1 database systems lecture #7 yan pan school of software, sysu 2011

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Database SystemsLecture #7

Yan Pan

School of Software, SYSU

2011

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Agenda Subqueries, etc.

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Relation operators Basic operators:

Selection: Projection: Cartesian Product:

Other set-theoretic ops: Union: Intersection: Difference: -

Additional operators: Joins (natural, equijoin, theta join, semijoin) Renaming: Grouping…

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New-style join types Cross joins (simplest):

FROM A CROSS JOIN B

Inner joins (regular joins): FROM A [INNER] JOIN B ON …

Natural join: FROM A NATURAL JOIN B; Joins on common fields and merges

Outer joins (later) No dangling rows

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SQL e.g. with tuple vars Reps(ssn, name, etc.) Clients(ssn, name, rssn)

Q: Who are George’s clients, in SQL?

Conceptually: Clients.name(Reps.name=“George” and Reps.ssn=rssn(Reps x

Clients))

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New topic: Subqueries Powerful feature of SQL: one clause can

contain other SQL queries Anywhere where a value or relation is allowed

Several ways: Selection single constant (scalar) in SELECT Selection single constant (scalar) in WHERE Selection relation in WHERE Selection relation in FROM

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Standard multi-table example Purchase(prodname, buyerssn, etc.) Person(name, ssn, etc.) Q: What did Christo buy?

As usual, need to AND on equality identifying ssn’s row and buyerssn’s row

SELECT Purchase.prodnameFROM Purchase, PersonWHERE buyerssn = ssn AND name = 'Christo'

SELECT Purchase.prodnameFROM Purchase, PersonWHERE buyerssn = ssn AND name = 'Christo'

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Subquery motivation Purchase(prodname, buyerssn, etc.) Person(name, ssn, etc.) Q: What did Christo buy?

Natural intuition: Go find Christo ’s ssn Then find purchases

SELECT ssnFROM PersonWHERE name = 'Christo'

SELECT ssnFROM PersonWHERE name = 'Christo'

SELECT Purchase.prodnameFROM PurchaseWHERE buyerssn = Christo’s-ssn

SELECT Purchase.prodnameFROM PurchaseWHERE buyerssn = Christo’s-ssn

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Subqueries Subquery: copy in Christo ’s selection for his ssn:

The subquery returns one value, so the = is valid If it returns more (or fewer), we get a run-time error

SELECT Purchase.prodnameFROM PurchaseWHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo')

SELECT Purchase.prodnameFROM PurchaseWHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo')

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Operators on subqueries Several new operators applied to (unary)

selections:1. IN R

2. EXISTS R

3. UNIQUE R

4. s > ALL R

5. s > ANY R

6. x IN R > is just an example op Each expression can be negated with NOT

Guifeng Zheng, DBMS, SS/SYSU 11

Subqueries with IN Product(prodname,maker), Person(name,ssn),

Purchase(buyerssn,product) Q: Find companies Martha bought products from Strategy:

1. Find Martha’s ssn2. Find products listed with that ssn as buyer3. Find company names of those products

SELECT DISTINCT Product.makerFROM ProductWHERE prodname IN (SELECT product FROM Purchase WHERE buyerssn =

(SELECT ssn FROM Person

WHERE name = 'Martha'))

SELECT DISTINCT Product.makerFROM ProductWHERE prodname IN (SELECT product FROM Purchase WHERE buyerssn =

(SELECT ssn FROM Person

WHERE name = 'Martha'))

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Subqueries returning relations Equivalent to:

Or:

SELECT DISTINCT Product.makerFROM Product, Purchase, PeopleWHERE prodname = product AND buyerssn = ssn AND name = 'Martha'

SELECT DISTINCT Product.makerFROM Product, Purchase, PeopleWHERE prodname = product AND buyerssn = ssn AND name = 'Martha'

SELECT DISTINCT Product.makerFROM Product JOIN Purchase ON prodname=product JOIN People ON buyerssn=ssnWHERE name = 'Martha'

SELECT DISTINCT Product.makerFROM Product JOIN Purchase ON prodname=product JOIN People ON buyerssn=ssnWHERE name = 'Martha'

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FROM subqueries Motivation for another way:

suppose we’re given Martha’s purchases Then could just cross with Products to get product makers

Substitute (named) subquery for Martha’s purchases

SELECT makerFROM Product, (SELECT product FROM Purchase WHERE buyerssn =

(SELECT ssn FROM Person WHERE name = 'Martha')) Marthas

WHERE Product.name = Marthas.product

SELECT makerFROM Product, (SELECT product FROM Purchase WHERE buyerssn =

(SELECT ssn FROM Person WHERE name = 'Martha')) Marthas

WHERE Product.name = Marthas.product

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Complex RA Expressions Scenario:

1. Purchase(pid, seller-ssn, buyer-ssn, etc.)

2. Person(ssn, name, etc.)

3. Product(pid, name, etc.)

Q: Who (give names) bought gizmos from Dick? Where to start? Purchase uses pid, ssn, so must get them…

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Complex RA Expressions

Person Purchase Person Product

name='Dick' name='Gizmo'

pid ssn

seller-ssn=ssn

pid=pid

buyer-ssn=Person.ssn

name

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Translation to SQL

We’re converting the tree on the last slide into SQL The result of the query should be the names indicated above One step at a time, we’ll make the query more complete, until

we’ve translated the English-language description to an actual SQL query

We’ll also simplify the query when possible

(the names of the people who bought gadgets from Dick)

(the names of the people who bought gadgets from Dick)

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Translation to SQL

Blue type = actual SQL Black italics = description of subquery

Note: the subquery above consists of purchase records, except with the info describing the buyers attached In the results, the column header for name will be 'buyer'

SELECT DISTINCT name buyer FROM

(the info, along with buyer names, for purchases of gadgets sold by Dick)


(the info, along with buyer names, for purchases of gadgets sold by Dick)

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Translation to SQL

Note: the subquery in this version is being given the name P2 We’re pairing our rows from Person with rows from P2


(SELECT *FROM Person, (the purchases of gadgets from Dick) P2

WHERE Person.ssn = P2.buyer-ssn)


(SELECT *FROM Person, (the purchases of gadgets from Dick) P2

WHERE Person.ssn = P2.buyer-ssn)

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Translation to SQL

We simplified by combining the two SELECTs

SELECT DISTINCT name buyer

FROM Person, (the purchases of gadgets from Dick) P2

WHERE Person.ssn = P2.buyer-ssn


FROM Person, (the purchases of gadgets from Dick) P2



Translation to SQL

P2 is still the name of the subquery It’s just been filled in with a query that contains two

subqueries Outer parentheses are bolded for clarity


FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn)

AND pid = (the id of gadget)) P2



FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn)

AND pid = (the id of gadget)) P2


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Translation to SQL

Now the subquery to find Dick’s ssn is filled in


FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2



FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2


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Translation to SQL

And now the subquery to find Gadget’s product id is filled in, too Note: the SQL simplified by using subqueries

Not used in relational algebra


FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2WHERE Person.ssn = P2.buyer-ssn


FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2WHERE Person.ssn = P2.buyer-ssn

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George’s-neighbors with subqueries People(ssn, name, street, city, state, state) Q: Who lives on George’s street?

A: First, find George: name='George'(People)

And get George’s street/city/state: street(name='George'(People))

Look up people on that street…

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Next: ALL op

Employees(name, job, divid, salary)Find which employees are paid more than all the programmers

SELECT nameFROM EmployeesWHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer')

SELECT nameFROM EmployeesWHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer')


ANY/SOME op

Employees(name, job, divid, salary)Find which employees are paid more than at least one vice president

SELECT nameFROM EmployeesWHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP')

SELECT nameFROM EmployeesWHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP')


ANY/SOME op

Employees(name, job, divid, salary)Find which employees are paid more than at least one vice president

SELECT nameFROM EmployeesWHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP')

SELECT nameFROM EmployeesWHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP')

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Existential/Universal ConditionsEmployees(name, job, divid, salary)

Division(name, id, head)

Find all divisions with an employee whose salary is > 100000

Existential: easy!

SELECT DISTINCT Division.nameFROM Employees, DivisionWHERE salary > 100000 AND divid=id

SELECT DISTINCT Division.nameFROM Employees, DivisionWHERE salary > 100000 AND divid=id

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Existential/Universal ConditionsEmployees(name, job, divid, salary)

Division(name, id, head)

Find all divisions in which everyone makes > 100000

Existential: easy!

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Existential/universal with IN

2. Select the divisions we didn’t find:

1. Find the other divisions: in which someone makes <= 100000:

SELECT nameFROM DivisionWHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000

SELECT nameFROM DivisionWHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000

SELECT nameFROM DivisionWHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000

SELECT nameFROM DivisionWHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000

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Next: correlated subqueries Acc(name,bal,type…) Q: Who has the largest balance?

Can we do this with subqueries?

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Acc(name,bal,type,…) Q: Find holder of largest account

SELECT nameFROM AccWHERE bal >= ALL (SELECT bal FROM Acc)

SELECT nameFROM AccWHERE bal >= ALL (SELECT bal FROM Acc)

Correlated Queries

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Correlated Queries So far, subquery executed once;

result used for higher query More complicated: correlated queries

“[T]he subquery… [is] evaluated many times, once for each assignment of a value to some term in the subquery that comes from a tuple variable outside the subquery” (Ullman, p286).

Q: What does this mean? A: That subqueries refer to vars from outer queries

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Acc(name,bal,type,…) Q2: Find holder of largest account of each type

SELECT name, typeFROM AccWHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type)

SELECT name, typeFROM AccWHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type)

Correlated Queries

correlation

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Acc(name,bal,type,…) Q2: Find holder of largest account of each type

Note:1. scope of variables

SELECT name, typeFROM Acc as a1WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type)

SELECT name, typeFROM Acc as a1WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type)

Correlated Queries

correlation

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New topic: R.A./SQL Set Operators Relations are sets have set-theoretic ops

Venn diagrams

Union: R1 R2 Example:

ActiveEmployees RetiredEmployees

Difference: R1 – R2 Example:

AllEmployees – RetiredEmployees = ActiveEmployees

Intersection: R1 R2 Example:

RetiredEmployees UnionizedEmployees

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Set operations - exampleName Address Gender Birthdate

Fisher 123 Maple F 9/9/99

Hamill 456 Oak M 8/8/88

Name Address Gender Birthdate


Ford 345 Palm M 7/7/77

R:

S:





R S:

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R:

S:

R S: Name Address Gender Birthdate


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R:

S:

R - S: Name Address Gender Birthdate


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Set ops in SQL UNION, INTERSECT, EXCEPT Oracle SQL uses MINUS rather than

EXCEPT These ops applied to queries:

(SELECT name FROM Person WHERE City = 'New York')

INTERSECT(SELECT custname FROM Purchase WHERE store='Kim''s')

(SELECT name FROM Person WHERE City = 'New York')

INTERSECT(SELECT custname FROM Purchase WHERE store='Kim''s')

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Boat examples Reserve(ssn,bmodel,color)

Q: Find ssns of sailors who reserved red boats or green boats

SELECT DISTINCT ssn

FROM reserve

WHERE color = 'red' OR color = 'green'

SELECT DISTINCT ssn

FROM reserve

WHERE color = 'red' OR color = 'green'

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Q: Find ssns of sailors who reserved red boats and green boats

SELECT DISTINCT ssn

FROM reserve

WHERE color = 'red' AND color = 'green'

SELECT DISTINCT ssn

FROM reserve

WHERE color = 'red' AND color = 'green'




SELECT DISTINCT r1.ssn

FROM reserve r1, reserve r2

WHERE r1.ssn = r2.ssn AND r1.color = 'red' AND r2.color = 'green'

SELECT DISTINCT r1.ssn

FROM reserve r1, reserve r2

WHERE r1.ssn = r2.ssn AND r1.color = 'red' AND r2.color = 'green'

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(SELECT DISTINCT ssn

FROM reserve

WHERE color = 'red') INTERSECT(SELECT DISTINCT ssn

FROM reserve

WHERE color = 'green')


FROM reserve

WHERE color = 'red') INTERSECT(SELECT DISTINCT ssn

FROM reserve


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Q: Find ssns of sailors who reserved red boats or green boats


FROM reserve

WHERE color = 'red') UNION (SELECT DISTINCT ssn

FROM reserve



FROM reserve

WHERE color = 'red') UNION (SELECT DISTINCT ssn

FROM reserve


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Q: Find ssns of sailors who reserved red boats but not green boats


FROM reserve

WHERE color = 'red') EXCEPT (SELECT DISTINCT ssn

FROM reserve



FROM reserve

WHERE color = 'red') EXCEPT (SELECT DISTINCT ssn

FROM reserve


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(SELECT name, address FROM Cust1)

UNION(SELECT name FROM Cust2)


UNION(SELECT name FROM Cust2)

Union-Compatibility Situation: Cust1(name,address,…), Cust2(name,…) Want: report of all customer names and addresses

(if known) Can’t do:

Both tables must have same sequence of types Applies to all set ops

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Union-Compatibility Situation: Cust1(name,address,…), Cust2(name,…) Want: report of all customer names and addresses

(if known) But can do:

Resulting field names taken from first table


UNION(SELECT name, '(N/A)' FROM Cust2)


UNION(SELECT name, '(N/A)' FROM Cust2)

Result(name, address)Result(name, address)

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New topic: Nulls in SQL If we don’t have a value, can put a NULL

Null can mean several things: Value does not exists Value exists but is unknown Value not applicable

But null is not the same as 0

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Null Values x = NULL 4*(3-x)/7 = NULL x = NULL x + 3 – x = NULL x = NULL 3 + (x-x) = NULL x = NULL x = 'Joe' is UNKNOWN

In general: no row using null fields appear in the selection test will pass the test With one exception

Pace Boole, SQL has three boolean values: FALSE = 0 TRUE = 1 UNKNOWN = 0.5

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Null values in boolean expressions C1 AND C2 = min(C1, C2) C1 OR C2 = max(C1, C2) NOT C1 = 1 – C1

height > 6 = UNKNOWN UNKNOWN OR weight > 190 = UNKOWN (age < 25) AND UNKNOWN = UNKNOWN

E.g.age=20height=NULLweight=180

SELECT *FROM PersonWHERE (age < 25) AND (height > 6 OR weight > 190)

SELECT *FROM PersonWHERE (age < 25) AND (height > 6 OR weight > 190)

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Comparing null and non-nulls The schema specifies whether null is allowed for

each attribute NOT NULL to forbid Nulls are allowed by default

Unexpected behavior:

Some Persons are not included! The “trichotomy law” does not hold!

SELECT *FROM PersonWHERE age < 25 OR age >= 25

SELECT *FROM PersonWHERE age < 25 OR age >= 25

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Testing for null values Can test for NULL explicitly:

x IS NULL x IS NOT NULL

But: x = NULL is never true

Now it includes all Persons

SELECT *FROM PersonWHERE age < 25 OR age >= 25 OR age IS NULL

SELECT *FROM PersonWHERE age < 25 OR age >= 25 OR age IS NULL

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Null/logic review TRUE AND UNKNOWN = ?

TRUE OR UNKNOWN = ?

UNKNOWN OR UNKNOWN = ?

X = NULL = ?

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Next: Outer join Like inner join except that dangling tuples are

included, padded with nulls

Left outerjoin: dangling tuples from left are included Nulls appear “on the right”

Right outerjoin: dangling tuples from right are included Nulls appear “on the left”

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Cross join - example


Hanks 123 Palm Rd M 01/01/60

Taylor 456 Maple Av F 02/02/40

Lucas 789 Oak St M 03/03/55

Name Address Networth

Spielberg 246 Palm Rd 10M

Taylor 456 Maple Av 20M

Lucas 789 Oak St 30M

MovieStar

MovieExec

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Name Address G. Birthdate Name Address Net


Taylor 456 Maple Av F 02/02/40 Taylor 456 Maple Av 20M

Lucas 789 Oak St M 03/03/55 Lucas 789 Oak St 30M



Outer Join - ExampleSELECT * FROM MovieStar LEFT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name

SELECT * FROM MovieStar LEFT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name

SELECT * FROM MovieStar RIGHT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name

SELECT * FROM MovieStar RIGHT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name


Hanks 123 Palm Rd M 01/01/60 Null Null Null



Null Null Null Null Spielberg 246 Palm Rd 10M






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Outer Join - Example



Taylor 456 Maple Av F 02/02/40

Lucas 789 Oak St M 03/03/55

Name Address Networth


Taylor 456 Maple Av 20M

Lucas 789 Oak St 30M

MovieStar MovieExec

SELECT * FROM MovieStar FULL OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name

SELECT * FROM MovieStar FULL OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name






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New-style outer joins Outer joins may be left, right, or full

FROM A LEFT [OUTER] JOIN B; FROM A RIGHT [OUTER] JOIN B; FROM A FULL [OUTER] JOIN B;

OUTER is optional If OUTER is included, then FULL is the default

Q: How to remember left v. right? A: It indicates the side whose rows are always

included

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Next: Grouping & Aggregation In SQL:

aggregation operators in SELECT, Grouping in GROUP BY clause

Recall aggregation operators: sum, avg, min, max, count

strings, numbers, dates Each applies to scalars Count also applies to row: count(*) Can DISTINCT inside aggregation op: count(DISTINCT x)

Grouping: group rows that agree on single value Each group becomes one row in result

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Aggregation functions Numerical: SUM, AVG, MIN, MAX Char: MIN, MAX

In lexocographic/alphabetic order Any attribute: COUNT

Number of values

SUM(B) = 10 AVG(A) = 1.5 MIN(A) = 1 MAX(A) = 3 COUNT(A) = 4

A B

1 2

3 4

1 2

1 2

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Straight aggregation In R.A. sum(x)total(R) In SQL:

Just put the aggregation op in SELECT NB: aggreg. ops applied to each non-null val

count(x) counts the number of nun-null vals in field x Use count(*) to count the number of rows

SELECT SUM(x) totalFROM R

SELECT SUM(x) totalFROM R

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Straight aggregation example COUNT applies to duplicates, unless otherwise stated:

Better:

Can we say:

same as Count(*), except excludes nulls

SELECT Count(category)FROM ProductWHERE year > 1995

SELECT Count(category)FROM ProductWHERE year > 1995

SELECT COUNT(DISTINCT category)FROM ProductWHERE year > 1995

SELECT COUNT(DISTINCT category)FROM ProductWHERE year > 1995

SELECT category, COUNT(category)FROM ProductWHERE year > 1995

SELECT category, COUNT(category)FROM ProductWHERE year > 1995

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Straight aggregation example Purchase(product, date, price, quantity)

Q: Find total sales for the entire database:

Q: Find total sales of bagels:

SELECT SUM(price * quantity)FROM Purchase

SELECT SUM(price * quantity)FROM Purchase

SELECT SUM(price * quantity)FROM PurchaseWHERE product = 'bagel'

SELECT SUM(price * quantity)FROM PurchaseWHERE product = 'bagel'

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Largest balance again Acc(name,bal,type) Q: Who has the largest balance? Q: Who has the largest balance of each

type?

Can we do these with aggregation functions?

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Straight grouping Group rows together by field values Produces one row for each group

I.e., by each (combin. of) grouped val(s) Don’t select non-grouped fields

Reduces to DISTINCT selections:

SELECT productFROM PurchaseGROUP BY product

SELECT productFROM PurchaseGROUP BY product

SELECT DISTINCT productFROM Purchase

SELECT DISTINCT productFROM Purchase

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Grouping & aggregation Sometimes want to group and compute

aggregations by group Aggregation op applied to rows in group, not to all rows in table

Purchase(product, date, price, quantity) Find total sales for products that sold for > 0.50:

SELECT product, SUM(price*quantity) totalFROM PurchaseWHERE price > .50GROUP BY product


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Illustrated G&A example

Product Date Price Quantity

Bagel 10/21 0.85 15

Banana 10/22 0.52 7

Banana 10/19 0.52 17

Bagel 10/20 0.85 20

Purchase

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Product Date Price Quantity

Banana 10/19 0.52 17

Banana 10/22 0.52 7

Bagel 10/20 0.85 20

Bagel 10/21 0.85 15

First compute the FROM-WHERE Then GROUP BY product:


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Product TotalSales

Bagel $29.75

Banana $12.48

Finally, aggregate and select:


SELECT product, SUM(price*quantity) totalFROM PurchaseWHERW price > .50GROUP BY product

SELECT product, SUM(price*quantity) totalFROM PurchaseWHERW price > .50GROUP BY product

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Illustrated G&A example GROUP BY may be reduced to (a possibly more

complicated) subquery:



SELECT DISTINCT x.product, (SELECT SUM(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.price > .50) totalFROM Purchase xWHERE x.price > .50

SELECT DISTINCT x.product, (SELECT SUM(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.price > .50) totalFROM Purchase xWHERE x.price > .50

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For every product, what is the total sales and max quantity sold?

Product SumSales MaxQuantity

Banana $12.48 17

Bagel $29.75 20

Multiple aggregations

SELECT product, SUM(price * quantity) SumSales, MAX(quantity) MaxQuantityFROM PurchaseWHERE price > .50GROUP BY product

SELECT product, SUM(price * quantity) SumSales, MAX(quantity) MaxQuantityFROM PurchaseWHERE price > .50GROUP BY product

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Another grouping/aggregation e.g. Movie(title, year, length, studioName)

Q: How many total minutes of film have been produced by each studio?

Strategy: Divide movies into groups per studio, then add lengths per group

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Another grouping/aggregation e.g.

Title Year Length Studio

Star Wars 1977 120 Fox

Jedi 1980 105 Fox

Aviator 2004 800 Miramax

Pulp Fiction 1995 110 Miramax

Lost in Translation

2003 95 Universal

SELECT studio, sum(length) totalLengthFROM MoviesGROUP BY studio


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Jedi 1980 105 Fox



Lost in Translation

2003 95 Universal

SELECT studio, sum(length) lengthFROM MoviesGROUP BY studio

SELECT studio, sum(length) lengthFROM MoviesGROUP BY studio

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Jedi 1980 105 Fox



Lost in Translation

2003 95 Universal

Studio Length

Fox 225

Miramax 910

Universal 95



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Grouping/aggregation example StarsIn(SName,Title,Year) Q: Find the year of each star’s first movie

Q: Find the span of each star’s career Look up first and last movies

SELECT sname, min(year) firstyearFROM StarsInGROUP BY sname

SELECT sname, min(year) firstyearFROM StarsInGROUP BY sname

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Account types again Acc(name,bal,type) Q: Who has the largest balance of each

type?

Can we do this with grouping/aggregation?

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G & A for constructed relations Movie(title,year,producerSsn,length) MovieExec(name,ssn,netWorth)

Can do the same thing for larger, non-atomic relations Q: How many mins. of film did each producer make?

What happens to non-producer movie-execs?

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY name

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY name

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HAVING clauses Sometimes want to limit which rows may be grouped Q: How many mins. of film did each rich producer

make? Rich = netWorth > 10000000

Q: Is HAVING necessary here? A: No, could just add rich req. to WHERE

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY nameHAVING netWorth > 10000000

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY nameHAVING netWorth > 10000000

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HAVING clauses Sometimes want to limit which rows may be

grouped Q: How many mins. of film did each rich producer

make? Old = made movies before 1930

Q: Is HAVING necessary here?

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY nameHAVING min(year) < 1930

SELECT name, sum(length) totalFROM Movie, MovieExecWHERE producerSsn = ssnGROUP BY nameHAVING min(year) < 1930

1 database systems lecture #7 yan pan school of software, sysu 2011

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