the relational model chapter 3
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The Relational Model Chapter 3. How is data represented in the relational model? What integrity constraints can be expressed? How can be data created and modified? How can data be manipulated and queried? How can we create, modify and query tables using SQL? - PowerPoint PPT PresentationTRANSCRIPT
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Unit 3/Fall04/Melikyan 1
The Relational ModelChapter 3
• How is data represented in the relational model?
• What integrity constraints can be expressed?
• How can be data created and modified?
• How can data be manipulated and queried?
• How can we create, modify and query tables using SQL?
• How do we obtain a relational db design from an ER diagram?
• What are views and where are they used?
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Why Study the Relational Model?
Most widely used model.
– Vendors: IBM, Informix, Microsoft, Oracle, Sybase, etc.
“Legacy systems” in older models
– E.G., IBM’s IMS
Recent competitor: object-oriented model
– ObjectStore, Versant, Ontos
– A synthesis emerging: object-relational model Informix Universal Server, UniSQL, O2, Oracle, DB2
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Data Definition Language (DDL)
The standard language for creating, manipulating, and
querying data in relational DBMS.
We will discuss the concept of relation and show how to
create relation using the SQL language
We will use DDL to specify conditions that must be satisfied
by the data (ICs) - integrity constraints
Then we turn to the mechanism for accessing data from DB
and introduce the querying features of SQL
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Relational Database: Definitions
Relational database: a set of relations
Relation: made up of 2 parts:Instance : a table, with rows and columns.
# Rows = cardinality, # Fields = degree / arity.
Schema :the relation schema specifies the relations name, the
name of each field and the domain of each field:
Students(sid: string, name: string, login: string, age: integer, gpa: real).
Can think of a relation as a set of rows or tuples (i.e., all rows are distinct).
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Example Instance of Students Relation
sid name login age gpa
53666 Jones jones@cs 18 3.4
53688 Smith smith@eecs 18 3.2
53650 Smith smith@math 19 3.8
Cardinality = 3, degree = 5, all rows distinct
Do all columns in a relation instance have to be distinct?
An instance of relation is a set of tuples, also called records. An instence of the students relation
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Creating Relations in SQL
Creates the Students relation. Observe that the type (domain) of
each field is specified, and enforced by the DBMS whenever
tuples are added or modified.
CREATE TABLE Students
(sid: CHAR(20), name:
CHAR(20), login:
CHAR(10), age: INTEGER, gpa: REAL)
CREATE TABLE Enrolled
(sid: CHAR(20), cid: CHAR(20), grade:
CHAR(2))
As another example, the Enrolled table holds information
about courses that students take.
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Adding and Deleting Tuples
Can insert a single tuple using:
INSERT INTO Students (sid, name, login, age, gpa)VALUES (53688, ‘Smith’, ‘smith@ee’, 18, 3.2)
Can delete all tuples satisfying some condition (e.g., name = Smith):
DELETE FROM Students SWHERE S.name = ‘Smith’
Powerful variants of these commands are available; more later!
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Destroying and Altering Relations
Destroys the relation Students. The schema information and the tuples are deleted.
DROP TABLE Students
The schema of Students is altered by adding a new field;
every tuple in the current instance is extended with a null
value in the new field.
ALTER TABLE Students ADD COLUMN firstYear: integer
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Update
We can modify the column values in existing row using the
update command
UPDATE Stidents S
SET S.age = S.age + 1, S.gpa = S.gpa – 1
WHERE S.sid = 53688
Where clause is applied first to determine which roe are to be modified.
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Relational Query Languages
A major strength of the relational model: supports simple, powerful querying of data.
Queries can be written intuitively, and the DBMS is responsible for efficient evaluation.
– The key: precise semantics for relational queries.
– Allows the optimizer to extensively re-order operations, and still ensure that the answer does not change.
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The SQL Query Language
Developed by IBM (system R) in the 1970s
Need for a standard since it is used by many vendors
Standards:
– SQL-86
– SQL-89 (minor revision)
– SQL-92 (major revision, current standard)
– SQL-99 (major extensions)
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The SQL Query Language
To find all 18 year old students, we can write:
SELECT *FROM Students SWHERE S.age=18
•To find just names and logins, replace the first line:
SELECT S.name, S.login
sid name login age gpa
53666 Jones jones@cs 18 3.4
53688 Smith smith@ee 18 3.2
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Querying Multiple Relations
What does the following query compute?
SELECT S.name, E.cidFROM Students S, Enrolled EWHERE S.sid = E.sid AND E.grade=“A”
Given the following instance of Enrolled (is this possible if the DBMS ensures referential integrity?):
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Who Got an A ?
sid cid grade 53831 COMP3300 C 53831 MATH1100 B 53650 Topology112 A 53666 History105 B
sid name login age gpa
53666 Jones jones@cs 18 3.4
53688 Smith smith@eecs 18 3.2
53650 Smith smith@math 19 3.8
S.name E.cid
Smith Topology112
Result
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Integrity Constraints (ICs)
IC: condition that must be true for any instance of the database; e.g., domain constraints.
– ICs are specified when schema is defined.
– ICs are checked when relations are modified.
A legal instance of a relation is one that satisfies all specified ICs.
– DBMS should not allow illegal instances.
If the DBMS checks ICs, stored data is more faithful to real-world meaning.
– Avoids data entry errors, too!
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Primary Key Constraints
A set of fields is a key for a relation if
1. No two distinct tuples can have same values in all key fields, and
2. This is not true for any subset of the key.– Part 2 false? A superkey.– If there’s >1 key for a relation, one of the keys is chosen (by
DBA) to be the primary key.
E.g., sid is a key for Students.
(What about name?)
The set {sid, gpa} is a superkey.
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Primary and Candidate Keys in SQLPossibly many candidate keys (specified using UNIQUE), one of
which is chosen as the primary key.
CREATE TABLE Students
(sid: CHAR(20), name:
CHAR(20), login:
CHAR(10), age: INTEGER, gpa: REAL,
UNIQUE (name, age),CONSTRAINT StudentsKey PRIMARY KEY (sid))
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More About Keys
For a given student and course, there is a single grade.” vs.“Students can take only one course, and receive a single grade for thatcourse; further, no two students in a course receive the same grade.” Used carelessly, an IC can prevent the storage of Database instances thatarise in practice!
CREATE TABLE Enrolled ( sid CHAR(20) cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid, cid) )
CREATE TABLE Enrolled (sid CHAR(20) cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid), UNIQUE (cid, grade) )
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Foreign Keys, Referential IntegrityForeign key : Set of fields in one relation that is used to `refer’
to a tuple in another relation. (Must correspond to primary key
of the second relation.) Like a `logical pointer’.
E.g. sid is a foreign key referring to Students:
Students(sid: string, name: string, login: string, age: integer, gpa: real).
Enrolled(sid: string, cid: string, grade: string) If all foreign key constraints are enforced, referential
integrity is achieved, i.e., no dangling references. Can you name a data model w/o referential integrity?
Links in HTML!
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Foreign Keys in SQL
Only students listed in the Students relation should be allowed to enroll for courses.
CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid), FOREIGN KEY (sid) REFERENCES Students )
sid name login age gpa
53666 Jones jones@cs 18 3.453688 Smith smith@eecs 18 3.253650 Smith smith@math 19 3.8
sid cid grade53666 Carnatic101 C53666 Reggae203 B53650 Topology112 A53666 History105 B
Enrolled Students
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Enforcing Referential Integrity
Consider Students and Enrolled; sid in Enrolled is a foreign keythat references Students.What should be done if an Enrolled tuple with a non-existentstudent id is inserted? (Reject it!)
What should be done if a Students tuple is deleted?– Also delete all Enrolled tuples that refer to it.– Disallow deletion of a Students tuple that is referred to.– Set sid in Enrolled tuples that refer to it to a default sid.– (In SQL, also: Set sid in Enrolled tuples that refer to it to a
special value null, denoting `unknown’ or `inapplicable’.)Similar if primary key of Students tuple is updated.
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Referential Integrity in SQL/92
SQL/92 supports all 4 options on deletes and updates.
Default is NO ACTION (delete/update is rejected)
CASCADE (also delete all tuples that refer to deleted tuple)
SET NULL / SET DEFAULT (sets foreign key value of referencing tuple)
CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid), FOREIGN KEY (sid) REFERENCES Students ON DELETE CASCADE ON UPDATE SET DEFAULT )
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Where do ICs Come From?
ICs are based upon the semantics of the real-world enterprise that is being described in the database relations.
We can check a database instance to see if an IC is violated, but we can NEVER infer that an IC is true by looking at an instance.
– An IC is a statement about all possible instances!
– From example, we know name is not a key, but the assertion that sid is a key is given to us.
Key and foreign key ICs are the most common; more general ICs supported too.
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Transactions and ConstraintsBy default, a constraint is checked at the end of SQL statement
CREATE TABLE Students
(sid: CHAR(20),
name: CHAR(20),
login: CHAR(10),
age: INTEGER,
honors CHAR(10) NOT NULL,
gpa: REAL,
PRIMARY KEY (sid),
FOREIGN KEY (honors)
References Courses (cid) )
CREATE TABLE Courses
(cid: CHAR(10),
cname: CHAR(10),
credits: INTEGER,
grader: CHAR(20) NOT NULL,
PRIMARY KEY (cid),
FOREIGN KEY (grader)
References Students (sid) )SQL allows a constraint to be in
DEFERRED and IMMEDIATE mode
SET CONSTRAINT Foo DEFERRED.It is checked at commit time
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Views
A view is just a relation, but we store a definition, rather than a set of tuples.
CREATE VIEW YoungActiveStudents (name, grade)
AS SELECT S.name, E.gradeFROM Students S, Enrolled EWHERE S.sid = E.sid and S.age<21
Views can be dropped using the DROP VIEW
command.How to handle DROP TABLE if there’s a view on the
table? DROP TABLE command has options to let the user
specify this.
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Views and Security
Views can be used to present necessary information (or a summary), while hiding details in underlying relation(s).
– Given YoungStudents, but not Students or Enrolled, we can find students s who have are enrolled, but not the cid’s of the courses they are enrolled in.
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Logical DB Design: ER to Relational
Entity sets to tables.
CREATE TABLE Employees (ssn CHAR(11), name CHAR(20), lot INTEGER, PRIMARY KEY (ssn))
Employees
ssnname
lot
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Relationship Sets to Tables (Without constraints)
In translating a relationship set to a relation, attributes of the relation must include: Keys for each participating
entity set (as foreign keys).
This set of attributes forms a superkey for the relation. All descriptive attributes.
CREATE TABLE Works_In( ssn CHAR(1), did INTEGER, since DATE, PRIMARY KEY (ssn, did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments)
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Review: Key Constraints
Each dept has at most one manager, according to the key constraint on Manages.
Translation to relational model?
Many-to-Many1-to-1 1-to Many Many-to-1
dname
budgetdid
since
lot
name
ssn
ManagesEmployees Departments
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Translating ER Diagrams with Key Constraints
Map relationship to a table:
Note that did is the key now!
Separate tables for Employees and Departments.
Since each department has a unique manager, we could instead combine
Manages and Departments.
CREATE TABLE Manages( ssn CHAR(11), did INTEGER, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments)
CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11), since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees)
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Review: Participation ConstraintsDoes every department have a manager? If so, this is a participation constraint: the participation of
Departments in Manages is said to be total (vs. partial). Every did value in Departments table must appear in a row of
the Manages table (with a non-null ssn value!)
lot
name dnamebudgetdid
sincename dname
budgetdid
since
Manages
since
DepartmentsEmployees
ssn
Works_In
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Participation Constraints in SQL
We can capture participation constraints involving one entity set in
a binary relationship, but little else (without resorting to CHECK
constraints).
CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11) NOT NULL, since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE NO ACTION)
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Review: Weak Entities
A weak entity can be identified uniquely only by considering the
primary key of another (owner) entity. Owner entity set and weak entity set must participate in a one-to-
many relationship set (1 owner, many weak entities). Weak entity set must have total participation in this identifying
relationship set.
lot
name
agepname
DependentsEmployees
ssn
Policy
cost
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Translating Weak Entity Sets
Weak entity set and identifying relationship set are translated into a single table.
– When the owner entity is deleted, all owned weak entities must also be deleted.
CREATE TABLE Dep_Policy ( pname CHAR(20), age INTEGER, cost REAL, ssn CHAR(11) NOT NULL, PRIMARY KEY (pname, ssn), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE)
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Review: ISA Hierarchies
Contract_Emps
namessn
Employees
lot
hourly_wages
ISA
Hourly_Emps
contractid
hours_worked
As in C++, or other PLs, attributes are inherited.If we declare A ISA B, every A entity is also considered to be a B entity. Overlap constraints: Can Joe be an Hourly_Emps as well as a Contract_Emps entity? (Allowed/disallowed)
Covering constraints: Does every Employees entity also have to be an Hourly_Emps or a Contract_Emps entity? (Yes/no)
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Translating ISA Hierarchies to Relations
General approach:
– 3 relations: Employees, Hourly_Emps and Contract_Emps. Hourly_Emps: Every employee is recorded in Employees.
For hourly emps, extra info recorded in Hourly_Emps (hourly_wages, hours_worked, ssn); must delete Hourly_Emps tuple if referenced Employees tuple is deleted).
Queries involving all employees easy, those involving just Hourly_Emps require a join to get some attributes.
Alternative: Just Hourly_Emps and Contract_Emps.
– Hourly_Emps: ssn, name, lot, hourly_wages, hours_worked.
– Each employee must be in one of these two subclasses.
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Unit 3/Fall04/Melikyan 37
Review: Binary vs. Ternary Relationships
If each policy is
owned by just 1
employee:
Key constraint on
Policies would mean
policy can only
cover 1 dependent!
What are the
additional
constraints in the
2nd diagram?
agepname
DependentsCovers
name
Employees
ssn lot
Policies
policyid cost
Beneficiary
agepname
Dependents
policyid cost
Policies
Purchaser
name
Employees
ssn lot
Bad design
Better design
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Binary vs. Ternary Relationships (Contd.)
The key constraints allow us to combine Purchaser with Policies and Beneficiary with Dependents.
Participation constraints lead to NOT NULL
constraints.
What if Policies is a weak entity set?
CREATE TABLE Policies ( policyid INTEGER, cost REAL, ssn CHAR(11) NOT NULL, PRIMARY KEY (policyid). FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE)
CREATE TABLE Dependents ( pname CHAR(20), age INTEGER, policyid INTEGER, PRIMARY KEY (pname, policyid). FOREIGN KEY (policyid) REFERENCES Policies, ON DELETE CASCADE)
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Relational Model: Summary
A tabular representation of data.
Simple and intuitive, currently the most widely used.
Integrity constraints can be specified by the DBA, based on application semantics. DBMS checks for violations.
– Two important ICs: primary and foreign keys
– In addition, we always have domain constraints.
Powerful and natural query languages exist.
Rules to translate ER to relational model
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Homework#2 Problem 2.2
A university database contains information about
Professors (identifed by social security number, or SSN)
and
Courses (identifed by courseid).
Professors teach courses; each of the following situations
concerns the Teaches relationship set.
For each situation, draw an ER diagram that describes it
(assuming that no further constraints hold).
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Solutions for HW2
1. Professors can teach the same course in several semesters, and each offering must be recorded.
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2. Professors can teach the same course in several semesters, and only the most recent such offering needs to be recorded. (Assume this condition applies in all subsequent questions.)
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3. Every professor must teach some course.
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4. Every professor teaches exactly one course (no more, no less).
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5. Every professor teaches exactly one course (no more, no less), and every course must be taught by some professor.
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Now suppose that certain courses can be taught by a team of professors jointly, but it is possible that no one professor in a team can teach the course. Model this situation, introducing additional entity sets and relationship sets if necessary.
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Solutions for 2.3
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Unit 3/Fall04/Melikyan 48
Solution for 2.4
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HW #3
HOMEWORK:
Answer the following questions from your textbook, page 95-96
Ex 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.8, 3.12
Assigned 09/ 16 /04 Due 09/23/04
SUBMIT: hard copy by the beginning of class