Project

• Phase I

• Phase II

• Due Monday, April 15

• Groups

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Joins

• Retrieve data from two or more tables• Common Attributes PK and FK Other attributes• Based on Cartesian Product • Implementation Nested loops Optimization

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JoinsList 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, BookingWhere Guest.Guest_no = Booking.Guest_no;

-- Run in Oracle-- Duplicate records

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Joins

List all guests who have at least one booking.

Select Guest.* From Guest, BookingWhere 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.*

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Joins

List all guests who have at least one booking.Remove duplicate records!

Select Distinct Guest.* From Guest, BookingWhere Guest.Guest_no = Booking.Guest_no;

-- Could use Unique

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Joins: New Style

Select Distinct Guest.* From GuestJoin Booking on Guest.Guest_no = Booking.Guest_no;

Select Distinct Guest.* From Guest, BookingWhere Guest.Guest_no = Booking.Guest_no;

We should use the new style!

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Using Short Table Names

List all guests who have at least one booking.

Select Distinct G.* From Guest GJoin Booking B on G.Guest_no = B.Guest_no;

-- Cannot use the original table name any moreSelect Distinct Guest.* From Guest GJoin Booking B on Guest.Guest_no = B.Guest_no

ERROR at line 1:ORA-00904: "GUEST"."GUEST_NO": invalid identifier

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JoinsList all guests who have at least one bookingwith the details of each booking.

-- No need for Distinct-- All columns from both tables Select *From Guest GJoin Booking B on G.Guest_no = B.Guest_no;

Set linesize 100Col Guest_name Format a14Col address format a21

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JoinsWhat 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 GJoin Booking B;-- on G.Guest_no = B.Guest_no;

ERROR at line 3:ORA-00905: missing keyword

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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 GJoin Hotel H on G.Address = H.Address;

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Joins: Three Tables List Hotel name, guest name and date_fromfor 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 HJoin Booking B on H.Hotel_no = B.Hotel_no Join Guest G on G.Guest_no = B.Guest_noOrder By H.Hotel_No, G.Guest_no desc;-- H.Hotel_No, G.Guest_no not selected

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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 HJoin Booking B on H.Hotel_no = B.Hotel_no Join Guest G on G.Guest_no = B.Guest_noOrder By H.name, G.Guest_name, Date_from;-- Sort by name

Select H.name, G.Guest_name, Date_from From Hotel HJoin Booking B on H.Hotel_no = B.Hotel_no Join Guest G on G.Guest_no = B.Guest_noOrder By H.Hotel_No, G.Guest_no;-- Sort by Hotel_No 12

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 HJoin Booking B on H.Hotel_no = B.Hotel_no Group by H.Hotel_No;

-- Will it work?

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Joins with Group By

-- For each hotel, display Hotel name and number of -- bookings of the hotel.-- Tables: -- Hotel, Booking

-- Must also Group by nameSelect name, count(*) From Hotel HJoin Booking B on H.Hotel_no = B.Hotel_no Group by H.Hotel_No, name;

-- What if group by name?

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Joins with Group By -- For each hotel, display Hotel name and -- number of bookings of the hotel.

Select name, count(*) From Hotel HJoin 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?

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Current Year with Assumption-- For each hotel, display Hotel name and -- number of bookings of the hotel for this year.-- Assume no booking is longer than 1 year.

Select name, count(*) From Hotel HJoin 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;

-- could use where after join condition-- they are the same

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Current Year without Assumption-- For each hotel, display Hotel name and -- number of bookings of the hotel for this year.-- No assumptions.

Select name, count(*) From Hotel HJoin 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')) or ((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;

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Current Year without Assumption-- For each hotel, display Hotel name and -- number of bookings of the hotel for this year.-- No assumptions.

Select name, count(*)From Hotel HJoin Booking B on H.Hotel_no = B.Hotel_no and 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’ instead of ‘yyyy’

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Group By and Having -- For each hotel with at least 3 bookings this -- year, display Hotel name and the number of -- bookings of the hotel this year.-- With assumption

Select name, count(*)From Hotel HJoin Booking B on H.Hotel_no = B.Hotel_no and (to_char(sysDate, 'yyyy') = to_char(date_to, 'yyyy') or to_char(sysDate, 'yyyy') = to_char(date_from,'yyyy'))Group by H.Hotel_No, nameHaving Count(*) >= 3;

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Group By and Having

-- For each hotel with more than 5 bookings, -- display Hotel name and the number of -- bookings of the hotel.

Select name, count(*)From Hotel HJoin Booking B on H.Hotel_no = B.Hotel_no Group by H.Hotel_No, nameHaving Count(*) > 5;

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Schedule

• Assignment 9: Join

Due Wednesday, April 24

• Quiz 3

Friday, April 19

No Join

Assignment 7 & 8

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