Data

Structure Query Language (SQL)
Rithik Raj Vaishya
Data Scientist | Corporate Trainer
Dec-22 SQL 1

SQL Introduction
2
Standard language for querying and manipulating data
Structured Query Language
Many standards out there:
• ANSI SQL, SQL92 (a.k.a. SQL2), SQL99 (a.k.a. SQL3), ….
• Vendors support various subsets: watch for fun discussions in class !

SQL
• Data Definition Language (DDL)
• Create/alter/delete tables and their attributes
• Following lectures...
• Data Manipulation Language (DML)
• Query one or more tables – discussed next !
• Insert/delete/modify tuples in tables
3

Tables in SQL
4
PName Price Category Manufacturer
Gizmo $19.99 Gadgets GizmoWorks
Powergizmo $29.99 Gadgets GizmoWorks
SingleTouch $149.99 Photography Canon
MultiTouch $203.99 Household Hitachi
Product
Attribute names
Table name
Tuples or rows

Tables Explained
• The schema of a table is the table name and its
attributes:
Product(PName, Price, Category, Manfacturer)
• A key is an attribute whose values are unique;
we underline a key
Dec-22 SQL 5

Data Types in SQL
• Atomic types:
• Characters: CHAR(20), VARCHAR(50)
• Numbers: INT, BIGINT, SMALLINT, FLOAT
• Others: MONEY, DATETIME, …
• Every attribute must have an atomic type
• Hence tables are flat
• Why ?
Dec-22 SQL 6

Tables Explained
• A tuple = a record
• Restriction: all attributes are of atomic type
• A table = a set of tuples
• Like a list…
• …but it is unorderd:
no first(), no next(), no last().
Dec-22 SQL 7

SQL Query
Dec-22 SQL 8
Basic form: (plus many many more bells and whistles)
SELECT <attributes>
FROM <one or more relations>
WHERE <conditions>

Simple SQL Query
Dec-22 SQL 9
SELECT *
FROM Product
WHERE category=‘Gadgets’
Product
“selection”

Simple SQL Query
Dec-22 SQL 10
SELECT PName, Price, Manufacturer
FROM Product
WHERE Price > 100
Product
PName Price Manufacturer
SingleTouch $149.99 Canon
MultiTouch $203.99 Hitachi
“selection” and
“projection”

Notation
Dec-22 SQL 11
Answer(PName, Price, Manfacturer)
Input Schema
Output Schema
SELECT PName, Price, Manufacturer
FROM Product
WHERE Price > 100

Details
• Case insensitive:
• Same: SELECT Select select
• Same: Product product
• Different: ‘Seattle’ ‘seattle’
• Constants:
• ‘abc’ - yes
• “abc” - no
Dec-22 SQL 12

The LIKE operator
• s LIKE p: pattern matching on strings
• p may contain two special symbols:
• % = any sequence of characters
• _ = any single character
Dec-22 SQL 13
SELECT *
FROM Products
WHERE PName LIKE ‘%gizmo%’

Eliminating Duplicates
Dec-22 SQL 14
SELECT DISTINCT category
FROM Product
Compare to:
SELECT category
FROM Product
Category
Gadgets
Gadgets
Photography
Household
Category
Gadgets
Photography
Household

Ordering the Results
Dec-22 SQL 15
SELECT pname, price, manufacturer
FROM Product
WHERE category=‘gizmo’ AND price > 50
ORDER BY price, pname
Ties are broken by the second attribute on the ORDER BY list, etc.
Ordering is ascending, unless you specify the DESC keyword.

SELECT Category
FROM Product
ORDER BY PName
?
FROM Product
ORDER BY category
FROM Product
ORDER BY PName
?
?
Dec-22 SQL 16

Keys and Foreign Keys
Dec-22 SQL 17
Product
Company
CName StockPrice Country
GizmoWorks 25 USA
Canon 65 Japan
Hitachi 15 Japan
Key
Foreign
key

Joins
Dec-22 SQL 18
Product (pname, price, category, manufacturer)
Company (cname, stockPrice, country)
Find all products under $200 manufactured in Japan;
return their names and prices.
SELECT PName, Price
FROM Product, Company
WHERE Manufacturer=CName AND Country=‘Japan’
AND Price <= 200
Join
between Product
and Company

Joins
Dec-22 SQL 19
Product
Company
Cname StockPrice Country
GizmoWorks 25 USA
Canon 65 Japan
Hitachi 15 Japan
PName Price
SingleTouch $149.99
SELECT PName, Price
WHERE Manufacturer=CName AND Country=‘Japan’
AND Price <= 200

More Joins
Dec-22 SQL 20
Find all Chinese companies that manufacture products both in the
‘electronic’ and ‘toy’ categories
SELECT cname
FROM
WHERE

A Subtlety about Joins
Dec-22 SQL 21
Find all countries that manufacture some product in the ‘Gadgets’
category.
SELECT Country
WHERE Manufacturer=CName AND Category=‘Gadgets’
Unexpected duplicates

A Subtlety about Joins
Dec-22 SQL 22
Name Price Category Manufacturer
Product
Company
Cname StockPrice Country
GizmoWorks 25 USA
Canon 65 Japan
Hitachi 15 Japan
Country
??
??
What is
the problem ?
What’s the
solution ?
SELECT Country
WHERE Manufacturer=CName AND Category=‘Gadgets’

Tuple Variables
Dec-22 SQL 23
SELECT DISTINCT pname, address
FROM Person, Company
WHERE worksfor = cname
Which
address ?
Person(pname, address, worksfor)
Company(cname, address)
SELECT DISTINCT Person.pname, Company.address
FROM Person, Company
WHERE Person.worksfor = Company.cname
SELECT DISTINCT x.pname, y.address
FROM Person AS x, Company AS y
WHERE x.worksfor = y.cname

Meaning (Semantics) of SQL Queries
Dec-22 SQL 24
SELECT a1, a2, …, ak
FROM R1 AS x1, R2 AS x2, …, Rn AS xn
WHERE Conditions
Answer = {}
for x1 in R1 do
for x2 in R2 do
…..
for xn in Rn do
if Conditions
then Answer = Answer  {(a1,…,ak)}
return Answer

SELECT DISTINCT R.A
FROM R, S, T
WHERE R.A=S.A OR R.A=T.A
An Unintuitive Query
Dec-22 SQL 25
Computes R  (S  T) But what if S = f ?
What does it compute ?

Subqueries Returning Relations
Dec-22 SQL 26
SELECT Company.city
FROM Company
WHERE Company.name IN
(SELECT Product.maker
FROM Purchase , Product
WHERE Product.pname=Purchase.product
AND Purchase .buyer = ‘Joe Blow‘);
Return cities where one can find companies that manufacture
products bought by Joe Blow
Company(name, city)
Product(pname, maker)
Purchase(id, product, buyer)

Dec-22 SQL 27
SELECT Company.city
FROM Company, Product, Purchase
WHERE Company.name= Product.maker
AND Product.pname = Purchase.product
AND Purchase.buyer = ‘Joe Blow’
Is it equivalent to this ?
Beware of duplicates !

Removing Duplicates
Dec-22 SQL 28
Now
they are
equivalent
SELECT DISTINCT Company.city
FROM Company
WHERE Company.name IN
(SELECT Product.maker
FROM Purchase , Product
WHERE Product.pname=Purchase.product
AND Purchase .buyer = ‘Joe Blow‘);
SELECT DISTINCT Company.city
FROM Company, Product, Purchase
WHERE Company.name= Product.maker
AND Product.pname = Purchase.product
AND Purchase.buyer = ‘Joe Blow’

Dec-22 SQL 29
SELECT name
FROM Product
WHERE price > ALL (SELECT price
FROM Purchase
WHERE maker=‘Gizmo-Works’)
Product ( pname, price, category, maker)
Find products that are more expensive than all those produced
By “Gizmo-Works”
You can also use: s > ALL R
s > ANY R
EXISTS R

Question for Database Fans
and their Friends
• Can we express this query as a single SELECT-
FROM-WHERE query, without subqueries ?
Dec-22 SQL 30

Question for Database Fans
and their Friends
•Answer: all SFW queries are monotone
(figure out what this means). A query
with ALL is not monotone
Dec-22 SQL 31

Correlated Queries
Dec-22 SQL 32
SELECT DISTINCT title
FROM Movie AS x
WHERE year <> ANY
(SELECT year
FROM Movie
WHERE title = x.title);
Movie (title, year, director, length)
Find movies whose title appears more than once.
Note (1) scope of variables (2) this can still be expressed as single SFW
correlation

Complex Correlated Query
Product ( pname, price, category, maker, year)
• Find products (and their manufacturers) that are more expensive
than all products made by the same manufacturer before 1972
Very powerful ! Also much harder to optimize.
Dec-22 SQL 33
SELECT DISTINCT pname, maker
FROM Product AS x
WHERE price > ALL (SELECT price
FROM Product AS y
WHERE x.maker = y.maker AND y.year < 1972);

Aggregation
Dec-22 SQL 34
SELECT count(*)
FROM Product
WHERE year > 1995
Except count, all aggregations apply to a single attribute
SELECT avg(price)
FROM Product
WHERE maker=“Toyota”
SQL supports several aggregation operations:
sum, count, min, max, avg

COUNT applies to duplicates, unless otherwise stated:
SELECT Count(category)
FROM Product
WHERE year > 1995
same as Count(*)
We probably want:
SELECT Count(DISTINCT category)
FROM Product
WHERE year > 1995
Aggregation: Count
Dec-22 SQL 35

Purchase(product, date, price, quantity)
More Examples
Dec-22 SQL 36
SELECT Sum(price * quantity)
FROM Purchase
FROM Purchase
WHERE product = ‘bagel’
What do
they mean ?

Simple Aggregations
Dec-22 SQL 37
Purchase
Product Date Price Quantity
Bagel 10/21 1 20
Banana 10/3 0.5 10
Banana 10/10 1 10
Bagel 10/25 1.50 20
FROM Purchase
WHERE product = ‘bagel’
50 (= 20+30)

Grouping and Aggregation
Dec-22 SQL 38
Purchase(product, date, price, quantity)
SELECT product, Sum(price*quantity) AS TotalSales
FROM Purchase
WHERE date > ‘10/1/2005’
GROUP BY product
Let’s see what this means…
Find total sales after 10/1/2005 per product.

Grouping and Aggregation
Dec-22 SQL 39
1. Compute the FROM and WHERE clauses.
2. Group by the attributes in the GROUPBY
3. Compute the SELECT clause: grouped attributes and aggregates.

1&2. FROM-WHERE-GROUPBY
Dec-22 SQL 40
Bagel 10/21 1 20
Bagel 10/25 1.50 20
Banana 10/3 0.5 10
Banana 10/10 1 10

3. SELECT
Dec-22 SQL 41
FROM Purchase
WHERE date > ‘10/1/2005’
GROUP BY product
Bagel 10/21 1 20
Bagel 10/25 1.50 20
Banana 10/3 0.5 10
Banana 10/10 1 10
Product TotalSales
Bagel 50
Banana 15

GROUP BY v.s. Nested Quereis
Dec-22 SQL 42
FROM Purchase
WHERE date > ‘10/1/2005’
GROUP BY product
SELECT DISTINCT x.product, (SELECT Sum(y.price*y.quantity)
FROM Purchase y
WHERE x.product = y.product
AND y.date > ‘10/1/2005’)
AS TotalSales
FROM Purchase x
WHERE x.date > ‘10/1/2005’

Another Example
Dec-22 SQL 43
SELECT product,
sum(price * quantity) AS SumSales
max(quantity) AS MaxQuantity
FROM Purchase
GROUP BY product
What does
it mean ?

HAVING Clause
Dec-22 SQL 44
SELECT product, Sum(price * quantity)
FROM Purchase
WHERE date > ‘10/1/2005’
GROUP BY product
HAVING Sum(quantity) > 30
Same query, except that we consider only products that had
at least 100 buyers.
HAVING clause contains conditions on aggregates.

General form of Grouping and Aggregation
SELECT S
FROM R1,…,Rn
WHERE C1
GROUP BY a1,…,ak
HAVING C2
S = may contain attributes a1,…,ak and/or any aggregates but NO OTHER ATTRIBUTES
C1 = is any condition on the attributes in R1,…,Rn
C2 = is any condition on aggregate expressions
Dec-22 SQL 45
Why ?

General form of Grouping and Aggregation
Dec-22 SQL 46
Evaluation steps:
1. Evaluate FROM-WHERE, apply condition C1
2. Group by the attributes a1,…,ak
3. Apply condition C2 to each group (may have aggregates)
4. Compute aggregates in S and return the result
SELECT S
FROM R1,…,Rn
WHERE C1
GROUP BY a1,…,ak
HAVING C2

Advanced SQLizing
1. Getting around INTERSECT and EXCEPT
2. Quantifiers
3. Aggregation v.s. subqueries
Dec-22 SQL 47

1. INTERSECT and EXCEPT:
Dec-22 SQL 48
(SELECT R.A, R.B
FROM R)
INTERSECT
(SELECT S.A, S.B
FROM S)
SELECT R.A, R.B
FROM R
WHERE
EXISTS(SELECT *
FROM S
WHERE R.A=S.A and R.B=S.B)
(SELECT R.A, R.B
FROM R)
EXCEPT
(SELECT S.A, S.B
FROM S)
SELECT R.A, R.B
FROM R
WHERE
NOT EXISTS(SELECT *
FROM S
WHERE R.A=S.A and R.B=S.B)
If R, S have no
duplicates, then can
write without
subqueries
(HOW ?)
INTERSECT and EXCEPT: not in SQL Server

2. Quantifiers
Dec-22 SQL 49
Product ( pname, price, company)
Company( cname, city)
Find all companies that make some products with price < 100
SELECT DISTINCT Company.cname
FROM Company, Product
WHERE Company.cname = Product.company and Product.price < 100
Existential: easy ! 

2. Quantifiers
Dec-22 SQL 50
Product ( pname, price, company)
Company( cname, city)
Find all companies s.t. all of their products have price < 100
Universal: hard ! 
Find all companies that make only products with price < 100
same as:

2. Quantifiers
Dec-22 SQL 51
2. Find all companies s.t. all their products have price < 100
1. Find the other companies: i.e. s.t. some product  100
FROM Company
WHERE Company.cname IN (SELECT Product.company
FROM Product
WHERE Produc.price >= 100
FROM Company
WHERE Company.cname NOT IN (SELECT Product.company
FROM Product
WHERE Produc.price >= 100

3. Group-by v.s. Nested Query
• Find authors who wrote  10 documents:
• Attempt 1: with nested queries
Dec-22 SQL 52
SELECT DISTINCT Author.name
FROM Author
WHERE count(SELECT Wrote.url
FROM Wrote
WHERE Author.login=Wrote.login)
> 10
This is
SQL by
a novice
Author(login,name)
Wrote(login,url)

• Find all authors who wrote at least 10 documents:
• Attempt 2: SQL style (with GROUP BY)
Dec-22 SQL 53
SELECT Author.name
FROM Author, Wrote
WHERE Author.login=Wrote.login
GROUP BY Author.name
HAVING count(wrote.url) > 10
This is
SQL by
an expert
No need for DISTINCT: automatically from GROUP BY

Dec-22 SQL 54
Find authors with vocabulary  10000 words:
SELECT Author.name
FROM Author, Wrote, Mentions
WHERE Author.login=Wrote.login AND Wrote.url=Mentions.url
GROUP BY Author.name
HAVING count(distinct Mentions.word) > 10000
Author(login,name)
Wrote(login,url)
Mentions(url,word)

Two Examples
Dec-22 SQL 55
Store(sid, sname)
Product(pid, pname, price, sid)
Find all stores that sell only products with price > 100
same as:
Find all stores s.t. all their products have price > 100)

SELECT Store.name
FROM Store, Product
WHERE Store.sid = Product.sid
GROUP BY Store.sid, Store.name
HAVING 100 < min(Product.price)
SELECT Store.name
FROM Store
WHERE Store.sid NOT IN
(SELECT Product.sid
FROM Product
WHERE Product.price <= 100)
SELECT Store.name
FROM Store
WHERE
100 < ALL (SELECT Product.price
FROM product
WHERE Store.sid = Product.sid)
Almost equivalent…
Why both ?
Dec-22 SQL 56

Two Examples
Dec-22 SQL 57
Store(sid, sname)
Product(pid, pname, price, sid)
For each store,
find its most expensive product

Two Examples
Dec-22 SQL 58
SELECT Store.sname, max(Product.price)
FROM Store, Product
WHERE Store.sid = Product.sid
GROUP BY Store.sid, Store.sname
SELECT Store.sname, x.pname
FROM Store, Product x
WHERE Store.sid = x.sid and
x.price >=
ALL (SELECT y.price
FROM Product y
WHERE Store.sid = y.sid)
This is easy but doesn’t do what we want:
Better:
But may
return
multiple
product names
per store

Two Examples
Dec-22 SQL 59
SELECT Store.sname, max(x.pname)
FROM Store, Product x
WHERE Store.sid = x.sid and
x.price >=
ALL (SELECT y.price
FROM Product y
WHERE Store.sid = y.sid)
GROUP BY Store.sname
Finally, choose some pid arbitrarily, if there are many
with highest price:

NULLS in SQL
• Whenever we don’t have a value, we can put a NULL
• Can mean many things:
• Value does not exists
• Value exists but is unknown
• Value not applicable
• Etc.
• The schema specifies for each attribute if can be null (nullable attribute) or not
• How does SQL cope with tables that have NULLs ?
Dec-22 SQL 60

Null Values
• If x= NULL then 4*(3-x)/7 is still NULL
• If x= NULL then x=“Joe” is UNKNOWN
• In SQL there are three boolean values:
FALSE = 0
UNKNOWN = 0.5
TRUE = 1
Dec-22 SQL 61

Null Values
• C1 AND C2 = min(C1, C2)
• C1 OR C2 = max(C1, C2)
• NOT C1 = 1 – C1
Rule in SQL: include only tuples that yield TRUE
Dec-22 SQL 62
SELECT *
FROM Person
WHERE (age < 25) AND
(height > 6 OR weight > 190)
E.g.
age=20
heigth=NULL
weight=200

Null Values
Unexpected behavior:
Some Persons are not included !
Dec-22 SQL 63
SELECT *
FROM Person
WHERE age < 25 OR age >= 25

Null Values
Can test for NULL explicitly:
• x IS NULL
• x IS NOT NULL
Now it includes all Persons
Dec-22 SQL 64
SELECT *
FROM Person
WHERE age < 25 OR age >= 25 OR age IS NULL

Outerjoins
Explicit joins in SQL = “inner joins”:
Product(name, category)
Purchase(prodName, store)
Dec-22 SQL 65
SELECT Product.name, Purchase.store
FROM Product JOIN Purchase ON
Product.name = Purchase.prodName
FROM Product, Purchase
WHERE Product.name = Purchase.prodName
Same as:
But Products that never sold will be lost !

Outerjoins
Left outer joins in SQL:
Purchase(prodName, store)
Dec-22 SQL 66
FROM Product LEFT OUTER JOIN Purchase ON

Name Category
Gizmo gadget
Camera Photo
OneClick Photo
ProdName Store
Gizmo Wiz
Camera Ritz
Camera Wiz
Name Store
Gizmo Wiz
Camera Ritz
Camera Wiz
OneClick NULL
Product Purchase
Dec-22 SQL 67

Application
Compute, for each product, the total number of sales in ‘September’
Purchase(prodName, month, store)
Dec-22 SQL 68
SELECT Product.name, count(*)
FROM Product, Purchase
WHERE Product.name = Purchase.prodName
and Purchase.month = ‘September’
GROUP BY Product.name
What’s wrong ?

Application
Compute, for each product, the total number of sales in ‘September’
Purchase(prodName, month, store)
Dec-22 SQL 69
SELECT Product.name, count(*)
FROM Product LEFT OUTER JOIN Purchase ON
and Purchase.month = ‘September’
GROUP BY Product.name
Now we also get the products who sold in 0 quantity

Outer Joins
• Left outer join:
• Include the left tuple even if there’s no match
• Right outer join:
• Include the right tuple even if there’s no match
• Full outer join:
• Include the both left and right tuples even if there’s no match
Dec-22 SQL 70

Modifying the Database
Three kinds of modifications
• Insertions
• Deletions
• Updates
Sometimes they are all called “updates”
Dec-22 SQL 71

Insertions
Dec-22 SQL 72
General form:
Missing attribute  NULL.
May drop attribute names if give them in order.
INSERT INTO R(A1,…., An) VALUES (v1,…., vn)
INSERT INTO Purchase(buyer, seller, product, store)
VALUES (‘Joe’, ‘Fred’, ‘wakeup-clock-espresso-machine’,
‘The Sharper Image’)
Example: Insert a new purchase to the database:

Insertions
Dec-22 SQL 73
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
Dec-22 SQL 74
prodName is foreign key in Product.name
Suppose database got corrupted and we need to fix it:
name listPrice category
gizmo 100 gadgets
prodName buyerName price
camera John 200
gizmo Smith 80
camera Smith 225
Task: insert in Product all prodNames from Purchase
Product
Product(name, listPrice, category)
Purchase(prodName, buyerName, price)
Purchase

Dec-22 SQL 75
INSERT INTO Product(name)
SELECT DISTINCT prodName
FROM Purchase
WHERE prodName NOT IN (SELECT name FROM Product)
gizmo 100 Gadgets
camera - -

Dec-22 SQL 76
INSERT INTO Product(name, listPrice)
SELECT DISTINCT prodName, price
FROM Purchase
WHERE prodName NOT IN (SELECT name FROM Product)
gizmo 100 Gadgets
camera 200 -
camera ?? 225 ?? - Depends on the implementation

Deletions
Dec-22 SQL 77
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.
Example:

Updates
Dec-22 SQL 78
UPDATE PRODUCT
SET price = price/2
WHERE Product.name IN
(SELECT product
FROM Purchase
WHERE Date =‘Oct, 25, 1999’);
Example:

Introduction to Relational
Databases
Rithik Raj Vaishya
Data Scientist | Corporate Trainer

Introduction
• Database – collection of persistent data
• Database Management System (DBMS) – software system that
supports creation, population, and querying of a database

Relational Database
• Relational Database Management System (RDBMS)
• Consists of a number of tables and single schema (definition of tables and
attributes)
• Students (sid, name, login, age, gpa)
Students identifies the table
sid, name, login, age, gpa identify attributes
sid is primary key

An Example Table
• Students (sid: string, name: string, login: string, age: integer,
gpa: real)
sid name login age gpa
50000 Dave dave@cs 19 3.3
53666 Jones jones@cs 18 3.4
53688 Smith smith@ee 18 3.2
53650 Smith smith@math 19 3.8
53831 Madayan madayan@music 11 1.8
53832 Guldu guldu@music 12 2.0

Another example: Courses
• Courses (cid, instructor, quarter, dept)
cid instructor quarter dept
Carnatic101 Jane Fall 06 Music
Reggae203 Bob Summer 06 Music
Topology101 Mary Spring 06 Math
History105 Alice Fall 06 History

Keys
• Primary key – minimal subset of fields that is unique identifier for a
tuple
• sid is primary key for Students
• cid is primary key for Courses
• Foreign key –connections between tables
• Students (sid, name, login, age, gpa)
• How do we express which students take each course?

Many to many relationships
• In general, need a new table
Enrolled(cid, grade, studid)
Studid is foreign key that references sid in Student
table
cid grade studid
Carnatic101 C 53831
Reggae203 B 53832
Topology112 A 53650
History 105 B 53666
sid name login
50000 Dave dave@cs
53666 Jones jones@cs
53688 Smith smith@ee
53650 Smith smith@math
53831 Madayan madayan@music
53832 Guldu guldu@music
Enrolled
Student
Foreign
key

Relational Algebra
• Collection of operators for specifying queries
• Query describes step-by-step procedure for computing answer (i.e.,
operational)
• Each operator accepts one or two relations as input and returns a
relation as output
• Relational algebra expression composed of multiple operators

Basic operators
• Selection – return rows that meet some condition
• Projection – return column values
• Union
• Cross product
• Difference
• Other operators can be defined in terms of basic operators

Example Schema (simplified)
• Students (sid, name, gpa)
• Enrolled (cid, grade, studid)

Selection
Select students with gpa higher than 3.3 from S1:
σgpa>3.3(S1)
sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
S1
sid name gpa
53666 Jones 3.4
53650 Smith 3.8

Projection
Project name and gpa of all students in S1:
name, gpa(S1)
S1
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
name gpa
Dave 3.3
Jones 3.4
Smith 3.2
Smith 3.8
Madayan 1.8
Guldu 2.0

Combine Selection and Projection
• Project name and gpa of students in S1 with gpa
higher than 3.3:
name,gpa(σgpa>3.3(S1))
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
name gpa
Jones 3.4
Smith 3.8

Set Operations
• Union (R U S)
• All tuples in R or S (or both)
• R and S must have same number of fields
• Corresponding fields must have same domains
• Intersection (R ∩ S)
• All tuples in both R and S
• Set difference (R – S)
• Tuples in R and not S

Set Operations (continued)
• Cross product or Cartesian product (R x S)
• All fields in R followed by all fields in S
• One tuple (r,s) for each pair of tuples r  R, s  S

Example: Intersection
sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
sid name gpa
53666 Jones 3.4
53688 Smith 3.2
53700 Tom 3.5
53777 Jerry 2.8
53832 Guldu 2.0
S1 S2
S1  S2 =
sid name gpa
53666 Jones 3.4
53688 Smith 3.2
53832 Guldu 2.0

Joins
• Combine information from two or more tables
• Example: students enrolled in courses:
S1 S1.sid=E.studidE
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
cid grade studid
Carnatic101 C 53831
Reggae203 B 53832
Topology112 A 53650
History 105 B 53666
S1
E

Joins
sid name gpa cid grade studid
53666 Jones 3.4 History105 B 53666
53650 Smith 3.8 Topology112 A 53650
53831 Madayan 1.8 Carnatic101 C 53831
53832 Guldu 2.0 Reggae203 B 53832
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
cid grade studid
Carnatic101 C 53831
Reggae203 B 53832
Topology112 A 53650
History 105 B 53666
S1
E

Relational Algebra Summary
• Algebras are useful to manipulate data types (relations in
this case)
• Set-oriented
• Brings some clarity to what needs to be done
• Opportunities for optimization
• May have different expressions that do same thing
• We will see examples of algebras for other types of data
in this course

Intro to SQL
• CREATE TABLE
• Create a new table, e.g., students, courses, enrolled
• SELECT-FROM-WHERE
• List all CS courses
• INSERT
• Add a new student, course, or enroll a student in a course

Create Table
• CREATE TABLE Enrolled
(studid CHAR(20),
cid CHAR(20),
grade CHAR(20),
PRIMARY KEY (studid, cid),
FOREIGN KEY (studid) references Students)

Select-From-Where query
• “Find all students who are under 18”
SELECT *
FROM Students S
WHERE S.age < 18

Queries across multiple tables (joins)
• “Print the student name and course ID where the student received an
‘A’ in the course”
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid = E.studid AND E.grade = ‘A’

Other SQL features
• MIN, MAX, AVG
• Find highest grade in fall database course
• COUNT, DISTINCT
• How many students enrolled in CS courses in the fall?
• ORDER BY, GROUP BY
• Rank students by their grade in fall database course

Views
• Virtual table defined on base tables defined by a query
• Single or multiple tables
• Security – “hide” certain attributes from users
• Show students in each course but hide their grades
• Ease of use – expression that is more intuitively
obvious to user
• Views can be materialized to improve query
performance

Views
• Suppose we often need names of students who got a
‘B’ in some course:
CREATE VIEW B_Students(name, sid, course)
AS SELECT S.sname, S.sid, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.studid and E.grade = ‘B’
name sid course
Jones 53666 History105
Guldu 53832 Reggae203

Indexes
• Idea: speed up access to desired data
• “Find all students with gpa > 3.3
• May need to scan entire table
• Index consists of a set of entries pointing to locations of each search
key

Types of Indexes
• Clustered vs. Unclustered
• Clustered- ordering of data records same as ordering of
data entries in the index
• Unclustered- data records in different order from index
• Primary vs. Secondary
• Primary – index on fields that include primary key
• Secondary – other indexes

Example: Clustered Index
• Sorted by sid
sid name gpa
50000 Dave 3.3
53650 Smith 3.8
53666 Jones 3.4
53688 Smith 3.2
53831 Madayan 1.8
53832 Guldu 2.0
50000
53600
53800

Example: Unclustered Index
• Sorted by sid
• Index on gpa
sid name gpa
50000 Dave 3.3
53650 Smith 3.8
53666 Jones 3.4
53688 Smith 3.2
53831 Madayan 1.8
53832 Guldu 2.0
1.8
2.0
3.2
3.3
3.4
3.8

Comments on Indexes
• Indexes can significantly speed up query execution
• But inserts more costly
• May have high storage overhead
• Need to choose attributes to index wisely!
• What queries are run most frequently?
• What queries could benefit most from an index?
• Preview of things to come: SDSS

Summary: Why are RDBMS useful?
• Data independence – provides abstract view of the
data, without details of storage
• Efficient data access – uses techniques to store and
retrieve data efficiently
• Reduced application development time – many
important functions already supported
• Centralized data administration
• Data Integrity and Security
• Concurrency control and recovery

So, why don’t scientists use them?
• “I tried to use databases in my project, but they were just too [slow |
hard-to-use | expensive | complex] . So I use files”.
• Gray and Szalay, Where Rubber Meets the Sky: Bridging the Gap Between
Databases and Science

Some other limitations of RDBMS
• Arrays
• Hierarchical data

Example: Taxonomy of Organisms
• Hierarchy of categories:
• Kingdom - phylum – class – order – family – genus - species
• How would you design a relational schema for this?
Animals
Chordates
Vertebrates
Arthropods
birds
insects spiders crustaceans
reptiles mammals

NoSQL
“T
owards the end of RDBMS?”

What is RDBMS
 RDBMS: the relational database
management system.
 Relation: a relation isa 2D table
which has the following features:
 Name
 Attributes
 Tuples
Name
2

Issueswith RDBMS- Scalability
 Issueswith scaling up when the dataset is
just too big e.g. Big Data.
 Not designed to be distributed.
 Looking at multi-node database solutions.
Known as ‘horizontal scaling’.
 Different approaches include:
 Master-slave
 Sharding
3

Scaling RDBMS
Master-Slave
 All writes are written to the master.
All reads are performed against
the replicated slave databases.
 Critical reads may be incorrect as
writes may not have been
propagated down.
 Large data sets can pose problems
as master needs to duplicate data
to slaves.
Sharding
 Scales well for both reads and
writes.
 Not transparent, application needs
to be partition-aware.
 Can nolonger have relationships or
joinsacrosspartitions.
 Loss of referential integrity across
shards.
4

What is NoSQL
 Standsfor Not Only SQL. T
ermwas redefined by Eric Evansafter Carlo
Strozzi.
 Class of non-relational data storage system
s.
 Do not require a fixed table schema nor do they use the concept of joins.
 Relaxation for oneor moreof the ACIDproperties (Atomicity,Consistency,
Isolation, Durability) using CAP theorem.
5

Need of NoSQL
 Explosion of social media sites (Facebook, Twitter, Google etc.) with large
data needs. (Sharding isa problem)
 Rise of cloud-based solutions such as Amazon S3 (simple storage solution).
 Just as moving to dynamically-typed languages (Ruby/Groovy), a shift to
dynamically-typed data with frequent schema changes.
 Expansion of Open-source community.
 NoSQL solution is more acceptable to a client now than a year ago.
6

NoSQL T
ypes
NoSQL database are classified into four types:
• Key Value pair based
• Column based
• Document based
• Graph based
7

Key Value P
air Based
• Designed for processing dictionary. Dictionaries contain a
collection of records having fields containing data.
• Records are stored and retrieved using a key that uniquely
identifies the record, and is used to quickly find the data
within the database.
Example: CouchDB, Oracle NoSQL Database, Riak etc.
We use it for storing session information, user profiles, preferences,
shopping cart data.
We would avoid it when we need to query data having relationships
between entities.
8

Column based
It store data as Column families containing rows that have
many columns associated with a row key. Each row can have
different columns.
Column families are groups of related data that is accessed
together.
Example: Cassandra, HBase, Hypertable, and Amazon
DynamoDB.
We use it for content management systems, blogging platforms, log aggregation.
We would avoid it for systems that are in early development, changing query patterns.
9

Document Based
Thedatabase stores and retrieves documents. It stores documents in
the value part of the key-value store.
Self- describing, hierarchical tree data structures consisting of maps,
collections, and scalar values.
Example: Lotus Notes, MongoDB, Couch DB,Orient DB,Raven DB.
We use it for content management systems, blogging platforms, web analytics, real-time analytics,
e- commerce applications.
We wouldavoid it for systemsthat need complextransactionsspanningmultiple operations or
queries against varying aggregate structures.
10

Graph Based
Store entities and relationships between these entities as nodes
and edges of a graph respectively. Entities have properties.
Traversing the relationships is very fast as relationship between
nodes is not calculated at query time but is actually persisted
as a relationship.
Example: Neo4J, Infinite Graph, OrientDB, FlockDB.
It is well suited for connected data, such as social networks,
spatial data, routing information for goods and supply.
11

CAP Theorem
 According to Eric Brewer a distributed system has3 properties :
 Consistency
 Availability
 Partitions
 We can have at most two of these three properties for any shared-data system
 Toscaleout,wehaveto partition. It leaves a choicebetween consistencyand
availability. ( In almost all cases, we would choose availability over consistency)
 Everyonewhobuilds big applications builds themonCAP:Google, Yahoo,
Facebook, Amazon, eBay, etc.
12

Advantages of NoSQL
 Cheap and easy to implement (open source)
 Data are replicated to multiple nodes (therefore identical and fault-
tolerant) and can be partitioned
 When data is written, the latest version is on at least one node and then
replicated to other nodes
 No single point of failure
 Easy to distribute
 Don't require a schema
13

What is not provided by NoSQL
 Joins
 Group by
 ACID transactions
 SQL
 Integration with applicationsthat are based on SQL
14

Where to use NoSQL
 NoSQL Data storage systems makes sense for applications that process very large
semi-structureddata –like LogAnalysis, Social Networking Feeds,Time-based
data.
 Toimprove programmer productivity by using a database that better matches an
application's needs.
 Toimprove data access performance via some combination of handling larger data
volumes, reducing latency, and improving throughput.
15

Conclusion
 All the choices provided by the rise of NoSQL databases does not mean the demiseof R
R
elational databases are a powerful tool.
 We are entering anera of Polyglot persistence,a technique that usesdifferent data stora
varying data storage needs.It canapply acrossan enterprise or within an individual app
16

R
eferences
1. “NoSQL Databases: An Overview”. Pramod Sadalage, thoughtworks.com(2014)
2. “Data managementincloud environments:NoSQLand NewSQLdata stores”.
Grolinger, K.; Higashino, W. A.; Tiwari, A.; Capretz, M. A. M. (2013). JoCCASA,
Springer.
3. “Making the Shift from Relational to NoSQL”. Couchbase.com(2014).
4. “NoSQL- Death to R
elational Databases”. Scofield, Ben (2010).
17

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