2.1Database System Concepts - 6 th Edition Chapter 2: Introduction to Relational Model Structure of...

2.1Database System Concepts - 6th Edition

Chapter 2: Introduction to Relational Chapter 2: Introduction to Relational ModelModel

Structure of Relational Databases

Fundamental Relational-Algebra-Operations


Example of the Example of the instructorinstructor Relation Relation

attributes(or columns)

tuples(or rows)


Attribute TypesAttribute Types

A relation is represented as a table. The term attribute ( 屬性 ) refers to a column of a table.

屬性一般指稱物件的特性 , 或欲處理的資料 .

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

Attribute values are (normally) required to be atomic; that is, indivisible ( 分割後沒有意義 )

multivalued attribute values are not atomic (see page 1.17)

For example: author = {{Smith, Jones}, {Jones, Frick}}

composite attribute values are not atomic

For example: publisher = {(McGraw-Hill, New York), (Oxford, London)}

The special value null is a member of every domain, which signifies that the value is unknown or does not exist.

The null value causes complications in the definition of many operations, and will be discussed later.


Relation Schema and InstanceRelation Schema and Instance

A1, A2, …, An are attributes

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

Example:

instructor = (ID, name, dept_name, salary)

Formally, given sets D1, D2, …. Dn, a relation r 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

Example:

D1 = {a, b, c}, D2 = {1, 2}, D1XD2 = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}

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

An element t of r is a tuple, represented by a row in a table.


Relations are UnorderedRelations are Unordered

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

資料的排序方式屬 physical level, 非 logical level

寫 query 時不知資料如何排序 ;

Example: instructor relation with unordered tuples


DatabaseDatabase

A database consists of multiple relations

Information about an enterprise is broken up into parts, where each relation storing one part of the information

The university database example:

instructor (ID, name, dept_name, salary)

department (dept_name, building, budget)

student (ID, name, dept_name, tot_cred)

course (course_id, title, dept_name, credits)

prereq (course_id, prereq_id)

Bad design: (c.f. page 1.15) univ (instructor_ID, name, dept_name, salary, student_Id, ..)Normalization theory (Chapter 8) deals with how to design “good” relational schemas.


KeysKeys Let K R

K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R)

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

(see page 2.5)

Superkey K is a candidate key if K is minimal

Example: {ID} , {name} are both candidate keys for Instructor

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

which one? 通常依一般使用習慣 , 找最有代表性的 .

Another example: { 系別 , 年級 , 班級 , 座號 }, { 學號 }, { 身分證字號 ) 都是同學的 candidate key, 選 { 學號 } 做為 primary key.

注意 :

Key 通常做為物件的代表性屬性屬性值的唯一性會隨表格表示的資料不同而改變 .,

如下一頁的 dept_name

SK

CKPK


Foreign KeysForeign Keys

instructor

department

•Foreign key constraint: Value in one relation must appear in another•Referencing relation: e.g., instructor•Referenced relation: e.g., departmentWill discuss this again in Chapter 3 and Chapter 4.


Schema Diagram for University DatabaseSchema Diagram for University Database


Relational Query LanguagesRelational Query Languages

Language in which user requests information from the database.

Categories of languages

Procedural

non-procedural, or declarative

“Pure” languages:

Relational algebra: procedural

Tuple (Domain) relational calculus: declarative

“Algebra” is based on operators.

Example of arithmetic algebra: 1 + 5*3

How to write a query

Determine which relations to use

Determine which operators to use


Relational AlgebraRelational Algebra

Relational operators

select: project: Natural join:

Cartesian product: x

union: Intersection: set difference: –

The operators take one or two relations as inputs and produce a new relation as a result.


Selection of tuplesSelection of tuples

Relation r

Selection

σ 限制式 (r)

Select tuples with

A=B and D > 5

σ A=B ^ D > 5 (r)


Selection of Columns (Attributes)Selection of Columns (Attributes)

Relation r:

Projection

屬性名稱 (r)

Select columns A and C

A, C (r)

-> duplicates are removed


More ExamplesMore Examples

Return those instructors whose salaries are more than 85000. (see page 2.5, page2.6)

σ salary>=85000 (instructor)

Output the attributes ID and Salary of instructors.

Π ID, salary (instructor)

Find the ID and salary for those instructors

who have salary greater than $85000.

Π ID, salary (σ salary>=85000 (instructor))

<- composition


Joining two relations – Cartesian ProductJoining two relations – Cartesian Product

Relations r, s:

r x s:

Example: instructor X department => 12*7 = 84 tuples

(see page 2.8) <- 會希望同一列的 dept_name 要一樣 !


Joining two relations – Natural JoinJoining 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

Example:

R = (A, B, C, D)

S = (E, B, D)

Result schema = (A, B, C, D, E)

r s is defined as:

r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r x s))


Natural Join ExampleNatural Join Example

Relations r, s:

Natural Join

r s


ExampleExample

Instructor department (c.f., page 2.8)

Example: Output the attributes ID and building of instructors. ID, building (Instructor department)


Union of two relationsUnion of two relations

Relations r, s:

r s:

Find the names of instructors and students.

name (instructor) name (student)


Set difference of two relationsSet difference of two relations

Relations r, s:

r – s:

Find the departments which do not hire instructors.

dept _ name (department) – dept _ name (instructor)


Set Intersection of two relationsSet Intersection of two relations

Relation r, s:

r s

Find the instructors who are also students.

name (instructor) name (student)

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