Relational Query Languages 10_13.pdf

Relational Query
Languages
Professor Navneet Goyal
Department of Computer Science & Information Systems
BITS, Pilani

© Prof. Navneet Goyal, BITS, Pilani
Relational Query
Languages
n A major strength of the relational
model: supports simple, powerful
querying of data.
n Queries can be written intuitively,
and the DBMS is responsible for
efficient evaluation.
n The key: precise semantics for
relational queries.
n Allows the optimizer to extensively re-
order operations, and still ensure that
the answer does not change.

Relational Query
Languages
n Query languages: Allow manipulation
and retrieval of data from a database.
n Relational model supports simple,
powerful QLs:
n Strong formal foundation based on logic.
n Allows for much optimization.
n Query Languages != programming
languages!
n QLs not expected to be “Turing complete”.
n QLs not intended to be used for complex
calculations.
n QLs support easy, efficient access to large data
sets.

Formal Relational
Query Languages
n Two mathematical Query
Languages form the basis for “real”
languages (e.g. SQL), and for
implementation:
n Relational Algebra: More operational,
very useful for representing execution
plans.
n Relational Calculus: Lets users
describe what they want, rather than
how to compute it. (Non-operational,
declarative.)

The SQL Query
Language
n Developed by IBM’s System-R Project
in the 1970s
n Need for a standard since it is used by
many vendors
n Standards:
n SQL-86
n SQL-89 (minor revision)
n SQL-92 (major revision)
n SQL-99 (major extensions, current
standard)
n Find Out what is SQL-MM!!

The SQL Query
Language
n SQL has been influenced by both
Relational Algebra (RA) & Relational
Calculus (RC)
n More so by RC, particularly Tuple
relational Calculus (TRC)
n The other variant of RC is Domain
Relational Calculus (DRC) which has
greatly influenced Query By Example
(QBE)

The SQL Query
Language
n SQL consists of:
n DDL (Data Definition Language)
• Create conceptual schema
n DML (Data Manipulation Language)
• Relational operators
• Insert, Delete, Update
n VDL (View Definition Language)
• Specify user views & their mapping to the
conceptual schema (in most DBMSs, done by
DDL)
n SDL (Storage Definition Language)
• File organization
• Indexes

The SQL Query
Language
n SDL is being removed from SQL
n DMLs
n High-level or nonprocedural (declarative)
• Can be entered at the SQL > or can be embedded in a
general purpose programming language
• Can specify & retrieve many records in a single DML
statement & are hence called Set-at-a-time DMLs
n Low-level or procedural
• Must be embedded in a general purpose
programming language
• Typically retrieves individual records or objects from
the DB & processes each separately
• Therefore it needs PL constructs like looping
• Record-at-a-time DMLs

The SQL Query
Language
DMLs
n Whenever DML statements are embedded in a PL,
that language is called as the host language and
the DML is called the Data Sublanguage
n In object DBs, the host language & data
sublanguage form one integrated language – for
eg. C++ with some extensions to support
database functionality
n Some RDBMSs also provide integrated languages –
for eg. ORACLE’s PL/SQL.

Operations on
Relations
n Restrict
n Project
n Join
n Divide
n Union
n Intersection
n Difference
n Product
Relational Operations
Set Operations

Relational Algebra
n Collection of operations on relations
n 8 operators + Rename operator
n Codd’s 8 operators do not form a minimal
set
n Some of them are not primitive
n Join, Intersect, & Divide can be defined in
terms of other 5
n None of these 5 can be defined in terms of
the remaining 4
n MINIMAL SET
n Join, Intersect, & Divide are important &
should be supported directly

Relational Algebra
n Some common DB requests can
not be performed with the RA
operations
n Additional operations are
developed
n Aggregate functions & additional
types of joins

Relational Algebra
IMPORTANCE OF RA
n Provides a formal foundation for
relational model operations
n Plays a pivotal role in query
optimization in RDBMSs
n Some of its concepts are
implemented in SQL
n Expressive power of RA is used as a
metric of how powerful a relational
DB query language is!!

Relational Completeness
n A relational DB query language is
said to be Relationally Complete, if
it can express all queries that we can
express in RA
n Relational DB query languages are
expected to be relationally complete
n Commercial query languages like
SQL, support features that allow us
to express some queries that are not
possible in RA

Restrict & Project

Union, Intersection &
Difference

Union, Intersection &
Difference
Union Compatibility: r U s is valid if:
n Relations r & s have the same arity
n Domains of the ith attribute of r is the
same as the domain of the ith attribute of
s, ⍱ i.
Note that r & s can be either database
relations or derived relations

Product & Divide
A
B
C
X
Y
A
A
B
B
C
C
X
Y
X
Y
X
Y
X
Z
A
PRODUCT
DIVIDE BY
A
A
A
B
C
X
Y
Z
X
Y

Divide
DIVIDE
P1
P2
P4
S1
S1
S2
S2
S2
S2
S3
S4
S4
P1
P2
P3
P4
P1
P2
P2
P2
P4
S1
S2
S3
S4
Shipment
P2
P4
P2
S1
S4
S2
Parts

Join
A1
A2
A3
B1
B1
B2
B1
B2
B3
C1
C2
C3
A1
A2
A3
B1
B1
B2
C1
C1
C2
NATURAL
JOIN

Closure
n A relation is closed under
relational and set operators
n The result of these operators on
relation(s) is another relation
n Output from one operation can
become input to another
n Nested Expressions possible

Example Schema
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
sid bid day
22 101 10/10/96
58 103 11/12/96
R1 S1
S2

Projection
sname rating
yuppy 9
lubber 8
guppy 5
rusty 10
πsname rating
S
,
( )
2
age
35.0
55.5
πage S
( )
2
n Deletes attributes that are not in
projection list.
n Schema of result contains exactly
the fields in the projection list,
with the same names that they
had in the (only) input relation.
n Projection operator has to
eliminate duplicates! (Why??)
n Note: real systems typically
don’t do duplicate elimination
unless the user explicitly asks
for it. (Why not?)

Selection
σrating
S
>8
2
( )
28 yuppy 9 35.0
58 rusty 10 35.0
sname rating
yuppy 9
rusty 10
π σ
sname rating rating
S
,
( ( ))
>8
2
n Selects rows that
satisfy selection
condition.
n No duplicates in
result! (Why?)
n Schema of result
identical to schema of
(only) input relation.
n Result relation can be
the input for another
relational algebra
operation! (Operator
composition.)

Union, Intersection,
Set-Difference
n All of these operations take
two input relations, which
must be union-compatible:
n Same number of fields
n `Corresponding’ fields
have the same
domain/data type
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
44 guppy 5 35.0
28 yuppy 9 35.0
31 lubber 8 55.5
58 rusty 10 35.0
S S
1 2
∪
S S
1 2
∩
22 dustin 7 45.0
S S
1 2
−

Cross-Product
n Each row of S1 is paired with each row of R1.
n Result schema has one field per field of S1 and R1, with
field names `inherited’ if possible.
n Conflict: Both S1 and R1 have a field called sid.
ρ ( ( , ), )
C sid sid S R
1 1 5 2 1 1
→ → ×
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 22 101 10/10/96
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 22 101 10/10/96
31 lubber 8 55.5 58 103 11/12/96
58 rusty 10 35.0 22 101 10/10/96
58 rusty 10 35.0 58 103 11/12/96
§ Renaming operator:

Joins
n Condition Join:
n Result schema same as that of cross-
product.
n Fewer tuples than cross-product, might
be able to compute more efficiently
n Sometimes called a theta-join.
R c S c R S
!
" = ×
σ ( )
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 58 103 11/12/96
S R
S sid R sid
1 1
1 1
!
"
. .
<

Joins
n Equi-Join: A special case of condition join
where the condition c contains only equalities
n Result schema similar to cross-product, but
only one copy of fields for which equality is
specified.
n Natural Join: Equijoin on all common fields.
sid sname rating age bid day
22 dustin 7 45.0 101 10/10/96
58 rusty 10 35.0 103 11/12/96
S R
sid
1 1
!
"

Examples of Division A/B
sno pno
s1 p1
s1 p2
s1 p3
s1 p4
s2 p1
s2 p2
s3 p2
s4 p2
s4 p4
pno
p2
pno
p2
p4
pno
p1
p2
p4
sno
s1
s2
s3
s4
sno
s1
s4
sno
s1
A
B1
B2
B3
A/B1 A/B2 A/B3

Find names of sailors who’ve
reserved boat #103
n Solution 1: π σ
sname bid
serves Sailors
(( Re ) )
=103
!
"
n Solution 2: )
Re
,
1
(
103
serves
Temp
bid =
σ
ρ
)
1
,
2
( Sailors
Temp
Temp !
"
ρ
π sname Temp
( )
2
n Solution 3: ))
Re
(
103
( Sailors
serves
bid
sname !
"
=
σ
π
Query
Optimizer

Find names of sailors who’ve
reserved a red boat
n Information about boat color only available
in Boats; so need an extra join:
π σ
sname color red
Boats serves Sailors
((
' '
) Re )
=
!
" !
"
v A more efficient solution:
)
)
Re
)
'
'
((
( Sailors
s
Boats
red
color
bid
sid
sname !
"
!
"
=
σ
π
π
π
A query optimizer can find this, given the first solution!
Query
Optimizer

Find sailors who’ve reserved a
red or a green boat
n Can identify all red or green boats, then find
sailors who’ve reserved one of these boats:
ρ σ
( ,(
' ' ' '
))
Tempboats
color red color green
Boats
= ∨ =
π sname Tempboats serves Sailors
( Re )
!
" !
"
v Can also define Tempboats using union! (How?)
v What happens if is replaced by in this query?
∨ ∧

Find sailors who’ve reserved a
red and a green boat
n Previous approach won’t work! Must
identify sailors who’ve reserved red boats,
sailors who’ve reserved green boats, then
find the intersection
ρ π σ
( , ((
' '
) Re ))
Tempred
sid color red
Boats serves
=
!
"
π sname Tempred Tempgreen Sailors
(( ) )
∩ !
"
))
Re
)
'
'
(
(
,
( serves
Boats
green
color
sid
Tempgreen !
"
=
σ
π
ρ

Find the names of sailors
who’ve reserved all boats
n Uses division; schemas of the input
relations to / must be carefully chosen:
))
(
/
)
Re
,
(
,
( Boats
bid
serves
bid
sid
Tempsids π
π
ρ
π sname Tempsids Sailors
( )
!
"
v To find sailors who’ve reserved all ‘Interlake’ boats:
/ (
' '
)
π σ
bid bname Interlake
Boats
=
.....

Summary
n The relational model has rigorously
defined query languages that are
simple and powerful.
n Relational algebra is more
operational; useful as internal
representation for query evaluation
plans.
n Several ways of expressing a given
query; a query optimizer should
choose the most efficient version.

Summary
n Role of RA in query optimization

Role of RA in QO
n Parsing and translation
n Translate the query into its internal form.
n Translation is similar to the work performed
by the parser of a compiler
n Parser checks syntax, verifies relations
n Parse tree representation
n This is then translated into RA expression

Role of RA in QO
n Query Execution Plan
n In SQL, a query can be expressed is several ways
n Each SQL query can itself be translated into RA expression
in many ways
n An RA expression only partially tells you how to evaluate a
query
n Several ways to evaluate RA expression
n Annotate RA expression with instructions specifying how to
evaluate each operation
n Annotation may state the algorithm to be used for a specific
operation or the particular index to use
n Annotated RAE is called an evaluation primitive
n Sequence of primitive operations is a QEP

Role of RA in QO
n Example
select balance
from account
where balance < 2500
n RAEs
n σbalance<2500(∏balance(account))
n ∏balance(σbalance<2500(account))
n E.g., we can use an index on balance to find
accounts with balance < 2500,
n or can perform complete relation scan and
discard accounts with balance ≥ 2500

Role of RA in QO
Query Execution Plan

Role of RA in QO

Summary
n What we studied about RA is not used in
commercial systems!!!
n We assumed a SET model
n Commercial systems use a BAG model
n What are BAGS?
n Why BAGS?

Relational Operations on
BAGS
n So far we discussed relational operations
on relations (sets of tuples)
n In commercial systems, a relation is
allowed to have duplicate tuples
n If a set is allowed to have multiple
occurrences of a member, it is called a
BAG or MULTISET
n Set – A relation without duplicate tuples
n Bag – A relation that may or may not have
duplicate tuples

WHY BAGS?
n Allowing relations to be bags
can speed up operations on
relations
n For Example, allowing the
result to be a bag could speed
up union of two relations
n Duplicate elimination not
required

WHY BAGS?
n Another Example:
Projection of relations
n Each projected tuple
must be compared with
other projected tuples
to make sure that each
projection tuple
appears only once
n If bags are accepted as
results, we simply need
to project each tuple &
add it to the result
A B C
1 2 5
3 4 6
1 2 7
1 2 8
A B
1 2
3 4
A B
1 2
3 4
1 2
1 2

WHY BAGS?
If we are
projecting a
relation to take
an aggregate
(say AVG), we
could not use
the set model
A B C
1 2 5
3 4 6
1 2 7
1 2 8
A B
1 2
3 4
A B
1 2
3 4
1 2
1 2
AVG=2
AVG=1.5

Union, Intersection, &
Difference of BAGS
UNION
n Add the no. of occurrences of each
tuple
n If R is a bag in which tuple t appears
n times, & S is a bag in which tuple t
appears m times, then in the bag
R∪S, t appears n+m times
n Either n or m or both could be zero

Difference of BAGS
INTERSECTION
n If R is a bag in which tuple t
appears n times, & S is a bag in
which tuple t appears m times,
then in the bag R∩S, t appears
min(n,m) times

Difference of BAGS
DIFFERENCE
n If R is a bag in which tuple t appears n
times, & S is a bag in which tuple t appears
m times, then in the bag R-S, tuple t
appears max(0, n-m) times
n If t appears in R more times than it
appears in S, then in R-S, t appears the no.
of times it appears in R, minus the no. of
times it appears in S
n If n=m, then t does not appear in R-S at all
n Occurrences of t in S each cancel one
occurrence in R

Difference of BAGS
A B
1 2
3 4
1 2
1 2
A B
1 2
3 4
3 4
5 6
R S
Write down R∪S, R∩S, R-S, & S-R

Difference of BAGS
2
1
4
3
2
1
4
3
6
5
2
1
4
3
2
1
B
A
4
3
2
1
B
A
R∪S R∩S R-S
2
1
2
1
B
A S-R
6
5
4
3
B
A

Projection of BAGS
A B C
1 2 5
3 4 6
1 2 7
1 2 8
R
Write down πA, B (R)
A B
1 2
3 4
1 2
1 2
πA, B (R)

Selection on BAGS
A B C
1 2 5
3 4 6
1 2 7
1 2 7
R
Write down σC>=6 (R)
A B C
3 4 6
1 2 7
1 2 7
σC>=6 (R)

Product of BAGS
A B
1 2
1 2
R
Write down R x S
B C
2 3
4 5
4 5
S
A R.B S.B C
1 2 2 3
1 2 2 3
1 2 4 5
1 2 4 5
1 2 4 5

Natural Join of BAGS
A B
1 2
1 2
B C
2 3
4 5
4 5
R
Write down R S
S
A B C
1 2 3
1 2 3
!
"

Theta-Join of BAGS
A B
1 2
1 2
B C
2 3
4 5
4 5
R
Write down R S
S
A R.B S.B C
1 2 4 5
1 2 4 5
1 2 4 5
1 2 4 5
!
"
R.B < S.B

Extended Operators of
RA
n Duplicate-Elimination Operator δ
n Aggregation Operators
• Sum
• Count
• Average
• Max
• Min
n Grouping Operator γ
n Sorting Operator τ
n Extended Projection
n Outerjoin Operator

Duplicate-Elimination
Operator δ
n Converts a bag to a set
A B
1 2
3 4
1 2
1 2
R A B
1 2
3 4
δ(R)

Aggregate Functions &
Operations
n 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
n Aggregate operation in relational algebra: General Form
E is any relational-algebra expression
n G1, G2 …, Gn is a list of attributes on which to group
(can be empty)
n Each Fi is an aggregate function
n Each Ai is an attribute name

Aggregate Operation –
Example
Relation r: A B
α
α
β
β
α
β
β
β
C
7
7
3
10
g sum(c) (r) sum(C )
27

Aggregate Operation –
Example
Relation account grouped by branch-name:
branch_name g sum(balance) (account)
branch_name account_number balance
Perryridge
Perryridge
Brighton
Brighton
Redwood
A-102
A-201
A-217
A-215
A-222
400
900
750
750
700
branch_name sum(balance)
Perryridge
Brighton
Redwood
1300
1500
700

Aggregate Functions
(Cont.)
n Result of aggregation does not
have a name
n Can use rename operation to give it a
name
n For convenience, we permit renaming
as part of aggregate operation
branch_name g sum(balance)as sum_balance(account)

Generalized Projection
n Extends the projection operation by allowing
arithmetic functions to be used in the projection
list.
n E is any relational-algebra expression
n Each of F1, F2, …, Fn are are arithmetic
expressions involving constants and attributes in
the schema of E.
n Given relation credit_info(customer_name, limit,
credit_balance), find how much more each person
can spend:
∏customer_name, limit – credit_balance (credit_info)

Generalized Projection
Given relation credit_info(customer_name, limit,
credit_balance), find how much more each person can spend:
∏customer_name, (limit – credit_balance) as credit_available (credit_info)
1750
1500
700
400
Cust_name Credit_bal
Curry
Hayes
Jones
Smith
limit
2000
1500
6000
2000
Cust_name
Curry
Hayes
Jones
Smith
Credit_available
250
0
5300
1600

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

Outer Join – Example
Relation loan
Relation borrower
customer_name loan_number
Jones
Smith
Hayes
L-170
L-230
L-155
3000
4000
1700
loan_number amount
L-170
L-230
L-260
branch_name
Downtown
Redwood
Perryridge

Inner Join
loan Borrower
loan_number amount
L-170
L-230
3000
4000
customer_name
Jones
Smith
branch_name
Downtown
Redwood
Jones
Smith
null
loan_number amount
L-170
L-230
L-260
3000
4000
1700
customer_name
branch_name
Downtown
Redwood
Perryridge
Left Outer Join
loan Borrower

loan_number amount
L-170
L-230
L-155
3000
4000
null
customer_name
Jones
Smith
Hayes
branch_name
Downtown
Redwood
null
loan_number amount
L-170
L-230
L-260
L-155
3000
4000
1700
null
customer_name
Jones
Smith
null
Hayes
branch_name
Downtown
Redwood
Perryridge
null
■ Full Outer Join
loan borrower
■ Right Outer Join
loan borrower

Outer Join
n Left join, or inner join, select rows
common to the participating tables to
a join
n Selecting elements in a table
regardless of whether they are
present in the second table
n OUTER JOIN is the solution
n In Oracle, we will place an "(+)" in
the WHERE clause on the other side
of the table for which we want to
include all the rows.

Outer Join
store_name Sales Date
Los Angeles $1500 Jan-05-1999
San Diego $250 Jan-07-1999
Los Angeles $300 Jan-08-1999
Boston $700 Jan-08-1999
- We want to find out the sales amount for all of the stores
- If we do a regular join, we will not be able to get what we want because we
will have missed "New York," since it does not appear in the
Store_Information table
SELECT A1.store_name, SUM(A2.Sales) SALES
FROM Geography A1, Store_Information A2
WHERE A1.store_name = A2.store_name (+)
GROUP BY A1.store_name
Store_Information Geography
store_name SALES
Boston $700
New York
Los Angeles $1800
San Diego $250
region_name store_name
East Boston
East New York
West Los Angeles
West San Diego

NVL Function
n In Oracle/PLSQL, the NVL function lets you substitutes a value when a null
value is encountered.
n NVL (string1, replace_with )
n string1 is the string to test for a null value. Replace_with is the value
returned if string1 is null.
n Example #1:
n select NVL (supplier_city, 'n/a')
from suppliers;
n The SQL statement above would return 'n/a' if the supplier_city field
contained a null value. Otherwise, it would return the supplier_city value.
n Example #2:
n select supplier_id,
NVL (supplier_desc, supplier_name)
from suppliers;
n This SQL statement would return the supplier_name field if the
supplier_desc contained a null value. Otherwise, it would return the
supplier_desc.
n Example #3:
n select NVL (commission, 0)
from sales;
n This SQL statement would return 0 if the commission field contained
a null value. Otherwise, it would return the commission field.

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

Null Values
n Comparisons with null values return the special truth
value: unknown
n If false was used instead of unknown, then not (A < 5)
would not be equivalent to A >= 5
n Three-valued logic using the truth value unknown:
n OR: (unknown or true) = true,
(unknown or false) = unknown
(unknown or unknown) = unknown
n AND: (true and unknown) = unknown,
(false and unknown) = false,
(unknown and unknown) = unknown
n NOT: (not unknown) = unknown
n In SQL “P is unknown” evaluates to true if predicate P
evaluates to unknown
n Result of select predicate is treated as false if it
evaluates to unknown

Problem Solving
n Consider the relational schemas:
Email (IDNO, email_id)
Student (IDNO, Name, Hostel, Room)
The EMAIL relation is maintained by IPC, where as the
Student relation is maintained by SWD. It is required
that the information in the two relations be combined
into a new relation with schema:
Student_email (IDNO, email_id, Name, Hostel, Room)
It is observed that there are some IDNOs in the
Student relation that are not in the Email relation.
Write a Relational Algebra expression to create the new
relation such that no IDNO from the Student relation is
left out.
ρ (Student_email, (student left outer join email))

Problem Solving
n The student_master table has the schema:
Student_Master_2004 (SUID, IDNO, NAME)
where SUID is a unique identifier for a student
(for eg. f2004123). After completing one year,
students are either transferred to another
discipline/degree or offered a dual degree. The
following table contains the changed IDNO of
students:
Student_Master_NEW_2004 (SUID, IDNO, NAME)
You need to find out all cases of transfer and dual
degree.
Write a relational algebra expression that will
create & populate the following schema:
Dual_Transfer_2004 (SUID, OLD_IDNO,
NEW_IDNO, NAME)

Relational Calculus
n Relational Algebra: More
operational, very useful for
representing execution plans.
n Relational Calculus: Lets users
describe what they want, rather
than how to compute it. (Non-
operational, declarative.)

Relational Query Languages 10_13.pdf

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Relational Query Languages 10_13.pdf