ch2-abbrev

download ch2-abbrev

of 25

Transcript of ch2-abbrev

  • 7/28/2019 ch2-abbrev

    1/25

    Database System Concepts, 6th Ed.

    Silberschatz, Korth and Sudarshan

    See www.db-book.com for conditions on re-use

    Chapter 2: Intro to Relational Model

    http://www.db-book.com/http://www.db-book.com/http://www.db-book.com/http://www.db-book.com/
  • 7/28/2019 ch2-abbrev

    2/25

    Silberschatz, Korth and Sudarshan2.2Database System Concepts - 6th Edition

    Example of a Relation

    attributes(or columns)

    tuples(or rows)

  • 7/28/2019 ch2-abbrev

    3/25

    Silberschatz, Korth and Sudarshan2.3Database System Concepts - 6th Edition

    Attribute Types

    The set of allowed values for each attribute is called thedomain of the attribute

    Attribute values are (normally) required to be atomic; thatis, indivisible

    The special valuenul l is a member of every domain

    The null value causes complications in the definition ofmany operations

  • 7/28/2019 ch2-abbrev

    4/25

    Silberschatz, Korth and Sudarshan2.4Database System Concepts - 6th Edition

    Relation Schema and Instance

    A1,A2, ,Anare attributes

    R = (A1,A2, ,An ) is a relation schema

    Example:

    instructor = (ID, name, dept_name, salary)

    Formally, given sets D1, D2, . Dn a relationr is a subset of

    D1 x D2 x x Dn

    Thus, a relation is a set of n-tuples (a1, a2, , an) where each ai Di

    The current values (relation instance) of a relation are specified by atable

    An element tofr is a tuple, represented by a rowin a table

  • 7/28/2019 ch2-abbrev

    5/25

    Silberschatz, Korth and Sudarshan2.5Database System Concepts - 6th Edition

    Relations are Unordered

    Order of tuples is irrelevant (tuples may be stored in an arbitrary

    order) Example: instructorrelation with unordered tuples

  • 7/28/2019 ch2-abbrev

    6/25

    Silberschatz, Korth and Sudarshan2.6Database System Concepts - 6th Edition

    Database

    A database consists of multiple relations

    Information about an enterprise is broken up into parts

    instructor

    student

    advisor

    Bad design:

    univ (instructor -ID, name, dept_name, salary, student_Id, ..)

    results in

    repetition of information (e.g., two students have the same

    instructor) the need for null values (e.g., represent an student with no

    advisor)

    Normalization theory (Chapter 7) deals with how to design good

    relational schemas

  • 7/28/2019 ch2-abbrev

    7/25Silberschatz, Korth and Sudarshan2.7Database System Concepts - 6th Edition

    Keys

    Let K R

    Kis a superkeyof R if values for Kare sufficient to identify aunique tuple of each possible relation r(R)

    Example: {ID} and {ID,name} are both superkeys of instructor.

    Superkey Kis a candidate key if Kis minimal

    Example: {ID} is a candidate key for Instructor

    One of the candidate keys is selected to be the primary key.

    which one?

    Foreign key constraint: Value in one relation must appear inanother

    Referencing relation

    Referenced relation

  • 7/28/2019 ch2-abbrev

    8/25Silberschatz, Korth and Sudarshan2.8Database System Concepts - 6th Edition

    Schema Diagram for University Database

  • 7/28/2019 ch2-abbrev

    9/25Silberschatz, Korth and Sudarshan2.9Database System Concepts - 6th Edition

    Relational Query Languages

    Procedural vs.non-procedural, or declarative

    Pure languages:

    Relational algebra

    Tuple relational calculus

    Domain relational calculus

    Relational operators

  • 7/28/2019 ch2-abbrev

    10/25Silberschatz, Korth and Sudarshan2.10Database System Concepts - 6th Edition

    Selection of tuples

    Relation r

    Select tuples with A=Band D > 5

    A=B and D > 5(r)

    Quiz Q1:A B OR D < 7 (r) has (1) 1 tuple (2) 2 tuples (3) 3 tuples (4) 4 tuples

  • 7/28/2019 ch2-abbrev

    11/25Silberschatz, Korth and Sudarshan2.11Database System Concepts - 6th Edition

    Selection of Columns (Attributes)

    Relation r:

    Select A and C

    Projection

    A, C(r)

    Quiz Q2:The projection operation (1) removes duplicates (2) does not remove duplicates

  • 7/28/2019 ch2-abbrev

    12/25Silberschatz, Korth and Sudarshan2.12Database System Concepts - 6th Edition

    Joining two relations Cartesian Product

    Relations r, s:

    rx s:

  • 7/28/2019 ch2-abbrev

    13/25Silberschatz, Korth and Sudarshan2.13Database System Concepts - 6th Edition

    Composition of Operations Can build expressions using multiple operations

    Example: A=C(r x s)

    r x s

    A=C(r x s)

  • 7/28/2019 ch2-abbrev

    14/25Silberschatz, Korth and Sudarshan2.14Database System Concepts - 6th Edition

    Union of two relations

    Relations r, s:

    r s:

  • 7/28/2019 ch2-abbrev

    15/25Silberschatz, Korth and Sudarshan2.15Database System Concepts - 6th Edition

    Set difference of two relations

    Relations r, s:

    r s:

  • 7/28/2019 ch2-abbrev

    16/25Silberschatz, Korth and Sudarshan2.16Database System Concepts - 6th Edition

    Set Intersection of two relations

    Relation r, s:

    rs

  • 7/28/2019 ch2-abbrev

    17/25Silberschatz, Korth and Sudarshan2.17Database System Concepts - 6th Edition

    Natural Join Example

    Relations r, s:

    Natural Join

    r s

    Quiz Q3: The natural join operation matches tuples (rows) whose valuesfor common attributes are (1) not equal (2) equal (3) weird Greek letters (4) null

  • 7/28/2019 ch2-abbrev

    18/25Silberschatz, Korth and Sudarshan2.18Database System Concepts - 6th Edition

    Joining two relations Natural Join

    Let rand s be relations on schemas R and S respectively.Then, the natural join of relations R and S is a relation onschema R S obtained as follows:

    Consider each pair of tuples trfrom rand ts from s.

    If trand ts have the same value on each of the attributes

    in RS, add a tuple t to the result, where

    thas the same value as tron r

    thas the same value as ts on s

    r s

  • 7/28/2019 ch2-abbrev

    19/25

  • 7/28/2019 ch2-abbrev

    20/25Silberschatz, Korth and Sudarshan2.20Database System Concepts - 6th Edition

    Aggregate Functions and Operations

    Aggregation function takes a collection of values and returns a single

    value as a result.avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values

    Aggregate operation in relational algebra

    Eis any relational-algebra expression

    G1, G2, Gn is a list of attributes on which to group (can be empty) Each Fiis an aggregate function

    EachAiis an attribute name

    Note: Some books/articles use (gamma) instead of (Calligraphic G)

    )( )(,,(),(,,, 221121 Ennn AFAFAFGGG

  • 7/28/2019 ch2-abbrev

    21/25Silberschatz, Korth and Sudarshan2.21Database System Concepts - 6th Edition

    Aggregate Operation Example

    Relation r:

    A B

    C

    7

    7

    3

    10

    sum(c

    ) (r)sum(c )

    27

  • 7/28/2019 ch2-abbrev

    22/25Silberschatz, Korth and Sudarshan2.22Database System Concepts - 6th Edition

    Aggregate Operation Example

    Find the average salary in each department

    dept_name avg(salary) (instructor)

  • 7/28/2019 ch2-abbrev

    23/25

    Silberschatz, Korth and Sudarshan2.23Database System Concepts - 6th Edition

    Aggregate Functions (Cont.)

    Result of aggregation does not have a name

    Can use rename operation to give it a name For convenience, we permit renaming as part of aggregate

    operation

    dept_name avg(salary) as avg_sal(instructor)

  • 7/28/2019 ch2-abbrev

    24/25

    Database System Concepts, 6th Ed.

    Silberschatz, Korth and Sudarshan

    See www.db-book.com for conditions on re-use

    End of Chapter 2

    http://www.db-book.com/http://www.db-book.com/http://www.db-book.com/http://www.db-book.com/
  • 7/28/2019 ch2-abbrev

    25/25

    Division Operator

    Given relations r(R) and s(S), such that S R, r s is the largest

    relation t(R-S) such thatt x s r

    E.g. let r(ID, course_id) = ID, course_id(takes ) and

    s(course_id) = course_id(dept_name=Biology(course )then r s gives us students who have taken all courses in the Biology

    department Can write rs as

    temp1R-S (r)

    temp2R-S ((temp1 x s )R-S,S (r))

    result= temp1 temp2

    The result to the right of the is assigned to the relation variable on

    the left of the .

    May use variable in subsequent expressions.