Idea Transcript
The Relational Model
Murali Mani
Why Relational Model? z
Currently the most widely used z
z
Older models still used z
z
IBM’s IMS (hierarchical model)
Recent competitions z
z
Vendors: Oracle, Microsoft, IBM
Object Oriented Model: ObjectStore
Implementation standard for relational Model z z
SQL (Structured Query Language) SQL 3: includes object-relational extensions Murali Mani
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Relational Model z
Structures z z
Relations (also called Tables) Attributes (also called Columns or Fields)
Note: Every attribute is simple (not composite or multi-valued) Constraints z
z
z
z
Key and Foreign Key constraints (More constraints later)
Eg: Student Relation (The following 2 relations are equivalent) Student Student
sNumber
sName
sNumber
sName
1
Dave
2
Greg
2
Greg
1
Dave
Cardinality = 2 Arity/Degree = 2
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Relational Model z
Schema for a relation z z
z
Schema for a database z
z
Schemas for all relations in the database
Tuples (Rows) z
z
Eg: Student (sNumber, sName) PRIMARY KEY (Student) =
The set of rows in a relation are the tuples of that relation
Note: Attribute values may be null Murali Mani
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Primary Key Constraints z
A set of attributes is a key for a relation if: z z
z z z z
No two distinct tuples can have the same values in all key fields A proper subset of the key attributes is not a key.
Superkey: A proper subset of a superkey may be a superkey If multiple keys, one of them is chosen to be the primary key. Eg: PRIMARY KEY (Student) = Primary key attributes cannot take null values Murali Mani
Candidate Keys (SQL: Unique) z
z z z
Keys that are not primary keys are candidate keys. Specified in SQL using UNIQUE Attribute of unique key may have null values ! Eg: Student (sNumber, sName) PRIMARY KEY (Student) = CANDIDATE KEY (Student) = Murali Mani
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Violation of key constraints z
A relation violates a primary key constraint if: z
z
z
There is a row with null values for any attribute of primary key. (or) There are 2 rows with same values for all attributes of primary key
Consider R (a, b) where a is unique. R violates the unique constraint if all of the following are true z
2 rows in R have the same non-null values for a Murali Mani
Keys: Example Student sNumber
sName
address
1
Dave
144FL
2
Greg
320FL
Primary Key: Candidate key: Some superkeys: {, , , }
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Foreign Key Constraints z
z
To specify an attribute (or multiple attributes) S1 of a relation R1 refers to the attribute (or attributes) S2 of another relation R2 Eg: Professor (pName, pOffice) Student (sNumber, sName, advisor) PRIMARY KEY (Professor) = FOREIGN KEY Student (advisor) REFERENCES Professor (pName) Murali Mani
Foreign Key Constraints z z z
z z
FOREIGN KEY R1 (S1) REFERENCES R2 (S2) Like a logical pointer The values of S1 for any row of R1 must be values of S2 for some row in R2 (null values are allowed) S2 must be a key for R2 R2 can be the same as R1 (i.e., a relation can have a foreign key referring to itself). Murali Mani
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Foreign Keys: Examples Dept (dNumber, dName) Person (pNumber, pName, dept) Persons working for Depts
PRIMARY KEY (Dept) = PRIMARY KEY (Person) = FOREIGN KEY Person (dept) REFERENCES Dept (dNumber)
Person (pNumber, pName, father) PRIMARY KEY (Person) = FOREIGN KEY Person (father) REFERENCES Person (pNumber)
Person and his/her father
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Violation of Foreign Key constraints z
z
Suppose we have: FOREIGN KEY R1 (S1) REFERENCES R2 (S2) This constraint is violated if z
z
Consider a row in R1 with non-null values for all attributes of S1 If there is no row in R2 which have these values for S2, then the FK constraint is violated.
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Relational Model: Summary z
Structures z z
z
Relations (Tables) Attributes (Columns, Fields)
Constraints z
Key z
z z
Primary key, candidate key (unique)
Super Key Foreign Key Murali Mani
ER schema → Relational schema Simple Algorithm z
z
Entity type E → Relation E’ z Attribute of E → Attribute as E’ z Key for E → Primary Key for E’ For relationship type R between E1, E2, …, En z Create separate relation R’ z Attributes of R’ are primary keys of E1, E2, …, En and attributes of R z Primary Key for R’ is defined as:
z
If the maximum cardinality of any Ei is 1, primary key for R’ = primary key for Ei Else, primary key for R’ = primary keys for E1, E2, …, En
Define “appropriate” foreign keys from R’ to E1, E2, …, En Murali Mani
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Simple algorithm: Example 1 pNumber
dNumber Person
(1, *)
Works For
(0, *)
Dept
pName
dName years
Person (pNumber, pName) Dept (dNumber, dName) WorksFor (pNumber, dNumber, years) PRIMARY KEY (Person) = PRIMARY KEY (Dept) = PRIMARY KEY (WorksFor) = FOREIGN KEY WorksFor (pNumber) REFERENCES Person (pNumber) FOREIGN KEY WorksFor (dNumber) REFERENCES Dept (dNumber) Murali Mani
Simple Algorithm: Example 2 Supplier (sName, sLoc) Consumer (cName, cLoc) Product (pName, pNumber) Supply (supplier, consumer, product, price, qty)
pNumber
pName
Product (0, *) sName
cName Supplier
(1, *)
Supply
(0, *)
Consumer
sLoc
cLoc price
qty
PRIMARY Key (Supplier) = PRIMARY Key (Consumer) = PRIMARY Key (Product) = PRIMARY Key (Supply) = FOREIGN KEY Supply (supplier) REFERENCES Supplier (sName) FOREIGN KEY Supply (consumer) REFERENCES Consumer (cName) FOREIGN KEY Supply (product) REFERENCES Product (pName) Murali Mani
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Simple Algorithm: Example 3 pNumber
pName
Part
superPart (0, *)
subPart (0, 1)
Part (pName, pNumber) Contains (superPart, subPart, quantity)
Contains
quantity
PRIMARY KEY (Part) = PRIMARY KEY (Contains) = FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber) FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber) Murali Mani
Decreasing the number of Relations Technique 1 z
If the relationship type R contains an entity type, say E, whose maximum cardinality is 1, then R may be represented as attributes of E. z
z
If the cardinality of E is (1, 1), then no “new nulls” are introduced If the cardinality of E is (0, 1) then “new nulls” may be introduced.
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Example 1 sNumber
pNumber Student
(1,1)
Has Advisor
(0, *)
Professor
sName
pName years
Student (sNumber, sName, advisor, years) Professor (pNumber, pName) PRIMARY KEY (Student) = PRIMARY KEY (Professor) = FOREIGN KEY Student (advisor) REFERENCES Professor (pNumber) Note: advisor will never be null for a student Murali Mani
Example 2 pNumber
dNumber Person
(0,1)
Works For
(0, *)
Dept
pName
dName years
Person (pNumber, pName, dept, years) Dept (dNumber, dName) PRIMARY KEY (Person) = PRIMARY KEY (Dept) = FOREIGN KEY Person (dept) REFERENCES Dept (dNumber) Dept and years may be null for a person Murali Mani
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Example 3 pNumber
pName
Part
superPart (0, *)
subPart (0, 1) Contains
quantity
Part (pNumber, pname, superPart, quantity) PRIMARY KEY (Part) = FOREIGN KEY Part (superPart) REFERENCES Part (pNumber) Note: superPart gives the superpart of a part, and it may be null Murali Mani
Decreasing the number of Relations Technique 2 (not recommended) z
z
If the relationship type R between E1 and E2 is 1:1, and the cardinality of E1 or E2 is (1, 1), then we can combine everything into 1 relation. Let us assume the cardinality of E1 is (1, 1). We have one relation for E2, and move all attributes of E1 and for R to be attributes of E2. z z
If the cardinality of E2 is (1, 1), no “new nulls” are introduced If the cardinality of E2 is (0, 1) then “new nulls” may be introduced. Murali Mani
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Example 1 sNumber
pNumber Student
(0,1)
Has Advisor
(1,1)
Professor
sName
pName years
Student (sNumber, sName, pNumber, pName, years) PRIMARY KEY (Student) = CANDIDATE KEY (Student) = Note: pNumber, pName, and years can be null for students with no advisor
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Example 2 sNumber
pNumber Student
(1,1)
Has Advisor
(1,1)
Professor
sName
pName years
Student (sNumber, sName, pNumber, pName, years) PRIMARY KEY (Student) = CANDIDATE KEY (Student) = Note: pNumber cannot be null for any student. Murali Mani
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Other details z
Composite attribute in ER z
z
Include an attribute for every component of the composite attribute.
Multi-valued attribute in ER z
z
We need a separate relation for any multi-valued attribute. Identify appropriate attributes, keys and foreign key constraints.
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Composite and Multi-valued attributes in ER sNumber
sName
major
Student
sAge address
street
city
state
Student (sNumber, sName, sAge, street, city, state) StudentMajor (sNumber, major) PRIMARY KEY (Student) = PRIMARY KEY (StudentMajor) = FOREIGN KEY StudentMajor (sNumber) REFERENCES Student (sNumber) Murali Mani
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Weak entity types z
Consider weak entity type E z z
z
z
A relation for E, say E’ Attributes of E’ = attributes of E in ER + keys for all indentifying entity types. Key for E’ = the key for E in ER + keys for all the identifying entity types. Identify appropriate FKs from E’ to the identifying entity types.
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Weak entity types: Example
Dept (dNumber, dName) Course (cNumber, dNumber, cName) PRIMARY KEY (Dept) = PRIMARY KEY (Course) = FOREIGN KEY Course (dNumber) REFERENCES Dept (dNumber)
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ISA Relationship types: Method 1 sNumber
sName
Student (sNumber, sName) UGStudent (sNumber, year) GradStudent (sNumber, program)
Student
year
ISA
UGStudent
ISA
PRIMARY KEY (Student) = PRIMARY KEY (UGStudent) = PRIMARY KEY (GradStudent) =
program
GradStudent
An UGStudent will be represented in both Student relation as well as UGStudent relation (similarly GradStudent)
FOREIGN KEY UGStudent (sNumber) REFERENCES Student (sNumber) FOREIGN KEY UGStudent (sNumber) REFERENCES Student (sNumber)
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ISA Relationship types: Method 2 sNumber
sName
Student
year
ISA
ISA
UGStudent
program
GradStudent
Student (sNumber, sName, year, program) PRIMARY KEY (Student) = Note: There will be null values in the relation. Murali Mani
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ISA Relationship types: Method 3 Student (sNumber, sName) UGStudent (sNumber, sName, year) GradStudent (sNumber, sName, program) UGGradStudent (sNumber, sName, year, program) sNumber
sName
PRIMARY KEY (Student) = PRIMARY KEY (UGStudent) = PRIMARY KEY (GradStudent) = PRIMARY KEY (UGGradStudent) =
Student
year
ISA
UGStudent
ISA
program
Any student will be represented in only one of the relations as appropriate.
GradStudent
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