資料庫概論

Database Management System(DBMS)

• Collection of interrelated data• Set of programs to access the data• DBMS contains information about a

particular enterprise• DBMS provides an environment that

it both convenient and efficient to use

Purpose of Database SystemsDatabase management systems were developed tohandle the following difficulties of typical file-processing systems supported by conventionaloperating systems.• Data redundancy and inconsistency• Difficulty in accessing data• Data isolation multiple files and formats• Concurrent access by multiple users• Security problems

Instances and Schemas

• Schema the logical structure of the database

• Instance the actual content of the database at a particular point in time Database Systems Concepts

E-R Model

• Entity-Relationship model– Ref. Peter Chen’s paper in 1976.– an abstract and conceptual

representation of data– Top-down design– To describe an ontology

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E-R Model

• Building blocks– Entity

• Strong vs. weak

– Attribute• Key• Multi-valued attribute

– Relationship • Associated with attributes

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E-R Model

• Entity– An entity may be defined as a thing which

is recognized as being capable of an independent existence and which can be uniquely identified. [wiki].

– An entity may be a physical object such as a house or a car, an event such as a house sale or a car service, or a concept such as a customer transaction or order. [wiki]

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E-R Model

• Entity– Entities can be thought of as nouns.

Examples: a computer, an employee, a song, a mathematical theorem. [wiki]

– Strong vs. weak

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account transactionslog

E-R Model

• Attribute– Entities and relationships can both have

attributes. Examples: an employee entity might have a Social Security Number (SSN) attribute; the proved relationship may have a date attribute. [wiki]

– Every entity (unless it is a weak entity) must have a minimal set of uniquely identifying attributes, which is called the entity's primary key. [wiki]

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N_or_S

E-R Model

• Attribute– (General) atomic

attribute– Key attribute– Multi-valued attribute– Composite attribute– Derived attribute

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Student

SID

name

phone

addr

city

street

no

E-R Model

• Relationship– A relationship captures how two or

more entities are related to one another.

– Relationships can be thought of as verbs, linking two or more nouns.

– Examples: an owns relationship between a company and a computer,

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E-R Model

• Relationship– Role

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Person Personmarriedhusband wife

Person married

husband

wife

E-R Model

• Relationship– Cardinality: 1-to-1, 1-to-N, M-to-N

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Person LocationBirthplaceN 1

[Chen 1976]

E-R Model

• Non-binary relationship

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A CR

B

A CR1

R2

E-R Model

• Example of E-R diagram

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Teacher Course Student

TID CID SIDName

Name

Title

chooseteaching

credit

Score

Schema and constraint

• Model diagram [wiki]

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Schema and constraint

• An example [wiki]:– Customer(Customer ID, Tax ID, Name,

Address, City, State, Zip, Phone)– Order(Order No, Customer ID, Invoice No,

Date Placed, Date Promised, Terms, Status)– Order Line(Order No, Order Line No, Product

Code, Qty)– Invoice(Invoice No, Customer ID, Order No,

Date, Status)– Invoice Line(Invoice No, Invoice Line

No, Product Code, Qty Shipped)– Product(Product Code, Product Description)

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Foreign key

(Primary) Key

Schema and constraint• A “customer” relation• Properties

– No repeated tuples– Unordered tuples / attributes– Atomic values

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(Primary) Key

Schema and constraint

• Key – Primary key: not null, never change– Candidate key

• Domain (attributes)– Atomic– NULL values: missing / unknown / not

applicable

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Schema and constraint

• Foreign key– Foreign keys are integrity constraints

enforcing that the value of the attribute set is drawn from a candidate key in another relation.

– For example in the Order relation the attribute Customer ID is a foreign key.

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Functional dependency (FD)

• In a given table, an attribute Y is said to have a functional dependency on a set of attributes X (written X → Y) if and only if each X value is associated with precisely one Y value.

• For example, in an "Employee" table that includes the attributes "Employee ID" and "Employee Date of Birth",

• The FD {Employee ID} → {Employee Date of Birth} would hold.

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Functional dependency (FD)

• Full functional dependency• Partial functional dependency• Transitive functional dependency• Armstrong’s Rule

– Reflexive rule– Augmented rule– Transitive rule

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Functional dependency (FD)

• Decomposition– Lossless-join decomposition– Dependency-preserving

decomposition

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Normalization

• Normalization is the process of efficiently organizing data in a database.

• There are two goals of the normalization process:– eliminating redundant data (for example,

storing the same data in more than one table)

– ensuring data dependencies make sense (only storing related data in a table).

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Normalization• The process of organizing data to minimize

redundancy is called normalization. [wiki] – The goal: to decompose relations with

anomalies in order to produce smaller, well-structured relations.

– Normalization usually involves dividing large tables into smaller (and less redundant) tables and defining relationships between them.

– The objective is to isolate data so that additions, deletions, and modifications of a field can be made in just one table and then propagated through the rest of the database via the defined relationships.

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NormalizationNormal form Defined by Brief definition

First normal form (1NF) Two versions: E.F. Codd (1970), C.J. Date (2003)[11]

Table faithfully represents a relation and has no repeating groups

Second normal form (2NF) E.F. Codd (1971)[2]

No non-prime attribute in the table is functionally dependent on a proper subset of acandidate key

Third normal form (3NF)E.F. Codd (1971);[2] see +also Carlo Zaniolo's equivalent but differently-expressed definition (1982)[12]

Every non-prime attribute is non-transitively dependent on every candidate key in the table

Boyce–Codd normal form(BCNF)

Raymond F. Boyce and E.F. Codd (1974)[13]

Every non-trivial functional dependency in the table is a dependency on a superkey

Fourth normal form (4NF) Ronald Fagin (1977)[14] Every non-trivial multivalued dependency in the table is a dependency on a superkey

Fifth normal form (5NF) Ronald Fagin (1979)[15] Every non-trivial join dependency in the table is implied by the superkeys of the table

Domain/key normal form(DKNF) Ronald Fagin (1981)[16]

Every constraint on the table is a logical consequence of the table's domain constraints and key constraints

Sixth normal form (6NF) C.J. Date, Hugh Darwen, and Nikos Lorentzos (2002)[4]

Table features no non-trivial join dependencies at all (with reference to generalized join operator)

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In summary [wiki], …In summary [wiki], …

Basic Structure of SQL• SQL is based on set and relational operations with certain modifications and

enhancements• A typical SQL query has the form:

select A 1 , A 2 , ..., A n

from r 1 , r 2 , ..., r m

where P– Ais represent attributes

– ris represent relations– P is a predicate.• This query is equivalent to the relational algebra expression:

A 1 , A 2 , ..., A n ( p(r 1 X r 2 X …X r m ))• The result of an SQL query is a relation.

The Select Clause

• The select clause corresponds to the projection operation of the relational algebra. It is used to list the attributes desired in the result of a query.

• Find the names of all branches in the loan relationselect branch-namefrom loan

An asterisk in the select clause denotes”all attributes”

select *from loan

The Select Clause(Cont.)

• SQL allows duplicates in relations as well as in query results.

• To force the elimination of duplicates, insert the keyword distinct after select.Find the names of all branches in the loan relation, and remove duplicates

select distinct branch-namefrom loan

• The keyword all specifies that duplicates not be removed.

Select all branch-namefrom loan

The Select Clause(Cont.)

• The select clause can contain arithmetic expressions involving the operators,+,-,*,/,and operating on constants or attributes of tuples.

• The query:select branch-name,loan-

number,amount * 100from loan

would return a relation which is the same as the loan relation, except that the attribute amount is multiplied by 100

The where Clause

• The where clause corresponds to the selection predicate of the relational algebra. It consists of a predicate involving attributes of the relations that appear in the from clause.

• Find all loan numbers for loans made at the Perryridge branch with loan amounts greater than $1200.

select loan-numberfrom loanwhere branch-name=“Perryridge” and

amount>1200

The where Clause(Cont.)

• SQL includes a between comparison operator in order to simplify where clauses that specify that a value be less than or equal to some value and greater than or equal to some other value.

• Find the loan number of those loans with loan amounts between $90,000 and $100,000(that is, $90,000 and $100,000)

select loan-numberfrom loanwhere amount between 90000 and

100000

The from Clause• The from clause corresponds to the Cartesian product operation of the

relational algebra. It lists the relations to be scanned in the evaluation of the expression.

• Find the Cartesian product borrower loanselect *from borrower, loan

• Find the name and loan number of all customers having a loan at the Perryridge branch.

select distinct customer-name,borrower.loan-numberfrom borrower,loanwhere borrower.loan-number=loan.loan-number and branch-name=“Perryridge”

The Rename Operation

• The SQL mechanism for renaming relations and attributes is accomplished through the as clause:

old-name as new-name• Find the name and loan number of all customers having a

loan at the Perryridge branch;replace the column name loan-number with the name loan-id.

select distinct customer-name,borrower.loan-number as loan-idfrom borrower,loanwhere borrower.loan-number=loan.loan-number and

branch-name=“Perryridge”

Tuple Variables• Tuple variables are defined in the from clause via the use of the as clause.• Find the customer names and their loan numbers for all customers having

a loan at some branch.select distinct customer-name,T.loan-numberfrom borrow as T,loan as Swhere T.loan-number=S.loan-number

• Find the names of all branches that have greater assets than some branch located in Brooklyn.

select distinct T.branch-namefrom branch as T,branch as Swhere T.assets > S.assets and S.branch-city=“Brooklyn”

Ordering the Display of Tuples• List in alphabetic order the names of all customers having a loan at

Perryridge branchselect distinct customer-namefrom borrower,loanwhere borrower.loan-number=loan.loan-number and

branch-name=“Perryridge”order by customer-name

• We may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default.

• SQL must perform a sort to fulfill an order by request. Since sorting a large number of tuples may be costly, it is desirable to sort only when necessary.

Set Operations• Find all customers who have a loan, an account, or both:

(select customer-name from depositor)union(select customer-name from borrower)

• Find all customers who have both a loan and an account.(select customer-name from depositor)intersect(select customer-name from borrower)

• Find all customers who have an account but no loan.(select customer-name from depositor)except(select customer-name from borrower)

Aggregate Operations

• These functions operate on the multiset of values of a column of a relation, and return a value

avg:average valuemin:minimum valuemax:maximum valuesum:sum of valuescount:count of values

Aggregate Functions(Cont.)• Find the average account balance at the Perryridge branch

select avg (balance)from accountwhere branch-name=“Perryridge”

• Find the number of tuples in the customer relation.select count (*)from customer

• Find the number of depositors in the bankselect count(distinct customer-name)from depositor

Aggregate Functions -- Group By

• Find the number of depositors for each branch.select branch-name,count(distinct customer-name)from depositor,accountwhere depositor.account-number=account.account-numbergroup by branch-name

Note : Attributes in select clause outside of aggregate functions must appear in group by list.

Aggregate Functions -- Having Clause

• Find the names of all branches where the average account balance is more than $1,200

select branch-name,avg(balance)from accountgroup by branch-namehaving avg(balance)>1200

• Note: predicates in the having clause are applied after the formation of groups

Null Values• It is possible for tuples to have a null value, denoted by null, for some of

their attributes; null signifies an unknown value or that a value does not exist.

• The result of any arithmetic expression involving null is null.• Roughly speaking, all comparisons involving null return false. More

precisely,– Any comparison with null returns unknown– (true or unknown) = true, (false or unknown) = unknown (unknown or

unknown) = unknown, (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown

– Result of where clause predicate is treated as false if it evaluates to unknown

– “P is unknown”evaluates to true if predicate P evaluates to unknown

Null Values(Cont.)• Find all loan numbers which appear in the loan relation with null values

for amount.select loan-numberfrom loanwhere amount is null

• Total all loan amountsselect sum(amount)from loan

Above statement ignores null amounts; result is null if there is no non-null amount.

• All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes.

Nested Subqueries

• SQL provides a mechanism for the nesting of subqueries.

• A subquery is a select-from-where expression that is nested within another query.

• A common use of subqueries is to perform tests for set membership, set comparisons, and set cardinality.

Example Query• Find all customers who have both an account and a loan at bank.

Select distinct customer-namefrom borrowerwhere customer-name in (select customer-name from depositor)

• Find all customers who have a loan at the bank but do not have an account at the bank.Select distinct customer-namefrom depositorwhere customer-name not in (select customer-name from depositor)

Example Query

• Find all customers who have both an account and a loan at the Perryridge branch.

select distinct customer-name from borrower, loan where borrower.loan-number = loan.loan-number and

branch-name =“Perryridge”and (branch-name, customer-name) in

(select branch-name, customer-name from depositor, account where depositor.account-number =

account.account-number)

Set Comparison

• Find all branches that have greater assets than some branch located in Brooklyn.

select distinct T.branch-name from branch as T, branch as S where T.assets > S.assets and S.branch-city =“Brooklyn”

Example Query

• Find all branches that have greater assets than some branch located in Brooklyn.

select branch-name from branch where assets > some

(select assets from branch where branch-city =“Brooklyn”)

Example Query

• Find the names of all branches that have greater assets than all branches located in Brooklyn.

select branch-name from branch where assets > all

(select assets from branch where branch-city = “Brooklyn”)

Example Query• Find all customers who have an account at all branches located in Brooklyn.

select distinct S.customer-name from depositor as S where not exists ( (select branch-name from branch where branch-city = “Brooklyn”) except (select R.branch-name from depositor as T, account as R where T.account-number = R.account-number and S.customer-name = T.customer-name))

Derived Relations• Find the average account balance of those branches where the average

account balance is greater than $1200.select branch-name, avg-balance from (select branch-name, avg (balance)

from account group by branch-name) as result (branch-name, avg-balance)

where avg-balance > 1200Note that we do not need to use the having clause, since we compute in the from clause the temporary relation result, and the attributes of result can be used directly in the where clause.

Modification of the Database -- Deletion• Delete all account records at the Perryridge branch

delete from account where branch-name = “Perryridge”

• Delete all accounts at every branch located in Needham.delete from account where branch-name in (select branch-name from branch where branch-city = “Needham”) delete from depositor where account-number in (select account-number from branch, account where branch-city = “Needham” and branch.branch-name = account.branch-name)

Example Query• Delete the records of all accounts with balances below the

average at the bankdelete from account where balance < (select avg (balance)

from account)– Problem: as we delete tuples from deposit, the average

balance changes– Solution used in SQL:

� First, compute avg balance and find all tuples to delete � Next, delete all tuples found above (without recomputing

avg or retesting the tuples)

Modification of the Database -- Insertion

• Add a new tuple to accountinsert into account

values (“Perryridge”, A-9732, 1200)or equivalentlyinsert into account (branch-name, balance, account-number)

values (“Perryridge”, 1200, A-9732)• Add a new tuple to account with balance set to null

insert into account values (“Perryridge”, A-777, null)

Modification of the Database -- Updates• Increase all accounts with balances over $10,000 by 6%, all other

accounts receive 5%.– Write two update statements:

update account set balance = balance * 1.06 where balance > 10000

update account set balance = balance * 1.05 where balance 1000– The order is important– Can be done better using the case statement