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Today’s Class

IS C332: Database Systems and Applications. Today’s Class. Data Models Relational Model. Keys. Let K  R (I.e., K is a set of attributes which is a subset of the schema of R) K is a superkey of R if K can identify a unique tuple in a given relation r ( R ).

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Today’s Class

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  1. IS C332: Database Systems and Applications Today’s Class Data Models Relational Model

  2. Keys • Let K R (I.e., K is a set of attributes which is a subset of the schema of R) • K is a superkey of R if K can identify a unique tuple in a given relationr(R) Customer(CusNo, Name, Address, …)where customers have unique customer numbers and unique names.Possible superkeys: CusNo {CusNo, Name} {CusNo, Name, Address} plus many others • K is a candidate key if K is minimal There are two candidate keys: CusNo and Name • Every relation is guarantee to (must) have at least one key. Why?

  3. Key(Candidate key) • A key can not be determined from any particular instance data • it is an intrinsic property of a scheme • it can only be determined from the meaning of attributes • A relation can have more than one key. • Superkey: A set of attributes that contains any key as a subset. • A key can also be defined as a minimal superkey • Primary Key: One of the candidate keys chosen for indexing purposes ( More details later…)

  4. Keys • Keys are a way to associate tuples in different relations • Keys are one form of integrity constraint (IC) Enrolled Students PRIMARY Key FORIEGN Key

  5. Foreign Keys, Referential Integrity • Foreign key : Set of fields in one relation that is used to `refer’ to a tuple in another relation. • Must correspond to the primary key of the other relation. • Like a `logical pointer’. • If all foreign key constraints are enforced, referential integrity is achieved (i.e., no dangling references.)

  6. Enrolled Students 11111 English102 A Foreign Keys in SQL • E.g. Only students listed in the Students relation should be allowed to enroll for courses. • sid is a foreign key referring to Students: CREATE TABLE Enrolled (sid CHAR(20),cid CHAR(20),grade CHAR(2), PRIMARY KEY (sid,cid), FOREIGN KEY (sid) REFERENCES Students )

  7. Relational Model • Sets • collections of items of the same type • no order • no duplicates • Mappings domain range 1:many many:1 1:1 many:many

  8. Exercise • What are the mapping cardinalities of the following 4 relationships? B C D A

  9. Schema Diagram for University Database

  10. Relational Query Languages • Procedural vs.non-procedural, or declarative • “Pure” languages: • Relational algebra • Tuple relational calculus • Domain relational calculus • Relational operators

  11. Relational Algebra Operators

  12. Select Operation – Example • Relation r A B C D         1 5 12 23 7 7 3 10 • Select tuples with A=B and D > 5 • A=B ^ D > 5 (r) A B C D     1 23 7 10

  13. Project Operation – Example Selection of Columns (Attributes) A B C     10 20 30 40 1 1 1 2 • Relation r: • Select A and C A C A C A,C (r)     1 1 1 2    1 1 2 =

  14. Joining two relations – Cartesian Product • Relations r, s: A B C D E   1 2     10 10 20 10 a a b b r s • r xs: A B C D E         1 1 1 1 2 2 2 2         10 10 20 10 10 10 20 10 a a b b a a b b

  15. Union of two relations    1 2 1   2 3 • Relations r, s: s r • r  s: A B     1 2 1 3

  16. Set difference of two relations    1 2 1   2 3 • Relations r, s: s r • r – s: A B   1 1

  17. Set Intersection of two relations A B A B • Relation r, s: • r s    1 2 1   2 3 r s A B  2

  18. Natural Join • r s Natural Join Example A B C D • Relations r, s: 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b s r A B C D E      1 1 1 1 2      a a a a b     

  19. Joining two relations – Natural Join • Let r and s be relations on schemas R and S respectively. Then, the “natural join” of relations R and S is a relation on schema R S obtained as follows: • Consider each pair of tuples tr from r and ts from s. • If tr and ts have the same value on each of the attributes in RS, add a tuple t to the result, where • t has the same value as tr on r • t has the same value as ts on s

  20. Natural Join • Example: R = (A, B, C, D) S = (E, B, D) • Result schema = (A, B, C, D, E) • rs is defined as:r.A, r.B, r.C, r.D, s.E (r.B = s.Br.D = s.D (r x s))

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