct
-- All columns from both tables
Select *
From Guest G
Join Booking B
on G.Guest_no = B.Guest_no;
Set linesize 100
Col Guest_name Format a14 Heading “Guest Name”
Col address format a21

10. Joins What if no common attribute is specified? Select G.*
From Guest G, Booking B;
-- Where G.Guest_no = B.Guest_no;
-- Cartesian product!
Select Distinct G.*
From Guest G
Join Booking B;
-- on G.Guest_no = B.Guest_no;
ERROR at line 3:
ORA-00905: missing keyword

11. Join not on PK/FK
List all guests who are in a city where there is also a hotel.
Assuming Address is City.
Select Distinct G.*
From Guest G
Join Hotel H
on G.Address = H.Address;

12. Joins: Three Tables List Hotel name, guest name and date_from
for all bookings, sorted by hotel_no in ascending order
and then by guest_no in descending order.
Which tables?
Hotel Name: Hotel
Guest Name: Guest
Date_From : Booking
Select H.name, G.Guest_name, Date_from
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Join Guest G
on G.Guest_no = B.Guest_no
Order By H.Hotel_No, G.Guest_no desc;
-- H.Hotel_No, G.Guest_no not selected

13. Joins: Three Tables -- List Hotel name, guest name and date_from
-- for all bookings.
-- Sort the result
Select H.name, G.Guest_name, Date_from
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Join Guest G
on G.Guest_no = B.Guest_no
Order By H.name, G.Guest_name, Date_from;
-- Sort by name, Guest_name, Date_from;
Order By H.Hotel_No, G.Guest_no;
-- Sort by Hotel_No, Guest_no;

14. Joins with Group By
-- For each hotel, display Hotel name and number of
-- bookings of the hotel.
-- Tables:
-- Hotel, Booking
Select name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Group by H.Hotel_No;
-- Will it work?

15. Joins with Group By
-- For each hotel, display Hotel name and number of
-- bookings of the hotel.
-- Tables:
-- Hotel, Booking
-- Must also Group by name
Select name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Group by H.Hotel_No, name;
-- What if group by name only?

16. Joins with Group By -- For each hotel, display Hotel name and
-- number of bookings of the hotel.
Select name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Group by H.Hotel_No, name;
-- Missing hotels without bookings
-- How to display a zero for such hotels?
Select *
From hotel;

17. Current Year with Assumption
-- For each hotel, display Hotel number with hotel name
-- and number of bookings of the hotel for this year.
-- Assume no booking is longer than 1 year.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
and ((to_char(Date_From, 'yyyy') = to_char(SysDate, 'yyyy')) or
(to_char(Date_To, 'yyyy') = to_char(SysDate, 'yyyy')))
Group by H.Hotel_No, name;

18. Be Careful about And/Or
-- For each hotel, display Hotel number with hotel name
-- and number of bookings of the hotel for this year.
-- Assume no booking is longer than 1 year.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
and to_char(Date_From, 'yyyy') = to_char(SysDate, 'yyyy') or
to_char(Date_To, 'yyyy') = to_char(SysDate, 'yyyy')
Group by H.Hotel_No, name;
-- and ((to_char(Date_From, 'yyyy') = to_char(SysDate, 'yyyy')) or
(to_char(Date_To, 'yyyy') = to_char(SysDate, 'yyyy')))

19. Join and Where -- For each hotel, display Hotel number with hotel name
-- and number of bookings of the hotel for this year.
-- Assume no booking is longer than 1 year.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Where to_char(Date_From, 'yyyy') = to_char(SysDate, 'yyyy') or
to_char(Date_To, 'yyyy') = to_char(SysDate, 'yyyy')
Group by H.Hotel_No, name;
-- could use where after join condition
-- Same result for the query
-- But one is Join condition
the other is selection condition
-- We should do it this way!

20. Current Year without Assumption
-- For each hotel, display Hotel number with hotel name
-- and number of bookings of the hotel for this year.
-- No assumptions: booking could be longer than one year.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Where to_char(Date_From, 'yyyy') = to_char(SysDate, 'yyyy') or
to_char(Date_To, 'yyyy') = to_char(SysDate, 'yyyy')
(to_char(Date_From, 'yyyy') < to_char(SysDate, 'yyyy')
and
to_char(Date_To, 'yyyy') > to_char(SysDate, 'yyyy'))
Group by H.Hotel_No, name;

21. Current Year without Assumption
-- For each hotel, display Hotel number with hotel name
-- and number of bookings of the hotel for this year.
-- No assumptions.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Where to_char(Date_From, 'yyyy') <= to_char(SysDate, 'yyyy')
and to_char(Date_To, 'yyyy') >= to_char(SysDate, 'yyyy')
Group by H.Hotel_No, name;
-- How to make it work for the current month?
Use ‘yyyy mm’
Not ‘yyyy’
Not ‘mm yyyy’

22. Group By and Having -- For each hotel with at least 3 bookings this
-- year, display Hotel number with the name and the number
-- of bookings of the hotel this year.
-- With assumption
Select H.Hotel_no, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Where to_char(sysDate, 'yyyy') = to_char(date_to, 'yyyy') or
to_char(sysDate, 'yyyy') = to_char(date_from,'yyyy')
Group by H.Hotel_No, name
Having Count(*) >= 3;

23. Group By and Having -- For each hotel with more than 5 bookings,
-- display Hotel number with name and the number
-- of bookings of the hotel.
Select H.Hotel_No, name, count(*)
From Hotel H
Join Booking B
on H.Hotel_no = B.Hotel_no
Group by H.Hotel_No, name
Having Count(*) > 5;

24. Joins Due Tuesday, April 25
Assignment 9
Joins Due Tuesday, April 25

25. Friday, April 14? Based on Assignment 8
Quiz 3
Friday, April 14?
Based on Assignment 8