Database System Concepts

Transcription

1 s Slides (fortemente) baseados nos slides oficiais do livro c Silberschatz, Korth and Sudarshan. Chapter 2: Model Departamento de Engenharia Informática Instituto Superior Técnico 1 st Semester 2009/2010

2 Outline 1 s s

3 Outline 1 s s

4 Example of a Relation s customer_name customer_street customer_city Jones Main Harrison Smith North Rye Curry North Rye Lindsay Park Pittsfield

5 Basic Structure s Formally, given sets D 1, D 2,..., D n, a relation r is a subset of D 1 D 2 D n Thus, a relation is a set of n-tuples (a 1,a 2,...,a n ), where each a i D i Example: If Then customer name ={Jones, Smith, Curry, Lindsay} customer street ={Main, North, Park} customer city ={Harrison, Rye, Pittsfield} r = {(Jones, Main, Harrison),(Smith, North, Rye), is a relation over (Curry, North, Rye),(Lindsay, Park, Pittsfield)} customer_name customer_street customer_city

6 Attribute Types s Each attribute of a relation has a name The set of allowed values for each attribute is called the domain attribute Attribute values are (normally) required to be atomic; that is, indivisible Note: multivalued attribute values are not atomic Note: composite attribute values are not atomic The special value null is a member of every domain The null value causes complications in the definition of many operations We shall ignore the effect of null values in our main presentation and consider their effect later

7 Relation Schema s A 1, A 2,...,A n are attributes R = (A 1, A 2,...,A n ) is a relation schema Example: Customer schema=(customer name,customer street,customer city) r(r) is a relation on the relation schema R Example: Or: customer(customer schema) customer(customer name,customer street,customer city)

8 Relation Instance s 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 attributes customer name customer street customer city Jones Main Harrison Smith North Rye Curry North Rye Lindsay Park Pittsfield (columns) tuples (rows) Order of tuples is irrelevant (tuples may be stored in an arbitrary order)

9 s A database consists of multiple relations Information about an enterprise is broken up into parts, with each relation storing one part information account: stores information about accounts depositor: stores information about which customer owns which account customer : stores information about customers Why not store all information as a single relation? bank(account number, balance, customer name,...)

10 The bank database s

11 Keys s 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) by possible r we mean a relation r that could exist in the enterprise we are modeling. Example: {customer name, customer street} and {customer name} are both superkeys of Customer, if no two customers can possibly have the same name. K is a candidate key if K is minimal Example: {customer name} is a candidate key for Customer, since it is a superkey (assuming no two customers can possibly have the same name), and no subset of it is a superkey. Primary Key is a candidate key that will be taken as the main identifier for the tuples in the relation

12 Keys in the customer Relation s Is {customer_name, customer_street, customer_city} a superkey? Is {customer_street} a candidate key? Is {customer_street, customer_city} a superkey?

13 Keys in the depositor Relation s What are the possible candidate keys?

14 Outline 1 s s

15 Reduction of Relation s Primary keys allow entity sets and relationship sets to be expressed uniformly as relation schemas that represent the contents database. A database which conforms to an E-R diagram can be represented by a collection of schemas. For each entity set and relationship set there is a unique schema that is assigned the name corresponding entity set or relationship set. Each schema has a number of columns (generally corresponding to attributes), which have unique names.

16 Representing Strong Entity Sets as s A strong entity set reduces to a schema with the same attributes. customer(costumer id, costumer name, costumer street, costumer city)

17 Representing Weak Entity Sets as s A weak entity set becomes a table that includes a column for the primary key identifying strong entity set payment(loan number, payment number, payment date, payment amount)

18 Representing Relationship Sets as s A many-to-many relationship set is represented as a schema with attributes for the primary keys two participating entity sets, and any descriptive attributes of the relationship set. borrower(customer id, loan number) What about a n-ary relationship?

19 Representing Relationship Sets as s Many-to-one and one-to-many relationship sets can be represented with attributes for the primary keys two participating entity sets, using the attribute on the many side as primary key. borrower(customer id, loan number) For one-to-one relationship sets, either side can be chosen to act as the many side

20 Representing Relationship Sets as s Many-to-one and one-to-many relationship sets that are total on the many-side can be represented by adding an extra attribute to the many side, containing the primary key one side customer(costumer id, costumer name, costumer street, costumer city, loan number)

21 Composite and Multivalued Attributes s Composite attributes are flattened out by creating a separate attribute for each component attribute A multivalued attribute of an entity is represented by a separate schema customer(costumer id, first name, last name,...) phone numbers(costumer id, phone number)

22 Representing Specialization as - Method 1 s Form a schema for the higher-level entity Form a schema for each lower-level entity set, include primary key of higher-level entity set and local attributes person(person id, name, street, city) employee(person id, salary) customer(person id, credit rating) Drawback: getting information about an employee requires accessing two relations

23 Representing Specialization as - Method 2 Form a schema for each entity set with all local and inherited attributes s person(person id, name, street, city) employee(person id, name, street, city, salary) customer(person id, name, street, city, credit rating) If specialization is total, the schema for the generalized entity set (person) is not required to store information Drawback: street and city may be stored redundantly for people who are both customers and employees

24 Corresponding to Aggregation s To represent aggregation, create a schema containing primary key aggregated relationship, the primary key associated entity set any descriptive attributes manages(manager name, employee id, branch name, title) Schema works on is redundant provided we are willing to store null values for attribute manager name manages

25 Outline 1 s s

26 Query Languages s Language in which user requests information from the database. Categories of languages Procedural Non-procedural, or declarative Pure languages: algebra Tuple relational calculus Domain relational calculus Pure languages form underlying basis of query languages that people use.

27 s Procedural language Six basic operators select: σ project: π union: set difference: Cartesian product: rename: ρ The operators take one or two relations as inputs and produce a new relation as a result.

28 The Select Operation s Notation: σ p (r) p is called the selection predicate Defined as: σ p (r) = {t t r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by : (and), (or), (not) Each term is one of: < attribute > op < attribute > or < constant > where op is one of: =,, >,, <, Example of selection: σ branch name= Perryridge (account)

29 The Select Operation - Example s σ branch name= Perryridge (account)

30 The Select Operation - Example s Relation r σ A=B D>5 (r) =? A B C D a a 1 7 a b 5 7 b b 12 3 b b 23 10

31 The Select Operation - Example s Relation r σ A=B D>5 (r) =? A B C D a a 1 7 a b 5 7 b b 12 3 b b 23 10

32 The Select Operation - Example s Relation r σ A=B D>5 (r) =? A B C D a a 1 7 a b 5 7 b b 12 3 b b A B C D a a 1 7 b b 23 10

33 The Project Operation s Notation: π A1,A 2,...,A k (r) where A 1, A 2 are attribute names and r is a relation name. The result is defined as the relation of k columns obtained by erasing the columns that are not listed Duplicate rows removed from result, since relations are sets Example: To eliminate the branch name attribute of account π account number,balance (account)

34 The Project Operation - Example s π account number,balance (account) =?

35 The Project Operation - Example s Relation r π A,C (r) =? A B C a 10 1 a 20 1 b 30 1 b 40 2

36 The Project Operation - Example s Relation r π A,C (r) =? A B C a 10 1 a 20 1 b 30 1 b 40 2

37 The Project Operation - Example s Relation r π A,C (r) =? A A B C a 10 1 a 20 1 b 30 1 b 40 2 C a 1 a 1 b 1 b 2

38 The Project Operation - Example s Relation r π A,C (r) =? A A B C a 10 1 a 20 1 b 30 1 b 40 2 C a 1 a 1 b 1 b 2 = A C a 1 b 1 b 2

39 The Union Operation s Notation: Defined as: For r s to be valid: r s r s = {t t r or t s} 1 r and s must have the same arity (same number of attributes) 2 The attribute domains must be compatible (example: 2nd column of r deals with the same type of values as does the 2nd column of s) Example: to find all customers with either an account or a loan π customer name (depositor) π customer name (borrower)

40 The Union Operation - Example s π customer name (depositor) =? π customer name (borrower) =? π customer name (depositor) π customer name (borrower) =? Incorrect: π account number (depositor) π customer name (borrower)

41 The Union Operation - Example s π customer name (depositor) =? π customer name (borrower) =? π customer name (depositor) π customer name (borrower) =? Incorrect: π account number (depositor) π customer name (borrower)

42 The Union Operation - Example Relations r and s s r s =? r = A B a 1 a 2 b 1 s = A B a 2 b 3

43 The Union Operation - Example Relations r and s s r s =? r = A B a 1 a 2 b 1 s = A B a 2 b 3

44 The Union Operation - Example Relations r and s s r s =? r = A B a 1 a 2 b 1 A B a 1 a 2 b 1 b 3 s = A B a 2 b 3

45 The Set Difference Operation s Notation: Defined as: r s r s = {t t r and t s} For r s to be valid: 1 r and s must have the same arity 2 The attribute domains must be compatible Example: to find all customers with an account but not a loan π customer name (depositor) π customer name (borrower)

46 The Set Difference Operation - Example s π customer name (depositor) =? π customer name (borrower) =? π customer name (depositor) π customer name (borrower) =? Incorrect: π account number (depositor) π customer name (borrower)

47 The Set Difference Operation - Example s π customer name (depositor) =? π customer name (borrower) =? π customer name (depositor) π customer name (borrower) =? Incorrect: π account number (depositor) π customer name (borrower)

48 The Set Difference Operation - Example s Relations r and s A r s =? r = B a 1 a 2 b 1 s = A B a 2 b 3

49 The Set Difference Operation - Example s Relations r and s A r s =? r = B a 1 a 2 b 1 s = A B a 2 b 3

50 The Set Difference Operation - Example s Relations r and s A r s =? r = B a 1 a 2 b 1 A B a 1 b 1 s = A B a 2 b 3

51 The Cartesian Product Operation s Notation: Defined as: r s r s = {tq t r and q s} Assume that attributes of r(r) and s(s) are disjoint. (That is, R S = ). If attributes of r(r) and s(s) are not disjoint, then renaming must be used.

52 The Cartesian Product Operation - Example s Relations r and s r s =? r = A B a 1 b 2 s = C D E a 10 x b 10 x b 20 y c 10 y

53 The Cartesian Product Operation - Example s Relations r and s r s =? r = A B a 1 b 2 s = C D E a 10 x b 10 x b 20 y c 10 y

54 The Cartesian Product Operation - Example s Relations r and s r s =? r = A B a 1 b 2 s = A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y C D E a 10 x b 10 x b 20 y c 10 y

55 The Rename Operation s Allows us to name, and therefore to refer to, the results of relational-algebra expressions. Allows us to refer to a relation by more than one name. Example: ρ X (E) returns the expression E under the name X If a relational-algebra expression E has arity n, then ρ X(A1,A 2,...,A n)(e) returns the result of expression E under the name X, and with the attributes renamed to A 1, A 2,..., A n. ρ dep (depositor) dep(customer name, account number) ρ dep(cn,an) (depositor) dep(cn, an)

56 Operation Composition s Can build expressions using multiple operations Example: σ A=C (r s) A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y 1 r s a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y 2 σ A=C 3 σ A=C (r s) A B C D E a 1 a 10 x b 2 b 10 x b 2 b 20 y

57 Operation Composition s Can build expressions using multiple operations Example: σ A=C (r s) A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y 1 r s a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y 2 σ A=C 3 σ A=C (r s) A B C D E a 1 a 10 x b 2 b 10 x b 2 b 20 y

58 Operation Composition s Can build expressions using multiple operations Example: σ A=C (r s) A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y 1 r s a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y 2 σ A=C 3 σ A=C (r s) A B C D E a 1 a 10 x b 2 b 10 x b 2 b 20 y

59 Operation Composition s Can build expressions using multiple operations Example: σ A=C (r s) A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y 1 r s a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y 2 σ A=C 3 σ A=C (r s) A B C D E a 1 a 10 x b 2 b 10 x b 2 b 20 y

60 Operation Composition s Can build expressions using multiple operations Example: σ A=C (r s) A B C D E a 1 a 10 x a 1 b 10 x a 1 b 20 y 1 r s a 1 c 10 y b 2 a 10 x b 2 b 10 x b 2 b 20 y b 2 c 10 y 2 σ A=C 3 σ A=C (r s) A B C D E a 1 a 10 x b 2 b 10 x b 2 b 20 y

61 Outline 1 s s

62 s We define additional operations that do not add any power to the relational algebra, but that simplify common queries. Set intersection: Natural join: Division: Assignment:

63 The Set Intersection Operation s Notation: Defined as: r s r s = {t t r and t s} Note: r s = r (r s) For r s to be valid: 1 r and s must have the same arity 2 The attribute domains must be compatible Example: to find all customers with both an account and a a loan π customer name (depositor) π customer name (borrower)

64 The Set Intersection Operation - Example s Relations r and s A r = B a 1 a 2 b 1 s = A B a 2 b 3 r s =?

65 The Set Intersection Operation - Example s Relations r and s A r = B a 1 a 2 b 1 s = A B a 2 b 3 r s =?

66 The Set Intersection Operation - Example s Relations r and s A r = B a 1 a 2 b 1 s = A B a 2 b 3 r s =? A B a 2

67 The Natural Join Operation s Example Notation: r s Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R S obtained as follows: Consider each pair of tuples t r from r and t s from s If t r and t s have the same value on each attributes in R S, add a tuple t to the result, where t has the same value as t r on r t has the same value as t s on s R = (A,B,C,D), S = (E,B,D) Result schema = (A,B,C,D,E) r s = π r.a,r.b,r.c,r.d,s.e (σ r.b=s.b r.d=s.d (r s))

68 The Natural Join Operation - Example Relations r and s A B C D B D E s r = a 1 a x b 2 c x c 4 b y a 1 c x d 2 b y s = 1 x a 3 x b 1 x c 2 y d 3 y e r s =?

69 The Natural Join Operation - Example Relations r and s A B C D B D E s r = a 1 a x b 2 c x c 4 b y a 1 c x d 2 b y s = 1 x a 3 x b 1 x c 2 y d 3 y e r s =?

70 The Natural Join Operation - Example s Relations r and s A B C D r = r s =? a 1 a x b 2 c x c 4 b y a 1 c x d 2 b y s = A B C D E a 1 a x a a 1 a x c a 1 c x a a 1 c x c d 2 b y d B D E 1 x a 3 x b 1 x c 2 y d 3 y e

71 The Division Operation s Notation: r s Let r and s be relations on schemas R and S respectively, where R = (A 1,...,A m,b 1,...,B n ) S = (B 1,...,B n ) Then, r s is a relation on schema R S obtained, such that: r s = {t t π R S (t) u s (tu r)} Where tu means the concatenation of tuples t and u to produce a single tuple Note: r s = π R S (r) π R S ((π R S (r) s) π R S,S (r))

72 The Division Operation - Example Relations r and s A B s r s =? r = a 1 a 2 a 3 b 1 c 1 d 1 d 3 d 4 e 6 e 1 b 2 s = B 1 2

73 The Division Operation - Example Relations r and s A B s r s =? r = a 1 a 2 a 3 b 1 c 1 d 1 d 3 d 4 e 6 e 1 b 2 s = B 1 2

74 The Division Operation - Example Relations r and s A B s r s =? r = a 1 a 2 a 3 b 1 c 1 d 1 d 3 d 4 e 6 e 1 b 2 A a b s = B 1 2

75 Another Division Example Relations r and s A B C D E s r = r s =? a x a x 1 a x c x 1 a x c y 1 b x c x 1 b x c y 3 c x c x 1 c x c y 1 c x b y 1 s = D E x 1 y 1

76 Another Division Example Relations r and s A B C D E s r = r s =? a x a x 1 a x c x 1 a x c y 1 b x c x 1 b x c y 3 c x c x 1 c x c y 1 c x b y 1 s = D E x 1 y 1

77 Another Division Example Relations r and s A B C D E s r = r s =? a x a x 1 a x c x 1 a x c y 1 b x c x 1 b x c y 3 c x c x 1 c x c y 1 c x b y 1 A B C a x c c x c s = D E x 1 y 1

78 The Assignement Operation s The assignment operation ( ) provides a convenient way to express complex queries. Write query as a sequential program consisting of a series of assignmets followed by an expression whose value is displayed as a result query. Assignment must always be made to a temporary relation variable. Example: write r s as temp1 π R S (r) temp2 π R S ((temp1 s) π R S,S (r)) result = temp1 temp2 The result to the right is assigned to the relation variable on the left May use variable in subsequent expressions.

79 Outline 1 s s

80 s Generalized projection Aggregate functions Outer join

81 Generalized Projection s Example Extends the projection operation by allowing arithmetic functions to be used in the projection list π F1,F 2,...,F n (E) E is any relational-algebra expression Each of F 1, F 2,...,F n are are arithmetic expressions involving constants and attributes in the schema of E. Given relation credit info(customer name, limit, credit balance) find how much more each person can spend. π customer name,limit credit balance (credit info)

82 Generalized Projection - Example s Relation r π A,C 10 (r) =? A B C a 10 1 a 20 1 b 30 1 b 40 2

83 Generalized Projection - Example s Relation r π A,C 10 (r) =? A B C a 10 1 a 20 1 b 30 1 b 40 2

84 Generalized Projection - Example s Relation r π A,C 10 (r) =? A B C a 10 1 a 20 1 b 30 1 b 40 2 A C*10 a 10 b 10 b 20

85 Aggregate Functions and s Aggregation function takes a collection of values and returns a single value as a result. avg: average value min: minimum value max: maximum value sum: sum of values count: number of values Aggregate operation in relational algebra G 1,G 2,...,G n G F1 (A 1 ),F 2 (A 2 ),...,F n(a n)(e) E is any relational-algebra expression G 1,G 2...,G n is a list of attributes on which to group (can be empty) Each F i is an aggregate function Each A i is an attribute name

86 Aggregate Operation - Example s Relation r G sum(c) (r) =? A B C a a 7 a b 7 b b 3 b b 10

87 Aggregate Operation - Example s Relation r G sum(c) (r) =? A B C a a 7 a b 7 b b 3 b b 10

88 Aggregate Operation - Example s Relation r G sum(c) (r) =? A B C a a 7 a b 7 b b 3 b b 10 sum(c) 27

89 Aggregate Operation - Example s Relation account grouped by branch name branch name account number balance Perryridge A Perryridge A Brighton A Brighton A Redwood A

90 Aggregate Operation - Example s Relation account grouped by branch name branch name account number balance Perryridge A Perryridge A Brighton A Brighton A Redwood A branch nameg sum(balance) (account)

91 Aggregate Operation - Example s Relation account grouped by branch name branch name account number balance Perryridge A Perryridge A Brighton A Brighton A Redwood A branch nameg sum(balance) (account) branch name sum(balance) Perryridge 1300 Brighton 1500 Redwood 700

92 Aggregate Functions (cont.) s 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 branch nameg sum(balance) as sum balance (account) branch name sum balance Perryridge 1300 Brighton 1500 Redwood 700

93 Outer Join s An extension join operation that avoids loss of information. Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result join. Uses null values: null signifies that the value is unknown or does not exist All comparisons involving null are (roughly speaking) false by definition. We shall study precise meaning of comparisons with nulls later

94 Outer Join - Example s Relations loan and borrower loan number branch name amount L-170 Downtown 3000 L-230 Redwood 4000 L-260 Perryridge 1700 customer name loan number Jones L-170 Smith L-230 Hayes L-155

95 Outer Join - Example s Relations loan and borrower loan number branch name amount L-170 Downtown 3000 L-230 Redwood 4000 L-260 Perryridge 1700 Inner join: loan borrower customer name loan number Jones L-170 Smith L-230 Hayes L-155 loan number branch name amount customer name L-170 Downtown 3000 Jones L-230 Redwood 4000 Smith

96 Outer Join - Example s Relations loan and borrower loan number branch name amount L-170 Downtown 3000 L-230 Redwood 4000 L-260 Perryridge 1700 Left outer join: loan borrower customer name loan number Jones L-170 Smith L-230 Hayes L-155 loan number branch name amount customer name L-170 Downtown 3000 Jones L-230 Redwood 4000 Smith L-260 Perryridge 1700 null

97 Outer Join - Example s Relations loan and borrower loan number branch name amount L-170 Downtown 3000 L-230 Redwood 4000 L-260 Perryridge 1700 Right outer join: loan borrower customer name loan number Jones L-170 Smith L-230 Hayes L-155 loan number branch name amount customer name L-170 Downtown 3000 Jones L-230 Redwood 4000 Smith L-155 null null Hayes

98 Outer Join - Example Relations loan and borrower s loan number branch name amount L-170 Downtown 3000 L-230 Redwood 4000 L-260 Perryridge 1700 customer name loan number Jones L-170 Smith L-230 Hayes L-155 Full outer join: loan borrower loan number branch name amount customer name L-170 Downtown 3000 Jones L-230 Redwood 4000 Smith L-260 Perryridge 1700 null L-155 null null Hayes

99 Outline 1 s s

100 s It is possible for tuples to have a null value, denoted by null, for some ir attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null Aggregate functions simply ignore null values (as in SQL) For duplicate elimination and grouping, null is treated like any other value, and two nulls are assumed to be the same (as in SQL)

101 s Comparisons with null values return the special truth value: unknown x y x OR y unknown true true unknown false unknown unknown unknown unknown x y x AND y unknown true unknown unknown false false unknown unknown unknown x unknown NOT x unknown Result of select predicate is treated as false if it evaluates to unknown

102 Outline 1 s s

103 s The content database may be modified using the following operations: Deletion Insertion Updating All these operations are expressed using the assignment operator.

104 Deletion s A delete request is expressed similarly to a query, except instead of displaying tuples to the user, the selected tuples are removed from the database. Can delete only whole tuples; cannot delete values on only particular attributes A deletion is expressed in relational algebra by: r r E where r is a relation and E is a relational algebra query.

105 Deletion Examples Delete all account records in the Perryridge branch. s Delete all loan records with amount in the range of 0 to 50 Delete all accounts at branches located in Needham

106 Deletion Examples s Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) Delete all loan records with amount in the range of 0 to 50 Delete all accounts at branches located in Needham

107 Deletion Examples s Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) Delete all loan records with amount in the range of 0 to 50 Delete all accounts at branches located in Needham

108 Deletion Examples s Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham

109 Deletion Examples s Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham

110 Deletion Examples Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) s Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham r 1 σ branch city= Needham (account branch)

111 Deletion Examples Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) s Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham r 1 σ branch city= Needham (account branch) r 2 π branch name,account number,balance (r 1 )

112 Deletion Examples Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) s Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham r 1 σ branch city= Needham (account branch) r 2 π branch name,account number,balance (r 1 ) account account r 2

113 Deletion Examples Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) s Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham r 1 σ branch city= Needham (account branch) r 2 π branch name,account number,balance (r 1 ) account account r 2 r 3 π customer name,account number (r 2 depositor)

114 Deletion Examples Delete all account records in the Perryridge branch. account account σ branch name= Perryridge (account) s Delete all loan records with amount in the range of 0 to 50 loan loan σ amount 0 and amount 50 (loan) Delete all accounts at branches located in Needham r 1 σ branch city= Needham (account branch) r 2 π branch name,account number,balance (r 1 ) account account r 2 r 3 π customer name,account number (r 2 depositor) depositor depositor r 3

115 Insertion s To insert data into a relation, we either: specify a tuple to be inserted write a query whose result is a set of tuples to be inserted In relational algebra, an insertion is expressed by: r r E where r is a relation and E is a relational algebra expression. The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.

116 Insertion Examples Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. s Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

117 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

118 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} depositor depositor {( Smith, A 973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

119 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} depositor depositor {( Smith, A 973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

120 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} depositor depositor {( Smith, A 973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r 1 σ branch name= Perryridge (borrower loan)

121 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} depositor depositor {( Smith, A 973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r 1 σ branch name= Perryridge (borrower loan) account account π branch name,loan number,200 (r 1 )

122 Insertion Examples s Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( Perryridge, A 973, 1200)} depositor depositor {( Smith, A 973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r 1 σ branch name= Perryridge (borrower loan) account account π branch name,loan number,200 (r 1 ) depositor depositor π customer name,loan number (r 1 )

123 Updating s A mechanism to change a value in a tuple without charging all values in the tuple Use the generalized projection operator to do this task r π F1,F 2,...,F n (r) Each F i is either the i-th attribute of r, if the i-th attribute is not updated, or, if the attribute is to be updated F i is an expression, involving only constants and the attributes of r, which gives the new value for the attribute

124 Updating Examples s Make interest payments by increasing all balances by 5 percent. Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent

125 Updating Examples s Make interest payments by increasing all balances by 5 percent. account π account number,branch name,balance 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent

126 Updating Examples s Make interest payments by increasing all balances by 5 percent. account π account number,branch name,balance 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent

127 Updating Examples s Make interest payments by increasing all balances by 5 percent. account π account number,branch name,balance 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent r 1 σ balance>10000 (account)

128 Updating Examples s Make interest payments by increasing all balances by 5 percent. account π account number,branch name,balance 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent r 1 σ balance>10000 (account) r 2 σ balance (account)

129 Updating Examples s Make interest payments by increasing all balances by 5 percent. account π account number,branch name,balance 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent r 1 σ balance>10000 (account) r 2 σ balance (account) account π account number,branch name,balance 1.06 (r 1 ) π account number,branch name,balance 1.05 (r 2 )

130 s End of Chapter 2