# relational query languages

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Relational Query Languages. Naveen Ashish Calit2 &ICS UC-Irvine. Relational Languages. Relational Algebra ( procedural ) defines operations on tables Relational Calculus ( declarative ) based on first-order predicate calculus - PowerPoint PPT PresentationTRANSCRIPT

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Relational Query Languages

Naveen Ashish

Calit2 &ICS

UC-Irvine

Relational Languages

Relational Algebra (procedural) defines operations on tables

Relational Calculus (declarative) based on first-order predicate calculus

Every relational algebra query can be translated to relational calculus

Every safe relational calculus query can be translated to relational algebra.

Any language that is at least as expressive as relational algebra is said to be relationally complete.

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Relational Algebra Operators

Select: given a relation R and a predicate P, select tuples from R that satisfy P.

Project: given a relation R and a subset of its attributes X, return a relation which is the same as R except that all columns not in X are left out.

Rename: given a relation R and a name N, return a relation that is exactly the same as R except that it has a name N.

Cartesian Product: Given 2 relations R1and R2,.return a relation R3 whose tuples are the concatenation of tuples in R1 and R2

Union: Given relations R1 and R2, return a relation R3 which contains all tuples in R1 and R2

Set Difference: Given relations R1 and R2, return a relation R3 containing all tuples in R1 that are not in R2

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Selection Operation

Example: Employee(name,dept,sal)

Name Dept Salery Jane pharmacy 30,000Jack hardware 30,000Jill pharmacy 75,000

Name Dept Sal J oe toy 20,000Bill toy 12,000

Name Dept Sal

Jane pharmacy 30,000Jack hardware 30,000Jill pharmacy 75,000Joe toy 20,000Bill toy 12,000

d]Rection conselect[selRcondselectionσ or )(

Employee

select [sal > 20,000] Employee

select [(dept = toy) or (sal < 20,000)]

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Projection

Proj [list of attr of R] (R) or Π list of attr of R (R)A B C

Jane Toy 10,000

Jim Toy 20,000

June Complaint 20,000

A C

Jane Toy

John Toy

A

Jane

Jim

June

C B

10,000 Toy

20,000 Complaint

20,000 Toy

Proj[CB]R

R: S:

Proj[A]R

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Cartesian Product

Denoted by R x S

R: A B C S: A D

joe toy 10K joe jill

jack com 20K jack jill

RxS: R.A B C S.A D

joe toy 10K joe jill

joe toy 10K jack jill

jack com 20K joe jill

jack com 20K jack jill

Notice attribute naming strategy to disambiguate attribute names

attributes get the name, R.A, where A is attrib name, and R is the relation name from which attrib originates. If there is no possible ambiguity, relation name is dropped!

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Set Difference

Denoted by R – S (Illegal if R & S have different numbers of attributes or if

respective domains mismatch!)

R: A B S: A CJane Toy Jane ToyJim Toy John ComplaintJune Complaint

R - S = A BJim ToyJune Complaint

Note attributes in resulting relation take name from the first relation

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Rename Operator

Strategy used to disambiguate attribute names: For union and set difference operators, the resulting relation

takes the attribute names of the first relation For cartesian product, attributes are named as Relation-

name.attribute-name, where Relation name refers to the relation from which the attribute originally came.

Strategy will not disambiguate when the same relation appears multiple times in the relational query.

Let R(A,B) be a relation. Consider R x R what to name the attributes of the resulting relation?

Define a rename operator: denoted by rename[N]R or by N(R) returns a relation which is exactly same as R except it has the

name N

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Rename Operator

Consider relation Employee(name, dept, sal) List all the employees who work in the same department as Jill We first find the department(s) for which Jill works

Proj[dept](select[name = Jill] Employee) ---list of departments for which Jill works

To find out about all Employees working for this department, we need to reference the Employee table again: select[P] ( Employee x Proj[dept](select[name = Jill] Employee) ) where P is a selection predicate which requires dept values to

be equal. If we use Employee.dept in P it is ambiguous which instance of

Employee relation in the query the attribute refers to. To disambiguate, use rename operator: Proj[Employee.name](select[Employee.dept = Employee2.dept]

Employee x (Proj[dept] (select[name = Jill]( rename[Employee2](Employee))))

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Formal Definition of Relational Algebra

Basic Expressions: Relation in a database Constant Relations

General Expressions: constructed from basic ones. Let E1 and E2 be relational algebra expressions. Then the following are also expressions: E1 U E2, if arity of E1 = E2 and corresponding attributes are of

the same type E1 - E2, if arity of E1 = E2 and corresponding attributes are of

the same type E1 x E2, if attributes in E1 and E2 have different names Select[P](E1), where P is a predicate on attributes in E1 Proj[S](E1), where S is a list of some attributes in E1 rename[X](E1), where X is a new name for relation E1

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Additional Operators

Basic Relational Algebra operators are like assembly language.

Define more powerful operators that make the task of writing relational algebra queries easier

Each of these operators can be expressed in relational algebra and do not increase the expressibility of the language

Example: Intersection

R S = R - (R - S)

= { t | t R & t S }

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R join condition S = select[ join condition] (R x S)

join condition is of the form:

<condition> AND <condition> AND <condition>

where each condiition is of the form Ai Bj, where Ai is attribute of R Bj is attribute of S is a comparison operator {=, <, >, <=, >=, <>}

Joins

Example:

E(emp, dept) M(dept, mgr)

List all employees and their managers.

Proj[emp, mgr](select[E.dept = M.dept] (ExM))

can be represented as:

Proj[emp,mgr] ( E E.dept = M.dept M )

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Types of Joins

Theta-Join: if a join condition uses some comparison operator other than equality. E.g., list names of all managers who manage departments other

than Jill’s Proj[mgr]( select[emp = Jill](E ) (E.dept M.dept) M)

Equi-Join: if join conditions use only equality operator. E.g., list the manager’s name of Jill’s department Proj[mgr]( select[emp = Jill](E ) (E.dept M.dept) M)

Natural Join: special type of equi-join.. Let R and S be relations. Let attributes of R and S be denoted

by R and S respectively. Denote by R U S the union of the list of attributes of R and S Let list of attributes common to both R and S be {A1, A2, …, An} Natural join of R and S (denoted R S) is: Proj[R U S ] (R R.A1 = S.A1 and R.A2 = S.A2 and … and R.An =

S.An S) E.g., Proj[mgr]( select[emp = Jill](E ) M)

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Assignment Operator

Lots of time convenient to write relational algebra expressions in parts using assignment to temporary relational variables.

For this purpose use assignment operator, denoted by := E.g., Who makes more than their manager?

E(emp, dept, sal) M(mgr, dept) ESM(emp, sal, mgr) := Proj[emp, sal, mgr] (E M)(Proj[ESM.emp](ESM [mgr = E.emp & ESM.sal >E.sal] E) )

With the assignment operator, a query can be written as a sequential program consisting of a series of assignments followed by an expression whose value is the result of the query.

Examples

A query is a composition of basic relational algebra operators

Consider relations: customer(ssno, name, street, city) account(acctno, custid, balance)

list account balance of Sharad

balance ssnocustid

customeraccount ))((

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Examples

Diag Pat Dis

Winslett Strep

Liu Mono

Harandi Meningitis

Harandi Hepatitis

Liu Hepatitis

Outcome Pat Test Outcome

Winslett A T

Winslett B F

Liu B T

Harandi F T

Winslett E F

Harandi E F

Harandi G F

Winslett E T

To Diag Dis Test

Strep A

Mono B

Meningitis C

Hepatitis D

Encephalitis E

Meningitis F

Meningitis G

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1. Who has what disease?Diag

2. Who has a disease they have been tested for? Proj[pat](Diag | ToDiag | Outcome)

3. Who has a disease they tested positively for?Proj[pat](Diag | ToDiag | (select[outcome = ‘T’])Outcome))

4. Who has a disease that they tested both positively & negatively for? Temp1(pat, dis) := Proj[pat,dis](Diag | ToDiag

select[outcome = ‘T’])Outcome)Temp2(pat, dis) : = Proj[pat,dis](Diag | ToDiag

select[outcome = ‘T’])Outcome)Proj[pat](Temp1 Temp2)

Use better names!!

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Example of Queries in Relational Algebra

5. Who tested both positively and negatively for a disease, whether or not they have it?Testpos(pat, dis) = Proj[pat,dis](ToDiag | select[outcome = ‘T’])

Outcome)Testneg(pat, dis) = Proj[pat,dis](ToDiag | select[outcome = ‘T’])

Outcome)

(Testpos Testneg)[pat]

6. Who tested both positively & negatively for the same test? (Winslett)Proj[pat](Outcome | condition (rename[Outcome2](Outcome))where condition is: [Outcome.pat = Outcome2.pat & Outcome.test = Outcome2.test

& Outcome.outcome = Outcome2. outcome]

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7. What testable disease does no one have? (encephalitis)

Proj[dis]ToDiag Proj[dis]DiagNote technique: compute opposite of what you want, take a difference.

Use in hard queries with negation (‘no one’)8. What disease does more than one person have?

Proj[dis](Diag condition rename[Diag2](Diag)) where, condition is [Diag.pat Diag2.pat & Diag.dis = Diag2dis]

9. What disease does everyone have? clue that query is very hard

Disease(dis) := diag[dis]Patients(pat) := diag[pat] DiseasesNotEveryoneHas(dis) := Proj[dis]((Patients x Disease) -

Diag)Disease - Diseases Not Everyone Has

Note technique used! A very hard query might require taking the difference several times.

Outer Joins

E emp dept sal M mgr dept

Jones Missiles 10K Mendez Tanks

Chu Tanks 20K Frank Explosive

Swami Tanks 50K Jones Missiles

Barth Revolver 100K

Right outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K Mendeznull Explosives null Frank

left outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K MendezSwami Tanks 50K MendezBarth Revolver 100K null

Full outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K MendezSwami Tanks 50K MendezBarth Revolver 100K nullnull Explosives null Frank

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Recursive Closure Queries

Consider parent relation

parent Child Mom Dad sam Anda Chuck Chuck Donna Harvey Harvey Betty Reggie Reggie Cristie John It may be interesting to query the relation for the following:retrieve all the female anscestors of sam.retrieve all male ansestors of chuckretrieve harvey’s family treeretrieve all the descendants of cristie

Such queries (in general) require an unbounded application ofjoins of the parent relation to itself and CANNOT be representedin relational algebra.