1. CS 440 Database Management Systems
Lecture 3: Review of Relational Model and SQL

2. Announcements
You may find the chapters related to each lecture in the schedule page of the course Web site.
Check out the textbook website for practice problems with solutions: pages.cs.wisc.edu/~dbbook/

3. Example schema Beers(name, manf) Bars(name, addr, license)
Drinkers( name, addr, phone)
Likes(drinker, beer)
Sells(bar, beer, price)
Frequents(drinker, bar)

4. Aggregation functions
Compute some value based on the values of an attribute.
Example functions: Count, Sum, Avg, Min, Max
Each RDBMS may define additional functions.
Example: Using Bars(name, addr, license), find the number of bars.
Select Count(name)
From Bars;

5. Aggregation functions
Using Distinct, aggregation functions ignore duplicates.
Example: Using Likes(drinker, beer), find the number of drinkers who like Bud Lite.
Select Count( Distinct drinker)
From Likes
Where beer =‘Bud Lite’;

6. Aggregation functions
Generally, aggregation functions do not consider NULL values.
Select Count(price)
From Sells
Where bar=‘Joe Bar’;
Select Count(beer)
Select Count(*)
The number of priced beers sold by Joe Bar.
The number of beers sold by Joe Bar.
The number of beers sold by Joe Bar.

7. Aggregation functions over groups
We want to aggregate values for groups of tuples.
Example: Using Sells(bar, beer, price) find the minimum price of each beer.
Group tuples in Sells based on beer.
Compute Min over the prices in each group of tuples.

8. Group by
Example: Using Sells(bar, beer, price) find the minimum price of each beer.
Select beer, Min(price) As minprice
From Sells
Group By beer;
optional

9. Group by Select beer, Min(price),bar
You may use multiple attributes for grouping.
The attributes in the Select clause are either aggregated values or attributes in the Group By clause.
Select beer, Min(price),bar
From Sells
Group By beer;
Exceptions in some RDBMS, e.g., MySQL 5.7.
Generally, Group By does not sort the groups.
There are exceptions, e.g., older versions of MySQL, but do not trust them!
error

10. Grouping attributes from different relations.
Example: Using Likes(drinker, beer) and Sells(bar, beer, price), for each drinker find the minimum price of every beer he/she likes.
Select drinker, beer, Min(price) As minprice
From Likes, Sells
Where Likes.beer = Sells.beer
Group By drinker, beer;

11. Filtering groups
We may filter out some groups using their attributes’ values.
Select beer, Min(price) As minprice
From Sells
Where bar=‘Red Lion’ or bar=‘Big Horse’
Group By beer;

12. Filtering groups based on aggregated values
Example: Using Sells(bar, beer, price), find the minimum price of each beer whose maximum price is less than 11.
Select beer, Min(price) As minprice
From Sells
Where Max(price) < 11
Group By beer
error

13. Having clause
We use Having clauses to filter out groups based on their aggregated values.
Select beer, Min(price) As minprice
From Sells
Group By beer
Having Max(price) < 11

14. Having clause
We may use aggregated values over attributes other than the ones in the Group By clause.
Example: Using Sells(bar, beer, price), find the minimum price of each beer sold in more than three bars.
Select beer, Min(price) As minprice
From Sells
Group By beer
Having Count(bar) > 3

15. Having clause may act as a Where clause
Example: Using Sells(bar, beer, price), find the minimum price of Bud or beers whose maximum price is less than 11.
Select beer, Min(price)
From Sells
Group By beer
Having (Max(price) < 11) Or (beer=‘Bud’)
It works only for the attributes in the Group By clause.
Having (Max(price) < 11) Or (bar=‘Red Lion’)
error

16. Sorting the output
Example: Using Sells(bar, beer, price) find the minimum price of each beer whose maximum price is at least 15 and sort the results according to beers’ names.
Select beer, Min(price) As minprice
From Sells
Group By beer
Having Max(price) >= 15
Order By beer;

17. Sorting the output One may use Desc to change the sort order.
Previous example in descending order of beers’ names:
Select beer, Min(price) As minprice
From Sells
Group By beer
Having Max(price) >= 15
Order By beer Desc;
You may use Order By without Group By and Having.
Example: Using Sells(bar, beer, price), provide a list of beer prices sorted by bars’ and beers’ names.
Select bar, beer, price
Order By bar, beer;

18. Review problems: people betting on OSU football games
Out(game, outcome) Bets(who, outcome, game, amt)
Some games have not been played yet, e.g., Arizona.
game
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120

19. Problem 1 List the completed games that nobody bet on. game USC W UCLA
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
game

20. Problem 1 From Out) Except From Bets) (Select Game
List the completed games that nobody bet on.
(Select Game
From Out)
Except
From Bets)

21. Problem 2 Who bet the most money on a single game? game USC W UCLA L
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
who
amt
Kevin
210

22. Problem 2 Who bet the most money on a single game? Select Who, Amt
From Bets
Where Amt >= All
(Select Amt
From Bets)

23. Problem 3 List the games that all bettors agree on. game USC W UCLA L
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
game
Stanford
Arizona UO

24. Problem 3 List the games that all bettors agree on. (Select game
From Bets)
Except
(Select Bets1.game
From Bets Bets1, Bets Bets2
Where (Bets1.game = Bets2.game)
And (Bets1.outcome <> Bets2.outcome))

25. Problem 4
For each game, the number of people betting on OSU to win and the number betting on OSU to lose.
game
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
game
outcome
num
Stanford W 1 UO L
UCLA
USC
Arizona

26. Problem 4
For each game, the number of people betting on OSU to win and the number betting on OSU to lose.
Select game, outcome, Count(who) As num
From Bets
Group By game, outcome

27. Problem 5
Find the people who have made two or more bets on OSU to lose.
game
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
who
Kevin

28. Problem 5
Find the people who have made two or more bets on OSU to lose.
Select who
From Bets
Where outcome = ‘L’
Group By who
Having Count(outcome) >= 2

29. Problem 6 Who bet the most money overall? game USC W UCLA L Stanford
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
who
sumAmt
John
450

30. Problem 6 Who bet the most money overall?
Select who, Sum(amt) As sumAmt
From Bets
Group By who
Having Sum(amt) >= ALL
(Select Sum(amt)
Group By who)

31. Problem 7 Who has bet on every game? game USC W UCLA L Stanford who
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
Who

32. Problem 7 Who has bet on every game? Create View AllGames As
(Select game From Out) Union
(Select game From Bets)
Select who
From Bets
Group By who
Having Count(Distinct game) =
(Select Count(Distinct game)
From AllGames)

33. Problem 8
What games have won the most money for the people who bet on OSU to win?
game
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
game
USC

34. Problem 8
What games have won the most money for the people who bet on OSU to win?
Create View Success-Win As
(Select Bets.game, Sum(Bets.amt) As SumAmt
From Bets, Out
Where out.game = Bets.game And
Bets.outcome = ‘W’ And Out.outcome = ‘W’
Group By Game)
Select Distinct game
From Success-Win
Where SumAmt >= All
(Select SumAmt
From Success-Win)

35. Problem 9 List the people who won some money so far. game USC W UCLA L
outcome
USC W
UCLA L
Stanford
who
outcome
game
amt
John W
USC
200
UCLA
100 L
Arizona
150
Kevin UO
210 50
Stanford
120
who
John
Kevin

36. Problem 9 List the people who won some money so far.
Create View Success As
(Select who, Sum(amt) As pAmt
From Bets, Out
Where Out.game = Bets.game And Bets.outcome = Out.outcome
Group By who)
Create View Failure As
(Select who, Sum(amt) As nAmt
Where Out.game = Bets.game And Bets.outcome <> Out.outcome

37. Problem 9 Union List the people who won some money so far.
(Select Success.who
From Success, Failure
Where Success.who = Failure.who And
Success.pAmt > Failure.nAmt )
Union
(Select who
From Success
Where who not in
From Failure) )

38. Data manipulation: insertion
Inserting new tuple(s) into a relation.
Insert into R(A1,…,An) Values (a1,…,an);
Example: insert tuple (‘Big Horse’,’Bud’,3) in Sells.
Insert into Sells(bar, beer, price)
Values (‘Big Horse’,‘Bud’,3);
The input of Insert can be the result of another query.
Insert into Beers(name)
Select beer From Sells;

39. Data manipulation: deletion
Removing tuple(s) from a relation.
Delete From R
Where R.A = ‘a’;
Remove all tuples from relation R.
Delete From R;
Example: John does not go to Old Horse anymore.
Delete From Frequents
Where drinker = ‘John’ and
bar = ‘Old Horse’;

40. Data manipulation: update
Updating the values of given attributes in certain tuples of a relation.
Update R
Set R.A = a
Where R.B = b;
Old Horse has changed the price of Bud to 3.
Update Sells
Set price = 3
Where bar = ‘Old Horse’ And
beer = ‘Bud’;

41. Is SQL enough?
Using IsParent(parent,child), find all descendants of a given person.
It is proved that it is not possible to write this query in SQL that we have discussed so far.
It (relational algebra/ calculus) express first order logic. We need more expressive to describe recursion.
There are generally three methods to make SQL stronger to express these queries.