database system concepts · keys •let k r •k is a superkey of r if values for k are sufficient...

Database System Concepts

Chapter 2: Intro to Relational Model 11-Sep-20

Outline

• Structure of Relational Databases

• Database Schema

• Keys

• Schema Diagrams

• Relational Query Languages

• Relational Operations

Relational Model

• Use a collection of tables to represent both data and relationships among those data

• Terminology (basic notions of the relational model)

– relation/table

– tuple/row

– attributes/column

4

Example of a Relationattributes

(or columns)

tuples

(or rows)

Attribute Types

• The set of allowed values for each attributeis called the domain of the attribute

• Attribute values are (normally) required tobe atomic; that is, indivisible

• The special value null is a member of everydomain. Indicated that the value is“unknown”

• The null value causes complications in thedefinition of many operations

Outline

7


• Database Schema

• Keys

• Schema Diagrams



Relation Schema and Instance

• Database Schema

– Logical design of the database

– Like type definition in programming-language

• Database Instance

– Snapshot of the data

– Like variable in programming-language

Relation Schema and Instance

• Attributes： A1, A2, …, An• Relation schema: R = (A1, A2, …, An )

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

– a set of n-tuples (a1, a2, …, an) where each ai Di

– 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 Unordered

• Order of tuples is irrelevant

– tuples may be stored in an arbitrary order

– example: instructor relation with unordered tuples

Outline

11


• Database Schema

• Keys

• Schema Diagrams



Keys

• How to distinguish the tuples in a given relation?

– the value of the attribute values of a tuple should be able to uniquely identify the tuple

• Terminology

– Superkey

– Candidate Key

– Primary Key

– Foreign Key

Keys• 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} and {ID,name} are both superkeys of instructor.

• Superkey K is a candidate key if K is 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– Example – dept_name in instructor is a foreign key from instructor

referencing department

Primary Key Constraint

Referential Integrity Constraint

选修 (学号, 课程号, 成绩)

外键

假设存在关系𝑟和𝑠：𝑟(A, B, C), 𝑠(B, D)，则在关系𝑟上的属性B称作参照𝑠的外码，𝑟也称为外码依赖的参照关系，𝑠叫做外码被参照关系例学生(学号,姓名,性别,专业号,年龄) - 参照关系

专业(专业号,专业名称) - 被参照关系 (目标关系)

其中属性专业号称为关系学生的外码

Instructor (ID, name, dept_name, salary) - 参照关系

Department (dept_name, building, budget) - 被参照关系

参照关系中外码的值必须在被参照关系中实际存在或为null14

Foreign Key

选修 (学号, 课程号, 成绩)课程 (课程号,课程名, 学分, 先修课号)

Outline

15


• Database Schema

• Keys

• Schema Diagrams



Schema Diagram for University Database

classroom (building, room_number, capacity)

department (dept_name, building, budget)

course (course_id, title, dept_name, credits)

instructor (ID, name, dept_name, salary)

section (course_id, sec_id, semester, year, building, room_number, time_slot_id)

teaches (ID, course_id, sec_id, semester, year)

student (ID, name, dept_name, tot_cred)

takes (ID, course_id, sec_id, semester, year, grade)

advisor (s_ID, i_ID)

time_slot (time_slot_id, day, start_time, end_time)

prereq (course_id, prereq_id)

Schema Diagram for University Database

被参照关系参照关系

Outline

18


• Database Schema

• Keys

• Schema Diagrams



19

Relational Query Language

• Query Languages– allow manipulation (操纵) and retrieval (检索) of

data from a database

• Relational Query Languages– query languages for Relational Database– “real”/ “practical” query languages

• e.g. SQL

– “pure”/ “mathematical” query languages• Relational Algebra• Relational Calculus

20

? ?

Formal Relational Query Languages

Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation:

Relational Algebra: More operational (过程化), very useful for representing execution plans.

Relational Calculus: Lets users describe what they want, rather than how to compute it. (Non-operational, declarative (陈述).)

Understanding Relational Algebra & Calculus is key to understanding SQL, query processing!

22

Declarative vs Procedural

• Procedural programming requires that theprogrammer tell the computer what to do.– how to get the output for the range of required

inputs

– the programmer must know an appropriatealgorithm.

• Declarative programming requires a moredescriptive style.– the programmer must know what relationships

hold between various entities.

Why do we need Query Languages anyway?

• Two key advantages– Less work for user asking query

– More opportunities for optimization

• Relational Algebra– Theoretical foundation for SQL

– Higher level than programming language

• but still must specify steps to get desired result

• Relational Calculus– Formal foundation for Query-by-Example

– A first-order logic description of desired result

– Only specify desired result, not how to get it

Relational Query Languages

• Procedural vs .non-procedural, or declarative• “Pure” languages:

– Relational algebra– Tuple relational calculus– Domain relational calculus

• The above 3 pure languages are equivalent incomputing power

• We will concentrate in this chapter onrelational algebra– Not turning-machine equivalent– consists of 6 basic operations

Outline

25


• Database Schema

• Keys

• Schema Diagrams



Relational Operations

在某种程度上是过程化(operational)语言

六个基本运算 Select 选择

Project 投影

Union 并

set difference 差（集合差）

Cartesian product 笛卡儿积

Rename 更名（重命名）

用户输入一个或两个关系，并得到新的关系

Relational Operations

附加运算

Set intersection 交

Natural join 自然连接

Division 除

Assignment 赋值

Select Operation –selection of rows (tuples)

Relation r

A=B ^ D > 5 (r)

注：执行选择时，选择条件必须是针对同一元组中的相应属性值代入进行比较。

Select Operation –selection of rows (tuples)

定义：𝜎𝑝(𝑟) = {𝑡|𝑡 ∈ 𝑟 ∧ 𝑝(𝑡)}，𝜎 读作sigma, 其中p为选择谓词

其中p是由逻辑连词 ⋀(与), ∨(或),¬(非)连接起来的公式。逻辑连词的运算对象可以是包含比较运算符=、=和>

Project Operation –selection of columns (Attributes)

Relation r:

Π𝐴,𝐶(𝑟)

Project Operation –selection of columns (Attributes)

例：关系r

选择属性Ａ、Ｃ

投影：Π𝐴,𝐶(𝑟)

Union of two relations

Relations r, s:

r s:

Union of two relations并运算

定义：𝑟 ∪ 𝑠 = {𝑡|𝑡 ∈ 𝑟 𝑜𝑟 𝑡 ∈ 𝑠 }

𝑟 ∪ 𝑠条件：

等目，同元，即他们的属性数目必须相同

对任意𝑖, 𝑟的第𝑖个属性域和𝑠的第𝑖个属性域相同

例如：Π 𝑛𝑎𝑚𝑒 𝑖𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑜𝑟 ∪ Π 𝑛𝑎𝑚𝑒 (𝑠𝑡𝑢𝑑𝑒𝑛𝑡)

Set difference of two relations

Relations r, s:

r – s:

Set difference of two relations

定义：𝑟 – 𝑠 = {𝑡|𝑡 ∈ 𝑟 𝑎𝑛𝑑 𝑡 ∉ 𝑠}

𝑟 − 𝑠条件：

等目，同元，即他们的属性数目必须相同

对任意𝑖，𝑟的第𝑖个属性域和𝑠的第𝑖个属性域相同

joining two relations -- Cartesian-product

Relations r, s:

r x s:

A B

12

C D E

10102010

aabb

A B C D E

1111

10102010

aabb

2222

10102010

aabb

r

s

B

s.Br.B

Cartesian-product – naming issue

Relations r, s:

r x s:

Renaming a Table

• Allows us to refer to a relation, (say E) by more than one name.

𝜌𝑥(𝐸)

returns the expression E under the name X Relations r

𝑟 × 𝜌𝑠(𝑟)α

α

β

β

1

1

2

2

α

β

α

β

1

2

1

2

r.A r.B s.A s.B

Composition of Operations• Can build expressions using multiple operations

• Example: 𝜎𝐴=C(𝑟 × 𝑠)

r s

𝜎𝐴=C(𝑟 × 𝑠)

𝑟 × 𝑠 =

A B

𝛼 1

𝛽 2

C D E

𝛼 10 a

𝛽 10 a

𝛽 20 b

𝛾 10 b

A B C D E

𝛼 1 𝛼 10 a

𝛼 1 𝛽 10 a

𝛼 1 𝛽 20 b

𝛼 1 𝛾 10 b

𝛽 2 𝛼 10 a

𝛽 2 𝛽 10 a

𝛽 2 𝛽 20 b

𝛽 2 𝛾 10 b

A B C D E

𝛼 1 𝛼 10 a

𝛽 2 𝛽 20 a

𝛽 2 𝛽 20 b

银行示例

40

branch (branch-name, branch-city, assets)

customer (customer-name, customer-street,customer-city)

account (account-number, branch-name, balance)

loan (loan-number, branch-name, amount)

depositor (customer-name, account-number)

borrower (customer-name, loan-number)

查询示例

41

例1：找出贷款额大于$1200的元组

𝜎𝑎𝑚𝑜𝑢𝑛𝑡> 1200(𝑙𝑜𝑎𝑛)

例2：找出贷款大于$1200的贷款号

loan (loan-number, branch-name, amount)

ෑ𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟

(𝜎𝑎mo𝑢𝑛𝑡>1200(𝑙𝑜𝑎𝑛))

例3：找出有贷款或有帐户或两者兼有的所有客户姓

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒 borrower ∪ Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑜𝑟)

例4：找出至少有一个贷款及一个账户的客户姓名

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒 borrower ∩ Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑜𝑟)

查询示例

depositor (customer-name, account-number)

borrower (customer-name, loan-number)

例5：找出在Perryridge 分支机构有贷款的顾客姓名

• 查询1：

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝜎𝑏𝑟𝑎𝑛𝑐ℎ−𝑛𝑎𝑚𝑒="𝑃𝑒𝑟𝑟𝑦𝑟𝑖𝑑𝑔𝑒"(𝜎𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟=𝑙𝑜𝑎𝑛.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟 bo𝑟𝑟𝑜𝑤𝑒𝑟 × 𝑙𝑜𝑎𝑛 ))

• 查询2：

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝜎𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟=𝑙𝑜𝑎𝑛.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟 × 𝜎𝑏𝑟𝑎𝑛𝑐ℎ−𝑛𝑎𝑚𝑒="𝑃𝑒𝑟𝑟𝑦𝑟𝑖𝑑𝑔𝑒" 𝑙𝑜𝑎𝑛

查询示例

查询2更好

loan (loan-number, branch-name, amount)borrower (customer-name, loan-number)

例6：找出在Perryridge分支机构有贷款，但在其他分支机构没有账号的顾客姓名

• 查询1：

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝜎𝑏𝑟𝑎𝑛𝑐ℎ−𝑛𝑎𝑚𝑒="𝑃𝑒𝑟𝑟𝑦𝑟𝑖𝑑𝑔𝑒"

(𝜎𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟=𝑙𝑜𝑎𝑛.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟(𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟 × 𝑙𝑜𝑎𝑛)))

−Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑜𝑟)

• 查询2：

Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝜎𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟=𝑙𝑜𝑎𝑛.𝑙𝑜𝑎𝑛−𝑛𝑢𝑚𝑏𝑒𝑟

𝑏𝑜𝑟𝑟𝑜𝑤𝑒𝑟 × 𝜎𝑏𝑟𝑎𝑛𝑐ℎ−𝑛𝑎𝑚𝑒="𝑃𝑒𝑟𝑟𝑦𝑟𝑖𝑑𝑔𝑒" 𝑙𝑜𝑎𝑛

− Π𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟−𝑛𝑎𝑚𝑒(𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑜𝑟)

查询示例loan (loan-number, branch-name, amount)borrower (customer-name, loan-number)

查询示例1234

1234

d

123

4

account (account-number, branch-name, balance)

例7：找出银行中最大的账户余额

将account关系重命名为d

第一步：找出由非最大余额构成的临时关系

ෑ𝑎𝑐𝑐𝑜𝑢𝑛𝑡.𝑏𝑎𝑙𝑎𝑛𝑐𝑒

(𝜎𝑎𝑐𝑐𝑜𝑢𝑛𝑡.𝑏𝑎𝑙𝑎𝑛𝑐𝑒

附加运算

定义一些附加运算，它们虽不能增加关系代数的表达能力，

但却可以简化一些常用的查询

集合交

自然连接

除

赋值

Set intersection of two relations

Relation r, s:

r s

Note: r s = r – (r – s)

定义：𝑟 ∩ 𝑠 = { 𝑡 | 𝑡 ∈ 𝑟 𝑎𝑛𝑑 𝑡 ∈ 𝑠 }

假设：

𝑟和𝑠同元

𝑟和𝑠的属性域是可兼容的

𝑟 ∩ 𝑠 = 𝑟 − (𝑟 − 𝑠)

Set intersection of two relations

Joining two relations – Natural Join

• Let r and s be relations on schemas R and S respectively.

• “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

𝑟 ⋈ 𝑠

例：𝑅 = (𝐴,𝐵,𝐶,𝐷) 𝑆 = (𝐸,𝐵,𝐷)

关系𝑟和𝑠自然连接的结果模式为：(𝐴,𝐵,𝐶,𝐷,𝐸)

𝑟 ⋈ 𝑠 = Π𝑟.𝐴,𝑟.𝐵,𝑟.𝐶,𝑟.𝐷,𝑠.𝐸(𝜎𝑟.𝐵=𝑠.𝐵∧𝑟.𝐷=𝑠.𝐷𝑟 × 𝑠)

设关系𝑟和𝑠分别代表模式𝑅和𝑆，那么𝑟 ⋈ 𝑠是对模式𝑅 ∪ 𝑆运算后的关系表示：

考虑𝑟的每一个元组𝑡𝑟，𝑠的每一个元组𝑡𝑠 如果𝑡𝑟和𝑡𝑠在𝑅 ∩ 𝑆的公共属性下有相同的值，那么向结果集中插入元组 𝑡：

– 元组𝑡与𝑡𝑟有相同的值

– 元组𝑡与𝑡𝑠有相同的值

Natural Join Example

例：关系𝑟，𝑠

A B C D

12412

aabab

r

A B C D E

11112

aaaa b

B D E

13123

aaabb

s r s

注：(1) 𝑟, 𝑠必须含有共同属性 (名, 域对应相同),(2) 连接二个关系中同名属性值相等的元组(3) 结果属性是二者属性集的并集，但消去重名属性。

Natural Join Example

Notes about Relational Languages

• Each Query input is a table (or set of tables)

• Each query output is a table.

• All data in the output table appears in one of the input tables

Summary of Relational Algebra Operators

Symbol (Name) Example of Use

(Selection) σ salary > = 85000 (instructor)σ

Return rows of the input relation that satisfy the predicate.

Π (Projection) Π ID, salary (instructor)

Output specified attributes from all rows of the input relation. Remove duplicate tuples from the output.

x (Cartesian Product) instructor x department

Output pairs of rows from the two input relations that have the same value on all attributes that have the same name.

∪ (Union) Π name (instructor) ∪ Π name (student)

Output the union of tuples from the two input relations.

(Natural Join) instructor ⋈ department

Output pairs of rows from the two input relations that have same value on all attributes that have same name.

⋈

- (Set Difference) Π name (instructor) -- Π name (student)

Output the set difference of tuples from two input relations.

• 𝑡ℎ𝑒𝑡𝑎连接：𝑟 ⋈𝜃 𝑠 = 𝜎𝜃 𝑟 × 𝑠

• 𝜃是模式 𝑅 ∪ 𝑆 属性上的谓词

• 𝑡ℎ𝑒𝑡𝑎连接是自然连接的扩展

theta 连接

除运算

𝑟 ÷ 𝑠适用于包含了“对所有的”此类短语的查询

例:查询选修了所有课程的学生的学号

Sno Cno Grade

95001 1 92

95001 2 85

95001 3 88

95002 2 90

95002 3 80

÷

course

=95001

Sno

Cno

1

2

3

Sno, Cno (enrolled) ÷ Cno (course)

enrolled

除运算

除运算

设r和s分别代表模式R和S的关系

R = (A1,…,Am,B1,…,Bn)

S = (B1,…,Bn)

r s 的结果代表模式 R–S 的关系：

–R –S = (A1,…,Am)

–r s = {t|tR-S(r) us(tur)}

注：商来自于R-S(r)，并且其元组t与s所有元组的拼接被r覆盖

除运算

除运算

例：关系r，s

r s =

A

(r s = {t|tR-S(r) [us(tur)]})

B

1

2

A B

1 2 3 1 1 1 3 4 6 1 2

r

s

除运算

A B C D E

a a 1 b a 1 b b 1 a a 1 a c 2 a b 3 a a 1 a b 1 c b 1

r s =

D

a b

E

11

A B C

ba

r

s

A B

bb

C D

E

a a 1a. 1b. 1

a a 1a b 3

a c 2 a a 1 a b 1

c b 1

1

2

3

4

5

除运算

例，关系r，s

除运算

除运算

A B C

a1 b1 c1

a2 b1 c1

a1 b2 c1

a1 b2 c2

a2 b1 c2

a1 b2 c3

a1 b2 c4

a1 b1 c5

RC

c1

A B

a1 b1

a2 b1

a1 b2

S QC

c1

c4

c2

c3

A B

a1 b2

S QB C D

b1 c1 d1

b2 c1 d2

SA

a1

Q

C

c1

c2

SA B

a1 b2

a2 b1

Q

例:从SC表中查询至少选修1号课程和3号课程的学生学号

Sno Cno Grade

95001 1 92

95001 2 85

95001 3 88

95002 2 90

95002 3 80

Cno

1

3

临时表K

÷ =Sno

95001

Sno, Cno (sc)÷K

SC

Q = R÷S

除运算

性质

若q = r s，则q 是满足q x s r的最大关系

用基本代数运算来定义除运算

设r(R)和s(S)已知，且S R ：

r s = R-S(r) – R-S((R-S(r) x s) – R-S,S(r))

为什么：

–R-S,S(r) 重新排列r的属性顺序

–R-S(R-S(r) x s) – R-S,S(r))找出R-S(r)中的元组t，得到某些元组u s, tu r

除运算

赋值运算()可以使复杂的查询表达变得简单

使用赋值运算，可以把查询表达为一个顺序程序，该程序包括：

– 一系列赋值

– 一个其值被作为查询结果显示的表达式

对关系代数查询而言，赋值必须是赋给一个临时关系变量

例：可以把r s写作，

temp1 R-S (r)

temp2 R-S ((temp1 x s) – R-S,S (r))

result = temp1 – temp2

将右侧的表达式的结果赋给左侧的关系变量，该关系变量可以在后续的表达式中使用

赋值运算

例1：找出至少拥有一个“市区”和“住宅区”分支机构的账户的客户姓名

查询1：CN(BN =“Downtown”(depositor

CN(BN =“Uptown”(depositor

account))

account))

其中，CN表示“customer-name”，BN表示“branch-name”

查询2：customer-name, branch-name (depositor account)

temp(branch-name) ({(“Downtown”), (“Uptown”)})

示例

例2：找出拥有布鲁克林市所有分支机构的帐户的客户姓名

customer-name, branch-name (depositor account)

branch-name (branch-city = “Brooklyn” (branch))

例3：查询选修了全部课程的学生学号和姓名涉及表: 课程信息Course(cno, cname, pre-cno, score), 选课信息

SC(sno, cno, grade) , 学生信息Student(sno, sname, sex, age) 当涉及到求“全部”之类的查询，常用“除法” 找出全部课程号： Cno(Course)

找出选修了全部课程的学生的学号： Sno, Cno(SC) ÷ Cno(Course)

与Student表自然连接（连接条件Sno）获得学号、姓名

( Sno, Cno(SC) ÷ Cno(Course)) Sno,Sname(Student)

示例

64

• 学生=(学号, 姓名, 年龄, 性别, 系别)

• 课程=(课程号,课程名, 学分,先行课号)

• 教师=(职工号,姓名, 职称,年龄, 性别, 系别)

• 教材=(教材号,书名,作者名,出版号,出版社)

• 系=(系别,系名, 办公电话, 位置, 职工号)

• 选课=(学号, 课程号, 成绩)

• 参考=(课程号,教师号,教材号)

示例

示例示例

找年龄不小于20的男学生

AGE≥20 ∧ SEX=‘male’（S）

给出所有学生的姓名和年龄

SN, AGE(S)

找001号学生所选修的课程号

C#( S#=001 （SC））

求选修了001号或002号课程的学生号

方案1：∏S#(C# = 001∨ C# = 002(SC))

方案2：∏S#(C# = 001 (SC))∪∏S#(C# = 002(SC))

示例示例

求选修了001号而没有选002号课程的学生号

∏S#(C# = 001 (SC)) －∏S#(C# = 002(SC))

求未选修c1号课程的学生号

方案1：∏S#(S)－∏S#(C# = c1 (SC))

方案2：∏S#(C# ≠ c1 (SC))

都正确吗？

示例求计算机系学生的选课情况

(sno,sname,cname,score)

思考：有几种写法？哪种效率更高？

S SC C

Sno Sname Dept Sno Cno Score Cno Cname

S1 甲计 S1 C1 80 C1 DS

S2 乙软 S1 C2 90 C2 DB

S3 丙软 S2 C1 70

S4 丁计 S3 C2 60

示例

示例求学了c1和c2的学生学号

求各门课的最高成绩

SC

Sno Cno Score

S1 C1 80

S1 C2 90

S2 C1 70

S3 C1 60

Exercises• Practice Exercises:

– 2.1 2.5 2.7 2.8

• Exercises:

– 2.9 2.12 2.13

Thank you!

Q&A

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