From E/R Diagrams to Relations

1. From E/R Diagrams to Relations

2. The Relational Data Model Database Model (E/R) Relational Schema Physical storage Complex file organization and index structures. Diagrams (E/R) Tables: row names: attributes rows: tuples

3. Terminology Attribute names Title Year Length FilmType Star Wars 1997 124 color Mighty Ducks 1991 104 color Wayne’s World 1992 95 color . . .. . .. . . . . . components of tuples tuples

4. More Terminology Every attribute has an atomic type. Relation Schema: relation name + attribute names + attribute types Relation instance: a set of tuples. Only one copy of any tuple! Database Schema: a set of relation schemas. Database instance: a relation instance for every relation in the schema.

5. From E/R Diagrams to Relations • Entity sets become relations with the same set of attributes. • Relationships become relations whose attributes are only: • The keys of the connected entity sets. • Attributes of the relationship itself.

6. year title Movies filmType length Example: Entity Sets to Relations Relation schema: Movies(title, year, length, filmtype ) A relation instance:

7. E/R Relationships to Relations (Renaming Attributes) • We rename attributes when: one entity set is involved several times in a relationship - or - the same attribute name appears in the keys of different entity sets participating in a relationship, - or - we like to avoid confusion and to be clearer in meaning

8. name name addr manf Drinkers Likes Beers husband 2 1 Favorite Buddies Likes(drinker, beer) Favorite(drinker, beer) wife Married Buddies(name1, name2) Married(husband, wife) Example (with Renaming)

9. Example (with Renaming) • The relationship Stars-Inbetween entity sets Moviesand Starsis represented by a relation with schema: • Stars-In(title, year, starName) • A sample instance is: • title year starName • Star Wars 1977 Carrie Fisher • Star Wars 1977 Mark Hamill • Star Wars 1977 Harrison Ford • Mighty Ducks 1991 Emilio Estevez • Wayne’s World 1992 Dana Carvey • Wayne’s World 1992 Mike Meyers We rename here for clarity.

10. Combining Relations • What about combining: • Drinkers(name, addr) and Favorite(drinker, beer) to make Drinkers(name, addr, favBeer). • OK to combine the relation for an entity-set E with the relation R for a many-one relationship fromE to another entity set. • Caveat • However, when there are many drinkers that don't have a favorite beer, then we don’t combine the Drinkers and Favorite relations.

11. name addr beer Sally 123 Maple Bud Sally 123 Maple Miller Redundancy Risk with Many-Many Relationships • Combining Drinkers with Likes would be a mistake. Why? • It leads to redundancy, as:

12. Handling Weak Entity Sets • Relation for a weak entity set must include attributes for its complete key (including those belonging to other entity sets), as well as its own, nonkey attributes. • A supporting (double-diamond) relationship is redundant and yields no relation.

13. Must be the same At becomes part of Logins Example name name Logins At Hosts location billTo Hosts(hostName, location) Logins(loginName, hostName, billTo) At(loginName, hostName, hostName2)

14. length title year filmType to Stars Movies Voices isa isa weapon Cartoons Murder- Mysteries Our Movie Example (with ISA)

15. Subclass Structures to Relations Two different approaches • OO Approach • An object belongs to exactlyone class. • An object inherits properties from all its super-classes but it is not a member of them. • E/R Approach • An “object” can be represented by entities belonging to several entity sets that are related by isarelationships. • Thus, the linked entities together represent the object and give that object its properties (attributes and relationships).

16. OO approach • Every subclass has its own relation. • All the properties of that subclass, including all its inherited properties, are represented in this relation. • Example: For our examplethe relational database schema would be: • Movies( title, year, length, filmType ) • Cartoons( title, year, length, filmType ) • MurderMysteries( title, year, length, filmType, weapon) • Cartoon-MurderMysteries( title, year, length, filmType, weapon)

17. Can we merge Cartoons with Movies? • If we do, we lose information about which moves are cartoons. • For the relationship Voices, we create: • Voices( title, year, starName ) • Is it necessary to create two relations one connecting cartoons with stars, and one connecting cartoon-murder-mysteries with stars? • Not, really. We can use the same relation (table).

18. E/R Approach • We will have the following relations: • Movies(title, year, length, filmType). • MurderMystery(title, year, weapon). • Cartoons(title, year). • Voices(title, year, name).

19. E/R approach (II) • Remark: • There is no relation corresponding to the class Cartoon-MurderMystery. • For a movie that is both, we obtain: • its voices from the Voices relation, • its weapon from the MurderMystery relation, • and all other information from the Movies relation. • The relation Cartoonshas a schema that is a subset of the schema for the relation Voices.Should we eliminate the relation Cartoons? • However there may be silent cartoons in our database. Those cartoons would have no voices and we would lose them.

20. Comparison of Approaches OO translation drawback: • Too many tables! Why? • In the OO approach if we have a root and n children we need 2^n different tables!!! E/R translation drawback: • We may have to look in several relations to gather information about a single object. • For example, if we want the length and weapon used for a murder mystery film, we have to look at Movies and MurderMysteries relations.

21. Comparison of Approaches (Continued) OO translation advantage: • The OO translation keeps all properties of an object together in one relation. E/R translation advantage: • The E/R translation allows us to find in one relation tuples from all classes in the hierarchy.

22. Examples • What movies of 1999 were longer than 150 minutes? • Can be answered directly in the E/R approach. • In the OO approach we have to examine all the relations. • What weapons were used in cartoons of over 150 minutes in length? • More difficult in the E/R approach. • We should access Movies to find those of over 150 mins. • Then, we have to access Cartoons to see if they are cartoons. • Then we should access MurderMysteries to find the weapon. • In OO approach we need only access the Cartoon-MyrderMysteries table.

23. Null Values to Combine Relations • If we are allowed to use NULL in tuples, we can handle a hierarchy of classes with a single relation. • This relation has attributes for all the properties possessed by objects in any of the classes of the hierarchy. • An object is represented by a single tuple. This tuple has NULL in each attribute corresponding to a property that does not belong to the object’s class. • For the Movie hierarchy, we would create a single relation: • Movie(title, year, length, filmType, studioName, starName, voice, weapon) • A movie like “Who Framed Roger Rabbit?”, being both a cartoon and a murdermystery, would be represented by a tuple that had no NULL’s. • The Little Mermaid, being a cartoon but not a murder-mystery, would have NULL in the weapon component. • This approach allows us to find all the information about an object in one relation. Drawback?

24. Create Table CREATE TABLE Movies ( title CHAR(40), year INT, length INT, type CHAR(2) ); CREATE TABLE Studios ( name CHAR(20), address VARCHAR(255), noOfEmp INT ); CREATE TABLE Stars ( name CHAR(30), address VARCHAR(255), gender CHAR(1) ); • CHAR(n) allocates a fixed space, and if the string that we store is shorter than n, then it is padded with blanks. • Differently, VARCHAR(n) denotes a string of up to n characters. • ORACLE uses also VARCHAR2(n), which is semantically the same as VARCHAR(n), which is deprecated. • VARCHAR(n) or VARCHAR2(n) allow for compression in order to save space. • Use CHAR(n) for frequently used fields, and use VARCHAR(n) otherwise.

25. Insert, Update, Delete, INSERT INTO Movies(title, year, length, type) VALUES('Godzilla', 1998, 120, 'C'); INSERT INTO Movies VALUES('Godzilla', 1998, 120, 'C'); UPDATE Movies SET title = 'Godzilla 2' WHERE title = 'Godzilla' AND year=1998; DELETE FROM Movies WHERE title='Godzilla 2';

26. Declaring primary keys DROP TABLE Movies; CREATE TABLE Movies ( title CHAR(40) PRIMARY KEY, year INT, length INT, type CHAR(2) ); DROP TABLE Movies; CREATE TABLE Movies ( title CHAR(40), year INT, length INT, type CHAR(2), PRIMARY KEY (title, year) );

27. Altering, Dropping ALTER TABLE Stars ADD phone CHAR(16); ALTER TABLE Stars DROP COLUMN phone; ALTER TABLE Stars MODIFY phone CHAR(26); DROP TABLE Stars; DROP TABLE Movies; DROP TABLE Studios;

28. Declaring foreign keys CREATE TABLE Studios ( name CHAR(20) PRIMARY KEY, address VARCHAR(255), noOfEmp INT ); CREATE TABLE Movies ( title CHAR(40) PRIMARY KEY, year INT, length INT, type CHAR(2), studioName CHAR(20), FOREIGN KEY (studioName) REFERENCES Studios(name) );