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# Prof. Sin-Min Lee - PowerPoint PPT Presentation

Final Revision 3. CS 157B Lecture 23. Prof. Sin-Min Lee. Modeling. A database can be modeled as: a collection of entities, relationship among entities. An entity is an object that exists and is distinguishable from other objects. Example: specific person, company, event, plant

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CS 157B Lecture 23

Prof. Sin-Min Lee

• A database can be modeled as:

• a collection of entities,

• relationship among entities.

• An entityis an object that exists and is distinguishable from other objects.

• Example: specific person, company, event, plant

• Entities have attributes

• Example: people have names and addresses

• An entity set is a set of entities of the same type that share the same properties.

• Example: set of all persons, companies, trees, holidays

Entity Sets customer and loan

customer_id customer_ customer_ customer_ loan_ amount name street city number

• A relationship is an association among several entities

Example:HayesdepositorA-102customer entity relationship set account entity

• A relationship set is a mathematical relation among n 2 entities, each taken from entity sets

{(e1, e2, … en) | e1  E1, e2  E2, …, en  En}where (e1, e2, …, en) is a relationship

• Example:

(Hayes, A-102)  depositor

• An attribute can also be property of a relationship set.

• For instance, the depositor relationship set between entity sets customer and account may have the attribute access-date

• Refers to number of entity sets that participate in a relationship set.

• Relationship sets that involve two entity sets are binary (or degree two). Generally, most relationship sets in a database system are binary.

• Relationship sets may involve more than two entity sets.

• Relationships between more than two entity sets are rare. Most relationships are binary. (More on this later.)

• Example: Suppose employees of a bank may have jobs (responsibilities) at multiple branches, with different jobs at different branches. Then there is a ternary relationship set between entity sets employee, job, and branch

• Express the number of entities to which another entity can be associated via a relationship set.

• Most useful in describing binary relationship sets.

• For a binary relationship set the mapping cardinality must be one of the following types:

• One to one

• One to many

• Many to one

• Many to many

One to one

One to many

Note: Some elements in A and B may not be mapped to any

elements in the other set

Many to one

Many to many

Note: Some elements in A and B may not be mapped to any

elements in the other set

• A super key of an entity set is a set of one or more attributes whose values uniquely determine each entity.

• A candidate key of an entity set is a minimal super key

• Customer_id is candidate key of customer

• account_number is candidate key of account

• Although several candidate keys may exist, one of the candidate keys is selected to be the primary key.

• The combination of primary keys of the participating entity sets forms a super key of a relationship set.

• (customer_id, account_number) is the super key of depositor

• NOTE: this means a pair of entity sets can have at most one relationship in a particular relationship set.

• Example: if we wish to track all access_dates to each account by each customer, we cannot assume a relationship for each access. We can use a multivalued attribute though

• Must consider the mapping cardinality of the relationship set when deciding what are the candidate keys

• Need to consider semantics of relationship set in selecting the primary key in case of more than one candidate key

• Rectangles represent entity sets.

• Diamonds represent relationship sets.

• Lines link attributes to entity sets and entity sets to relationship sets.

• Ellipses represent attributes

• Double ellipses represent multivalued attributes.

• Dashed ellipses denote derived attributes.

• Underline indicates primary key attributes (will study later)

Roles Attributes

• Entity sets of a relationship need not be distinct

• The labels “manager” and “worker” are called roles; they specify how employee entities interact via the works_for relationship set.

• Roles are indicated in E-R diagrams by labeling the lines that connect diamonds to rectangles.

• Role labels are optional, and are used to clarify semantics of the relationship

Cardinality Constraints Attributes

• We express cardinality constraints by drawing either a directed line (), signifying “one,” or an undirected line (—), signifying “many,” between the relationship set and the entity set.

• One-to-one relationship:

• A customer is associated with at most one loan via the relationship borrower

• A loan is associated with at most one customer via borrower

One-To-Many Relationship Attributes

• In the one-to-many relationship a loan is associated with at most one customer via borrower, a customer is associated with several (including 0) loans via borrower

Many-To-One Relationships Attributes

• In a many-to-one relationship a loan is associated with several (including 0) customers via borrower, a customer is associated with at most one loan via borrower

Many-To-Many Relationship Attributes

• A customer is associated with several (possibly 0) loans via borrower

• A loan is associated with several (possibly 0) customers via borrower

• Total participation (indicated by double line): every entity in the entity set participates in at least one relationship in the relationship set

• E.g. participation of loan in borrower is total

• every loan must have a customer associated to it via borrower

• Partial participation: some entities may not participate in any relationship in the relationship set

• Example: participation of customer in borrower is partial

• Cardinality limits can also express participation constraints

• We allow at most one arrow out of a ternary (or greater degree) relationship to indicate a cardinality constraint

• E.g. an arrow from works_on to job indicates each employee works on at most one job at any branch.

• If there is more than one arrow, there are two ways of defining the meaning.

• E.g a ternary relationship R between A, B and C with arrows to B and C could mean

1. each A entity is associated with a unique entity from B and C or

2. each pair of entities from (A, B) is associated with a unique C entity, and each pair (A, C) is associated with a unique B

• Each alternative has been used in different formalisms

• To avoid confusion we outlaw more than one arrow

Binary Vs. Non-Binary Relationships Attributes

• Some relationships that appear to be non-binary may be better represented using binary relationships

• E.g. A ternary relationship parents, relating a child to his/her father and mother, is best replaced by two binary relationships, father and mother

• Using two binary relationships allows partial information (e.g. only mother being know)

• But there are some relationships that are naturally non-binary

• Example: works_on

• In general, any non-binary relationship can be represented using binary relationships by creating an artificial entity set.

• Replace R between entity sets A, B and Cby an entity set E, and three relationship sets:

1. RA, relating E and A 2.RB, relating E and B

3. RC, relating E and C

• Create a special identifying attribute for E

• Add any attributes of R to E

• For each relationship (ai , bi , ci) in R, create

1. a new entity eiin the entity set E 2. add (ei , ai ) to RA

3. add (ei , bi) to RB 4. add (ei , ci ) to RC

• Also need to translate constraints

• Translating all constraints may not be possible

• There may be instances in the translated schema thatcannot correspond to any instance of R

• Exercise: add constraints to the relationships RA, RB and RC to ensure that a newly created entity corresponds to exactly one entity in each of entity sets A, B and C

• We can avoid creating an identifying attribute by making E a weak entity set (described shortly) identified by the three relationship sets

Mapping Cardinalities affect ER Design Attributes

• Can make access-date an attribute of account, instead of a relationship attribute, if each account can have only one customer

• That is, the relationship from account to customer is many to one, or equivalently, customer to account is one to many

Weak Entity Sets Attributes

• An entity set that does not have a primary key is referred to as a weak entity set.

• The existence of a weak entity set depends on the existence of a identifying entity set

• it must relate to the identifying entity set via a total, one-to-many relationship set from the identifying to the weak entity set

• Identifying relationship depicted using a double diamond

• The discriminator(or partial key) of a weak entity set is the set of attributes that distinguishes among all the entities of a weak entity set.

• The primary key of a weak entity set is formed by the primary key of the strong entity set on which the weak entity set is existence dependent, plus the weak entity set’s discriminator.

Weak Entity Sets (Cont.) Attributes

• We depict a weak entity set by double rectangles.

• We underline the discriminator of a weak entity set with a dashed line.

• payment_number – discriminator of the payment entity set

• Primary key for payment – (loan_number, payment_number)

Weak Entity Sets (Cont.) Attributes

• Note: the primary key of the strong entity set is not explicitly stored with the weak entity set, since it is implicit in the identifying relationship.

• If loan_number were explicitly stored, payment could be made a strong entity, but then the relationship between payment and loan would be duplicated by an implicit relationship defined by the attribute loan_number common to payment and loan

More Weak Entity Set Examples Attributes

• In a university, a course is a strong entity and a course_offering can be modeled as a weak entity

• The discriminator of course_offering would be semester (including year) and section_number (if there is more than one section)

• If we model course_offering as a strong entity we would model course_number as an attribute.

Then the relationship with course would be implicit in the course_number attribute

Extended E-R Features: Specialization Attributes

• Top-down design process; we designate subgroupings within an entity set that are distinctive from other entities in the set.

• These subgroupings become lower-level entity sets that have attributes or participate in relationships that do not apply to the higher-level entity set.

• Depicted by a triangle component labeled ISA (E.g. customer “is a” person).

• Attribute inheritance – a lower-level entity set inherits all the attributes and relationship participation of the higher-level entity set to which it is linked.

Extended ER Features: Generalization Attributes

• A bottom-up design process – combine a number of entity sets that share the same features into a higher-level entity set.

• Specialization and generalization are simple inversions of each other; they are represented in an E-R diagram in the same way.

• The terms specialization and generalization are used interchangeably.

• Can have multiple specializations of an entity set based on different features.

• E.g. permanent_employee vs. temporary_employee, in addition to officer vs. secretary vs. teller

• Each particular employee would be

• a member of one of permanent_employee or temporary_employee,

• and also a member of one of officer, secretary, or teller

• The ISA relationship also referred to as superclass - subclassrelationship

• Constraint on which entities can be members of a given lower-level entity set.

• condition-defined

• Example: all customers over 65 years are members of senior-citizen entity set; senior-citizen ISA person.

• user-defined

• Constraint on whether or not entities may belong to more than one lower-level entity set within a single generalization.

• Disjoint

• an entity can belong to only one lower-level entity set

• Noted in E-R diagram by writing disjoint next to the ISA triangle

• Overlapping

• an entity can belong to more than one lower-level entity set

Design AttributesConstraintson a Specialization/Generalization (Cont.)

• Completenessconstraint -- specifies whether or not an entity in the higher-level entity set must belong to at least one of the lower-level entity sets within a generalization.

• total: an entity must belong to one of the lower-level entity sets

• partial: an entity need not belong to one of the lower-level entity sets

Aggregation Attributes

• Consider the ternary relationship works_on, which we saw earlier

• Suppose we want to record managers for tasks performed by an employee at a branch

3) Student (ufid, major) AttributesQuiz(Q_num, point_pos)Q_score(q_num,ufid,points_scored) (in TRC)

A) Which quizzes did a “CE” major score 100% on?

{ a[Q_num]: a  Quiz ,s STUDENT ; t  Q_score, s[ufid]=t[ufid]  s[ufid]=“CE”

 a[Q_num]=t[q_num]

t[points_scored]=100% }

Relational Algebra:

q_num [ Quiz |x| (major=“CE” Student) |x| (points_scored=“100” Q_score))

• Consider the following relational database schema: Attributes

• Pizza(pid, pname, size)

• Store(sname, phone, quality)

• Soldby(pid, sname, price)

• Express each of the following queries in the relational algebra.

• Find the name of all stores that sell both veggie and cheese pizza.

• sname(pname = ‘veggie’ Pizza |X| Soldby)

• sname(pname = ‘cheese’ Pizza |X| Soldby)

Pizza(pid, pname, size)

Store(sname, phone, quality)

Soldby(pid, sname, price)

Express each of the following queries in the relational algebra.

b) Find the names and phone numbers of all stores that sell good or excellent veggie pizza under \$10.

sname, phone((pname = ‘veggie’ Pizza) |X| (quality = ‘good’ Store) |X| (price < 10 Soldby))

sname, phone((pname = ‘veggie’ Pizza) |X| (quality = ‘excellent’ Store) |X| (price < 10 Soldby))

Student (ufid, major) AttributesQuiz(Q_num, point_pos)Q_score(q_num,ufid,points_scoredb) Who are the “CE” major who missed Quiz 3?

{ s [ufid]:s  STUDENT. t Q_SCORE

 s[ufid]=“CE”

t[q_num ]=“3”  t[points_scored] =“null”

 s[ufid]=t[ufid] }

Student (ufid, major) AttributesQuiz(Q_num, point_pos)Q_score(q_num,ufid,points_scoredc) Which students have a 0 score for 2 different quizzes?

{ s[ufid]: s  STUDENT t1, t2 Q_SCORE

s[ufid]=t1[ufid]= t2[ufid]

 t1[q_num] t2 [q_num]

 t1 [point_score]=t2 [point_score]=“0” }

7) Author (aid, name, phone) with key =(aid) Attributeswrote(aid, isbn, order) with key =(aid, Isbn)book(isbn, title, publisher,dte) with key=(isbn)

a) Find the names of authors who wrote or co wrote books published in 1995

{a [name]: a Author w  WROTE b  Book

 a.aid=w.aid  w.isbn=b.isbn  b.date=1995}

Author (aid, name, phone) with key =(aid) Attributeswrote(aid, isbn, order) with key =(aid, Isbn)book(isbn, title, publisher,dte) with key=(isbn)b) Find the names of authors who were always the first author of books they wrote

{ a[name]: a  Author,  w  Wrote, b  Book

 a.aid=w.aid  w.isbn = b.isbn  w.order=1 }

Author (aid, name, phone) with key =(aid) Attributeswrote(aid, isbn, order) with key =(aid, Isbn)book(isbn, title, publisher,dte) with key=(isbn) c) Find the AID of author (if any) who have written or co written books published by every publisher of book?

Let Publisher = {b[publisher]: b  Book }

{ a[aid]: a  Author  p  Publisher  w  Wrote & b*  Book

& a [aid]=w[aid] & w[isbn]= b*[isbn] & b*[publisher]=p }

6 Attributesc) Author (aid, name, phone) with key =(aid)wrote(aid, isbn, order) with key =(aid, Isbn)book(isbn, title, publisher,dte) with key=(isbn

Find the aid of authors (if any) who have written or co-written books published by every publisher of books? (Relational Algebra, TRC)

Relational Algebra:

Let T = publisherBook,

aid [ (Author |x| Wrote|x| Book) T ]

SQL Queries Attributes

• Principal form:

SELECT desired attributes

FROM tuple variables –– range over relations

WHERE condition about tuple variables;

Running example relation schema:

Beers(name, manf)

Likes(drinker, beer)

Sells(bar, beer, price)

Frequents(drinker, bar)

Example Attributes

What beers are made by Anheuser-Busch?

Beers(name, manf)

SELECT name

FROM Beers

WHERE manf = 'Anheuser-Busch';

• Note: single quotes for strings.

name

Bud

Bud Lite

Michelob

Union, Intersection, Difference Attributes

(SELECT name

FROM Person

WHERE City=“Seattle”)

UNION

(SELECT name

FROM Person, Purchase

WHERE buyer=name AND store=“The Bon”)

Similarly, you can use INTERSECT and EXCEPT.

You must have the same attribute names (otherwise: rename).

Formal Semantics Attributesof Single-Relation SQL Query

• Start with the relation in the FROM clause.

• Apply (bag) , using condition in WHERE clause.

• Apply (extended, bag)  using attributes in SELECT clause.

Equivalent Operational Semantics

Imagine a tuple variable ranging over all tuples of the relation. For each tuple:

• Check if it satisfies the WHERE clause.

• Print the values of terms in SELECT, if so.

Star as List of All Attributes Attributes

Beers(name, manf)

SELECT *

FROM Beers

WHERE manf = 'Anheuser-Busch';

name manf

Bud Anheuser-Busch

Bud Lite Anheuser-Busch

Michelob Anheuser-Busch

Renaming columns Attributes

Beers(name, manf)

SELECT name AS beer

FROM Beers

WHERE manf = 'Anheuser-Busch';

beer

Bud

Bud Lite

Michelob

Expressions as Values in Columns Attributes

Sells(bar, beer, price)

SELECT bar, beer,

price*120 AS priceInYen

FROM Sells;

bar beer priceInYen

Joe’s Bud 300

Sue’s Miller 360

… … …

• Note: no WHERE clause is OK.

Example each row, use that constant as an expression.

• Find the price Joe's Bar charges for Bud.

Sells(bar, beer, price)

SELECT price

FROM Sells

WHERE bar = 'Joe''s Bar' AND

beer = 'Bud';

• Note: two single-quotes in a character string represent one single quote.

• Conditions in WHERE clause can use logical operators AND, OR, NOT and parentheses in the usual way.

• Remember: SQL is case insensitive. Keywords like SELECT or AND can be written upper/lower case as you like.

• Only inside quoted strings does case matter.

Updates each row, use that constant as an expression.

UPDATE relation SET list of assignments WHERE condition.

Example

Drinker Fred's phone number is 555-1212.

UPDATE Drinkers

SET phone = '555-1212'

WHERE name = 'Fred';

Example

Make \$4 the maximum price for beer.

• Updates many tuples at once.

Sells(bar, beer, price)

UPDATE Sells

SET price = 4.00

WHERE price > 4.00;

Modifying the Database each row, use that constant as an expression.

Three kinds of modifications

• Insertions

• Deletions

Sometimes they are all called “updates”

Deleting or Modifying a Table each row, use that constant as an expression.

Deleting:

Exercise with care !!

Example:

DROP Person;

Altering: (adding or removing an attribute).

ALTER TABLE Person

ALTER TABLE Person

DROP age;

Example:

What happens when you make changes to the schema?

Default Values each row, use that constant as an expression.

Specifying default values:

CREATE TABLE Person(

name VARCHAR(30),

social-security-number INT,

age SHORTINT DEFAULT 100,

city VARCHAR(30) DEFAULT ‘Seattle’,

gender CHAR(1) DEFAULT ‘?’,

Birthdate DATE

The default of defaults: NULL

Conserving Duplicates each row, use that constant as an expression.

(SELECT name

FROM Person

WHERE City=“Seattle”)

UNIONALL

(SELECT name

FROM Person, Purchase

WHERE buyer=name AND store=“The Bon”)

Insertions each row, use that constant as an expression.

General form:

INSERT INTO R(A1,…., An) VALUES (v1,…., vn)

Example: Insert a new purchase to the database:

INSERT INTO Purchase(buyer, seller, product, store)

VALUES (‘Joe’, ‘Fred’, ‘wakeup-clock-espresso-machine’, ‘The Sharper Image’)

Missing attribute  NULL.

May drop attribute names if give them in order.

Insertions each row, use that constant as an expression.

INSERT INTO PRODUCT(name)

SELECT DISTINCT Purchase.product

FROM Purchase

WHERE Purchase.date > “10/26/01”

The query replaces the VALUES keyword.

Here we insert many tuples into PRODUCT

Insertion: an Example each row, use that constant as an expression.

Product(name, listPrice, category)

prodName is foreign key in Product.name

Suppose database got corrupted and we need to fix it:

corrupted

Purchase

Product

Task: insert in Product all prodNames from Purchase

Insertion: an Example each row, use that constant as an expression.

INSERT INTO Product(name)

SELECT DISTINCT prodName

FROM Purchase

WHERE prodName NOT IN (SELECT name FROM Product)

Insertion: an Example each row, use that constant as an expression.

INSERT INTO Product(name, listPrice)

SELECT DISTINCT prodName, price

FROM Purchase

WHERE prodName NOT IN (SELECT name FROM Product)

Depends on the implementation

Deletions each row, use that constant as an expression.

Example:

DELETE FROM PURCHASE

WHERE seller = ‘Joe’ AND

product = ‘Brooklyn Bridge’

Factoid about SQL: there is no way to delete only a single

occurrence of a tuple that appears twice

in a relation.

Deletion each row, use that constant as an expression.

DELETE FROM relation WHERE condition.

• Deletes all tuples satisfying the condition from the named relation.

Example

Sally no longer likes Bud.

Likes(drinker, beer)

DELETE FROM Likes

WHERE drinker = 'Sally' AND

beer = 'Bud';

Example

Make the Likes relation empty.

DELETE FROM Likes;

Updates each row, use that constant as an expression.

Example:

UPDATE PRODUCT

SET price = price/2

WHERE Product.name IN

(SELECT product

FROM Purchase

WHERE Date =‘Oct, 25, 1999’);

Patterns each row, use that constant as an expression.

• % stands for any string.

• _ stands for any one character.

• “Attribute LIKE pattern” is a condition that is true if the string value of the attribute matches the pattern.

• Also NOT LIKE for negation.

Example

Find drinkers whose phone has exchange 555.

SELECT name

FROM Drinkers

WHERE phone LIKE '%555-_ _ _ _’;

• Note patterns must be quoted, like strings.

Example each row, use that constant as an expression.

bar beer price

Joe's bar Bud NULL

SELECT bar

FROM Sells

WHERE price < 2.00 OR price >= 2.00;

UNKNOWN UNKNOWN

UNKNOWN

• Joe's Bar is not produced, even though the WHERE condition is a tautology.

Multi-relation Queries each row, use that constant as an expression.

• List of relations in FROM clause.

• Relation-dot-attribute disambiguates attributes from several relations.

Example

Find the beers that the frequenters of Joe's Bar like.

Likes(drinker, beer)

Frequents(drinker, bar)

SELECT beer

FROM Frequents, Likes

WHERE bar = 'Joe''s Bar' AND

Frequents.drinker = Likes.drinker;

Subqueries each row, use that constant as an expression.

Result of a select-from-where query can be used in the where-clause of another query.

Simplest Case: Subquery Returns a Single, Unary Tuple

Find bars that serve Miller at the same price Joe charges for Bud.

Sells(bar, beer, price)

SELECT bar

FROM Sells

WHERE beer = 'Miller' AND price =

(SELECT price

FROM Sells

WHERE bar = 'Joe''s Bar' AND

beer = 'Bud');

• Notice the scoping rule: an attribute refers to the most closely nested relation with that attribute.

• Parentheses around subquery are essential.

The each row, use that constant as an expression.IN Operator

“Tuple IN relation” is true iff the tuple is in the relation.

Example

Find the name and manufacturer of beers that Fred likes.

Beers(name, manf)

Likes(drinker, beer)

SELECT *

FROM Beers

WHERE name IN

(SELECT beer

FROM Likes

WHERE drinker = 'Fred’);

• Also: NOT IN.

EXISTS each row, use that constant as an expression.

“EXISTS(relation)” is true iff the relation is nonempty.

Example

Find the beers that are the unique beer by their manufacturer.

Beers(name, manf)

SELECT name

FROM Beers b1

WHERE NOT EXISTS

(SELECT *

FROM Beers

WHERE manf = b1.manf AND

name <> b1.name);

• Note scoping rule: to refer to outer Beers in the inner subquery, we need to give the outer a tuple variable, b1 in this example.

• A subquery that refers to values from a surrounding query is called a correlated subquery.

Quantifiers each row, use that constant as an expression.

ANY and ALL behave as existential and universal quantifiers, respectively.

• Beware: in common parlance, “any” and “all” seem to be synonyms, e.g., “I am fatter than any of you” vs. “I am fatter than all of you.” But in SQL:

Example

Find the beer(s) sold for the highest price.

Sells(bar, beer, price)

SELECT beer

FROM Sells

WHERE price >= ALL(

SELECT price

FROM Sells);

Class Problem

Find the beer(s) not sold for the lowest price.

Union, Intersection, Difference each row, use that constant as an expression.

“(subquery) UNION (subquery)” produces the union of the two relations.

• Similarly for INTERSECT, EXCEPT = intersection and set difference.

• But: in Oracle set difference is MINUS, not EXCEPT.

Example

Find the drinkers and beers such that the drinker likes the beer and frequents a bar that serves it.

Likes(drinker, beer)

Sells(bar, beer, price)

Frequents(drinker, bar)

(SELECT * FROM Likes)

INTERSECT

(SELECT drinker, beer

FROM Sells, Frequents

WHERE Frequents.bar = Sells.bar

);

Example each row, use that constant as an expression.

Find the different prices charged for beers.

Sells(bar, beer, price)

SELECT DISTINCT price

FROM Sells;

Join-Based Expressions each row, use that constant as an expression.

A number of forms are provided.

• Can be used either stand-alone (in place of a select-from-where) or to define a relation in the FROM-clause.

R NATURAL JOIN S

R JOIN S ON condition

e.g., condition:R.B=S.B

R CROSS JOIN S

R OUTER JOIN S

• Outerjoin can be modified by:

1. Optional NATURAL in front.

2. Optional ON condition at end.

3. Optional LEFT, RIGHT, or FULL (default) before OUTER.

• LEFT = pad (with NULL) dangling tuples of R only; RIGHT = pad dangling tuples of S only.

Aggregations each row, use that constant as an expression.

Sum, avg, min, max, and count apply to attributes/columns. Also, count(*) applies to tuples.

• Use these in lists following SELECT.

Example

Find the average price of Bud.

Sells(bar, beer, price)

SELECT AVG(price)

FROM Sells

WHERE beer = 'Bud';

• Counts each tuple (presumably each bar that sells Bud) once.

Class Problem

What would we do if Sells were a bag?

Eliminating Duplicates each row, use that constant as an expression.Before Aggregation

Find the number of different prices at which Bud is sold.

Sells(bar, beer, price)

SELECT COUNT(DISTINCT price)

FROM Sells

WHERE beer = 'Bud';

• DISTINCT may be used in any aggregation, but typically only makes sense with COUNT.

Grouping each row, use that constant as an expression.

Follow select-from-where by GROUP BY and a list of attributes.

• The relation that is the result of the FROM and WHERE clauses is grouped according to the values of these attributes, and aggregations take place only within a group.

Example

Find the average sales price for each beer.

Sells(bar, beer, price)

SELECT beer, AVG(price)

FROM Sells

GROUP BY beer;

Example each row, use that constant as an expression.

Find, for each drinker, the average price of Bud at the bars they frequent.

Sells(bar, beer, price)

Frequents(drinker, bar)

SELECT drinker, AVG(price)

FROM Frequents, Sells

WHERE beer = 'Bud' AND

Frequents.bar = Sells.bar

GROUP BY drinker;

• Note: grouping occurs after the  and  operations.

Restriction on each row, use that constant as an expression.SELECT Lists With Aggregation

If any aggregation is used, then each element of a SELECT clause must either be aggregated or appear in a group-by clause.

Example

• The following might seem a tempting way to find the bar that sells Bud the cheapest:

Sells(bar, beer, price)

SELECT bar, MIN(price)

FROM Sells

WHERE beer = 'Bud';

• But it is illegal in SQL.

Problem

How would we find that bar?

HAVING each row, use that constant as an expression. Clauses

HAVING clauses are selections on groups, just as WHERE clauses are selections on tuples.

• Condition can use the tuple variables or relations in the FROM and their attributes, just like the WHERE can.

• But the tuple variables range only over the group.

• And the attribute better make sense within a group; i.e., be one of the grouping attributes.

Example each row, use that constant as an expression.

Find the average price of those beers that are either served in at least 3 bars or manufactured by Anheuser-Busch.

Beers(name, manf)

Sells(bar, beer, price)

SELECT beer, AVG(price)

FROM Sells

GROUP BY beer

HAVING COUNT(*) >= 3 OR

beer IN (

SELECT name

FROM Beers

WHERE manf = 'Anheuser-Busch'

);