Post on 14-Jan-2016
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
2.2Database System Concepts - 6th Edition
Example of the Example of the instructorinstructor Relation Relation
attributes(or columns)
tuples(or rows)
2.3Database System Concepts - 6th Edition
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.
2.4Database System Concepts - 6th Edition
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.
2.5Database System Concepts - 6th Edition
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
2.6Database System Concepts - 6th Edition
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.
2.7Database System Concepts - 6th Edition
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
2.8Database System Concepts - 6th Edition
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.
2.9Database System Concepts - 6th Edition
Schema Diagram for University DatabaseSchema Diagram for University Database
2.10Database System Concepts - 6th Edition
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
2.11Database System Concepts - 6th Edition
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.
2.12Database System Concepts - 6th Edition
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)
2.13Database System Concepts - 6th Edition
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
2.14Database System Concepts - 6th Edition
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
2.15Database System Concepts - 6th Edition
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 !
2.16Database System Concepts - 6th Edition
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))
2.17Database System Concepts - 6th Edition
Natural Join ExampleNatural Join Example
Relations r, s:
Natural Join
r s
2.18Database System Concepts - 6th Edition
ExampleExample
Instructor department (c.f., page 2.8)
Example: Output the attributes ID and building of instructors. ID, building (Instructor department)
2.19Database System Concepts - 6th Edition
Union of two relationsUnion of two relations
Relations r, s:
r s:
Find the names of instructors and students.
name (instructor) name (student)
2.20Database System Concepts - 6th Edition
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)
2.21Database System Concepts - 6th Edition
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)