Basic SQL

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BASIC SQL

Joe Meehean

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WHAT WE ARE GOING TO LEARN

How do we access the data stored in a DBMS?

What happens when we access data in a DBMS?

How do we add or modify data in a DBMS?

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STRUCTURED QUERY LANGUAGE (SQL)

Often pronounced “sequel” Standardized interface for interacting

with DBMSs as standardized as any CS interface gets Oracle, DB2, SQL Server, MySQL, PostgreSQL,…

Used to create/define tables destroy tables look up data insert/modify data delete data

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STRUCTURED QUERY LANGUAGE (SQL)

Declarative language states what answer should look like NOT how to compute it e.g. “Who are the students with GPAs < 2.0?”

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QUERYING TABLES

Looking up information in relational tables Use the SQL SELECT statement SELECT <list of columns>

FROM table Result of a SELECT is another table

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EXTRACTING COLUMNS

SELECT LastName, MajorFROM Students

StdNo Last Name

First Name

Major GPA

732 Arnett Will Theatre 3.74

967 Crumb Robert Art 2.3

764 Vogelzang

Anna Music 3.5

568 Park Phil CS 3.6

587 Park Anna CS 3.9

Students

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EXTRACTING COLUMNS


StdNo Last Name

First Name

Major GPA

732 Arnett Will Theatre 3.74

967 Crumb Robert Art 2.3

764 Vogelzang

Anna Music 3.5

568 Park Phil CS 3.6

587 Park Anna CS 3.9

Students

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EXTRACTING COLUMNS


Last Name

Major

Arnett Theatre

Crumb Art

Vogelzang

Music

Park CS

Park CS

Result

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EXTRACTING COLUMNS


Note the duplicate rows in the result

Last Name

Major

Arnett Theatre

Crumb Art

Vogelzang

Music

Park CS

Park CS

Result

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EXTRACTING COLUMNS

SELECT DISTINCT LastName, MajorFROM Students

DISTINCT removes duplicate rows

Last Name

Major

Arnett Theatre

Crumb Art

Vogelzang

Music

Park CS

Result

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GETTING ALL THE COLUMNS

SELECT * returns all columns

Student ID

LastName

FirstName GPA

S01 Daniels Rush 3.4

S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

SELECT *FROM Students

Students

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GETTING ALL THE COLUMNS

SELECT * returns all columns

Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

SELECT *FROM Students

Result

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CALCULATED FIELDS

Create new columns using arithmetic Can use +, −, *, /, ( ) Should name new column Use the AS keyword

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CALCULATED FIELDS

SELECT PartNo, Pdesc, Pqty * Wholsale AS StockCostFROM Parts

PartsPartNo PQty PDesc Wholesale

P1 2 5mm bolt $0.55

P2 4 10mm nut $0.23

P3 2 5mm wrench $2.50

P4 4 8mm washer $0.67

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CALCULATED FIELDS

SELECT PartNo, Pdesc, Pqty * Wholsale AS StockCostFROM Parts

ResultPartNo StockCost PDesc

P1 $1.10 5mm bolt

P2 $0.92 10mm nut

P3 $5.00 5mm wrench

P4 $2.68 8mm washer

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EXTRACTING SPECIFIC ROWS

Add the WHERE clause to the SELECT SELECT <columns,…>

FROM <table>WHERE <condition>

WHERE conditions similar to C++ conditions

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Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

Students

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Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

SELECT LastName, FirstNameFROM StudentsWHERE GPA >= 3.5

Students

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Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5


Students

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Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5


Students

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LastName

FirstName

Blem Emily

Park Phil


Result

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SQL supports following comparison operators = : equals <, <= : less than, less than or equal to >, >= : greater than, greater than or equal to <> : not equal

Combination of conditions OR: inclusive AND NOT

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SELECT PartNo, Pdesc, PqtyFROM PartsWHERE PDesc = ‘5mm wrench’ OR Pdesc = ‘10mm nut’


P1 2 5mm bolt $0.55

P2 4 10mm nut $0.23



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SELECT PartNo, Pdesc, PqtyFROM PartsWHERE PDesc = ‘5mm wrench’ OR Pdesc = ‘10mm nut’

ResultPartNo PQty PDesc

P2 4 10mm nut

P3 2 5mm wrench

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LIKE operator used for inexact string matching like regular expressions % wild card for any series of characters

e.g., ‘CS%’ matches strings prefixed with CS e.g., ‘%Algebra%’ matches strings that contain Algebra

_ matches a single character e.g., ‘CS24_’ matches CS241 and CS242, but not

CS242A

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SELECT PartNo, Pdesc, PqtyFROM PartsWHERE PDesc LIKE ‘5mm %’


P1 2 5mm bolt $0.55

P2 4 10mm nut $0.23



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SELECT PartNo, Pdesc, PqtyFROM PartsWHERE PDesc LIKE ‘5mm %’

ResultPartNo PQty PDesc

P1 2 5mm bolt

P3 2 5mm wrench

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Matching a range e.g., parts that cost between $0.50 and $1.00 use the BETWEEN-AND operator

SELECT *FROM PartsWHERE Wholesale BETWEEN 0.50 AND 1.00


P1 2 5mm bolt $0.55

P2 4 10mm nut $0.23



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Matching a range e.g., parts that cost between $0.50 and $1.00 use the BETWEEN-AND operator

SELECT *FROM PartsWHERE Wholesale BETWEEN 0.50 AND 1.00

ResultPartNo PQty PDesc Wholesale

P1 2 5mm bolt $0.55


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Testing for null IS NULL IS NOT NULL

SectionNo

<CourseNo> Term Faculty ID

A CS242 Spring null

B CS242 Spring F01

A CS370 Fall F03

A CS375 Fall F02

Offering

SELECT CourseNo, SectionNoFROM OfferingWHERE CourseNo = ‘CS242’

AND FacultyID IS NOT NULL

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Testing for null IS NULL IS NOT NULL

SectionNo

<CourseNo>

B CS242

Result

SELECT CourseNo, SectionNoFROM OfferingWHERE CourseNo = ‘CS242’

AND FacultyID IS NOT NULL

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QUESTIONS?

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SORTING THE RESULTS

Want sorted results ORDER BY clause Ascending order by default Add DESC for descending order More than 1 column

sorts by 1st column, then 2nd, 3rd, etc…

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SORTING THE RESULTS

Student ID

LastName FirstName GPA Major Class

S01 Daniels Rush 3.4 CS JR

S02 Blem Emily 3.9 Math JR

S03 Cook Matt 2.4 Math SR

S04 Park Phil 3.5 Math JR

Students

SELECT LastName, GPAFROM StudentsORDER BY GPA

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SORTING THE RESULTS

LastName GPA

Cook 2.4

Daniels 3.4

Park 3.5

Blem 3.9

Results

SELECT LastName, GPAFROM StudentsORDER BY GPA

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SORTING THE RESULTS

Student ID






Students

SELECT LastName, GPAFROM StudentsORDER BY GPA DESC

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SORTING THE RESULTS

LastName GPA

Blem 3.9

Park 3.5

Daniels 3.4

Cook 2.4

Results

SELECT LastName, GPAFROM StudentsORDER BY GPA DESC

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SORTING THE RESULTS

Student ID






Students

SELECT Major, GPAFROM StudentsORDER BY Major, GPA DESC

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SORTING THE RESULTS

Major GPA

CS 3.4

Math 3.9

Math 3.5

Math 2.4

Students

SELECT Major, GPAFROM StudentsORDER BY Major, GPA DESC

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DATA AGGREGATION

Summarizing data e.g., average GPA of all students called data aggregation

COUNT(*) computes the total number of rows

COUNT(column) counts the non-null values in a column counts duplicates

COUNT( DISTINCT column) counts the non-null unique values in a column

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DATA AGGREGATION

AVG computes the average of a numeric column excludes null values adding DISTINCT computes average of unique

column values SUM

computes the sum of a numeric column excludes null values adding DISTINCT computes sum of unique

column values

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DATA AGGREGATION

MIN finds the smallest value works on both numeric and string columns

MAX finds the largest value works on both numeric and string columns

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DATA AGGREGATION

Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

Students

SELECT AVG(GPA) AS AvgGPAFROM Students

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DATA AGGREGATION

AvgGPA

3.3

Result

SELECT AVG(GPA) AS AvgGPAFROM Students

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DATA AGGREGATION

Student ID

LastName

FirstName GPA


S02 Blem Emily 3.9

S03 Cook Matt 2.4

S04 Park Phil 3.5

Students

SELECT MAX(GPA) AS MaxGPAFROM Students

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DATA AGGREGATION

MaxGPA

3.9

Result

SELECT MAX(GPA) AS MaxGPAFROM Students

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DATA AGGREGATION

OfferID <CourseNo>

SectionNo

Term Faculty ID

O1 CS242 A Spring null

O2 CS242 B Spring F01

O3 CS370 A Spring F03

O4 CS375 A Fall F02

Offering

SELECT COUNT(*) AS NumOfferings,COUNT(DISTINCT CourseNo) AS NumCourse

FROM OfferingWHERE Term = ‘Spring’

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DATA AGGREGATION

Result

SELECT COUNT(*) AS NumOfferings,COUNT(DISTINCT CourseNo) AS NumCourses

FROM OfferingWHERE Term = ‘Spring’

NumOfferings

NumCourses

3 2

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JOINING TABLES

Combining multiple tables into a single query Join

most common way to query multiple tables combines related rows across several tables rows combined must have common value in

column

OfferingCourseNo

CourseDesc

CS242 DataStructures II

CS370 Databases

CS375 Digital Systems

CourseOfferID

<CourseNo>

SectionNo

Term Faculty ID




O4 CS375 A Fall F02

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JOINING TABLES

Combining multiple tables into a single query Join

most common way to query multiple tables combines related rows across several tables rows combined must have common value in

column

OfferID

CourseNo

SectionNo

Term Faculty ID

CourseDesc

O1 CS242 A Spring null DataStructures II

O2 CS242 B Spring F01 DataStructures II

O3 CS370 A Fall F03 Databases

O4 CS375 A Fall F02 Digital Systems

Result

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JOINING TABLES

Joining tables in the WHERE clause Table1.column = Table2.column both tables must be in FROM clause

Joining tables in the FROM clause more explicit option uses JOIN and ON keywords

FROM Table1 JOIN Table2 ON Table1.column = Table2.column

or JOIN and USING keywords only if both columns have the same name FROM Table1 JOIN Table2 USING column

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JOINING TABLES

Using syntactic sugar to introduce brevity Can rename a table in the FROM clause

From Table1 t1, Table2 t2 Then use that name later

WHERE t1.column = t2.column FROM Table1 t1, Table2 t2

ON t1.column = t2.column

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JOINING TABLESOffering

CourseNo

CourseDesc


CS370 Databases


Course

SELECT CourseNo, SectionNo, CourseDescFROM Offering o, Course cWHERE Term = ‘Spring’

AND CourseDesc LIKE ‘%Structures%’AND o.CourseNo = c.CourseNo

OfferID

<CourseNo>

SectionNo

Term Faculty ID




O4 CS375 A Fall F02

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JOINING TABLES

SectionNo CourseNo CourseDesc

A CS242 DataStructures II

B CS242 DataStructures II

Result

SELECT CourseNo, SectionNo, CourseDescFROM Offering o, Course cWHERE Term = ‘Spring’

AND CourseDesc LIKE ‘%Structures%’AND o.CourseNo = c.CourseNo

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JOINING TABLES

SELECT CourseNo, SectionNo, CourseDescFROM Offering JOIN Course USING CourseNoWHERE Term = ‘Spring’

AND CourseDesc LIKE ‘%Structures%’

OfferingCourseNo

CourseDesc


CS370 Databases


CourseOfferID

<CourseNo>

SectionNo

Term Faculty ID




O4 CS375 A Fall F02

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JOINING TABLESResultSectionNo CourseNo CourseDesc

A CS242 DataStructures II

B CS242 DataStructures II

SELECT CourseNo, SectionNo, CourseDescFROM Offering JOIN Course USING CourseNoWHERE Term = ‘Spring’


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JOINING TABLES

SELECT CourseNo, SectionNo, CourseDescFROM Offering o JOIN Course c

ON o.CourseNo = c.CourseNoWHERE Term = ‘Spring’


OfferingCourseNo

CourseDesc


CS370 Databases


CourseOfferID

<CourseNo>

SectionNo

Term Faculty ID




O4 CS375 A Fall F02

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JOINING TABLES

SectionNo

CourseNo CourseNo CourseDesc

A CS242 CS242 DataStructures II

B CS242 CS242 DataStructures II

Result

SELECT CourseNo, SectionNo, CourseDescFROM Offering o JOIN Course c

ON o.CourseNo = c.CourseNoWHERE Term = ‘Spring’


QUESTIONS?

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SUMMARIZING DATA BY CATEGORY

e.g., average GPA for each major GROUP BY statement

columns in SELECT must be in GROUP BY OR part of the data aggregation

GROUP BY multiple columns comma separated list

GROUP BY joined tables join happens first then the grouping

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Student ID






Students

SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major

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Major AvgGPA

Math 3.3

CS 3.4

Result

SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major

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Student ID






Students

SELECT LastName, Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major

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Student ID






Students

SELECT LastName, Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major

LastName is not an aggregate or in GROUP BY

ERROR

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Student ID






Students

SELECT Major, Class, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major, Class

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Major Class AvgGPA

Math JR 3.7

Math SR 2.4

CS JR 3.4

Result

SELECT Major, Class, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY Major, Class

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SELECT Faculty.LastName, COUNT(*) AS ClassesFROM Offering JOIN Faculty USING FacultyIDGROUP BY Faculty.LastName

OfferingFacultyID

LastName

F01 Miller

F02 DeWitt

F03 Turing

FacultyOfferID

<CourseNo>

SectionNo

Term <Faculty ID>

O1 CS242 A Spring

F01

O2 CS242 B Spring

F01

O3 CS370 A Spring

F03

O4 CS375 A Fall F02

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SELECT Faculty.LastName, COUNT(*) AS ClassesFROM Offering JOIN Faculty USING FacultyIDGROUP BY Faculty.LastName

Result

LastName Classes

Miller 2

DeWitt 1

Turing 1

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Limiting the groups that are included like WHERE, but for groups e.g., Majors with an avg GPA >= 3.4

HAVING clause selects groups created by GROUP BY GROUP BY clause musts be listed first

Different from WHERE WHERE removes rows HAVING removes whole groups

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Student ID






Students

SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY MajorHAVING AvgGPA >= 3.4

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SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY MajorHAVING GPA >= 3.4

Result

Major AvgGPA

Math 3.3

CS 3.4

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SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsGROUP BY MajorHAVING GPA >= 3.4

Major AvgGPA

Math 3.3

CS 3.4

Result

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Student ID






Students

SELECT Major, AVG(GPA) AS AvgGPAFROM StudentsWHERE GPA >= 3.4GROUP BY Major

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Student ID






Students


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Major AvgGPA

Math 3.7

CS 3.4

Result

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CONCEPTUAL QUERY EVALUATION

1. FROM tables (with joins)2. exclude rows using WHERE clause3. aggregate rows using GROUP BY4. exclude groups using the HAVING clause5. sort resulting rows by ORDER BY6. remove columns not listed in SELECT

QUESTIONS?

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EFFECTS OF NULL ON QUERIES

WHERE clause includes rows that evaluate true rows evaluating false or null are excluded

Simple conditions conditions without AND, OR, or NOT operators e.g., Salary > 20,000 null values cause simple conditions to evaluate

null

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Compound conditions conditions with AND, OR, or NOT operators e.g., Salary > 20,000 AND LastName LIKE “Mc%” null will cause part of condition to be null

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Compound conditions conditions with AND, OR, or NOT operators e.g., Salary > 20,000 AND LastName LIKE “Mc*” null will cause part of condition to be null

AND True False Null

True True False Null

False False False False

Null Null False Null

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OR True False Null

True True True True

False True False Null

Null True Null Null

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NOT True False Null

False True Null

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Aggregation COUNT

COUNT(*) counts all rows (even ones with nulls) COUNT(column) counts non-null rows

SUM, AVG, MIN, MAX discard nulls

Expressions including a null are null 500 + null = null

Therefore SUM(C1) + SUM(C2) ≠ SUM(C1 + C2) if C2 has nulls it can make C1 + C2 null

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GROUP BY All null values are placed into 1 group

blank value represents null e.g., Get # of classes each faculty is teaching

SELECT FacultyNo, COUNT(*) AS NumRowsFROM OfferingGROUP BY FacultyNo

FacultyNo NumRows

5

F5 6

F12 1

F3 2

F4 3

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MODIFYING DATA

Database data is modified using: INSERT

adds a new row to the database UPDATE

changes column(s) in one or more rows DELETE

remove one or more rows from database

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INSERT STATEMENT

Adding a single row INSERT INTO table_name

(column1, column2, column3, …)VALUES (value1, value2, value3, …)

e.g., add a new faculty member INSERT INTO Faculty

(FacultyNo, FirstName, LastName, Department)VALUES( ‘F346’, ‘Joe’, ‘Meehean’, ‘CS’)

Formatting for literals in VALUES is different for every DBMS check your own DBMS documentation

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INSERT STATEMENT

Adding a set of rows uses a nested SELECT

INSERT INTO table_nameSELECT column1, column2, …FROM other_tableWHERE condition

e.g., adding students to an honor student table HnrStudents and Students have same columns INSERT INTO HnrStudents

SELECT * FROM Students WHERE GPA >= 3.5

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UPDATE STATEMENT

Updating a single column UPDATE table_name

SET column = expressionWHERE conditions

e.g., Updating vacation time for employees with over 2 years of employment UPDATE Employees

SET VacationDays = VacationDays + 14WHERE Years > 2

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UPDATE STATEMENT

Updating a multiple columns UPDATE table_name

SET column1 = expr1, column2 = expr2WHERE conditions

e.g., Change Phil Parks major and class UPDATE Student

SET Major = ‘CS’, Class = ‘SO’WHERE FirstName = ‘Phil’ AND LastName = ‘Park’

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DELETE STATEMENT

Delete selected rows DELETE FROM table

WHERE condition e.g., remove all the non-honor students from

the honor students table DELETE FROM HnrStudents

WHERE GPA < 2.0

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DELETE STATEMENT

Delete all rows DELETE FROM table e.g., remove all of the honor students from

the honor students table DELETE FROM HnrStudents

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DELETE STATEMENT

Delete using a join DELETE table1.*

FROM table1 JOIN table2 ON join_conditionWHERE condition

e.g., remove failing athletes from Athletes(StudentId, TeamId) DELETE Athletes.*

FROM Students s JOIN Athletes a ON s.StudentId = a.StudentIdWHERE Students.GPA < 2.0

QUESTIONS?

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CREATING TABLES

Relational database consists of tables: data organized into rows and columns integrity constraints: rules limiting data

Create tables using CREATE TABLE statement need to give each table a unique name need to give each column a name and data type

CREATE TABLE TableName(column1 data_type1,column2 data_type2,column3 data_type3

)

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DATA TYPES

CHAR (L) fixed L-length strings used for strings of known length e.g., zip codes, state abbreviations full length is allocated even if unused

VARCHAR (L) variable length strings up to L in size, many DBs limit to 255 allocation length is determined by data stored

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DATA TYPES

INTEGER integers, whole numbers

FLOAT (P) decimal values with P significant digits most DBMSs have an upper limit on P e.g., 2.75

DECIMAL (W,R) decimal values with W significant digits R to the right of the decimal point data with fixed precision e.g., $, GPA

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DATA TYPES

BOOLEAN true/false or yes/no

DATE/TIME not standard across DBMSs commonly support DATE, TIME, and TIMESTAMP TIMESTAMP is combined date and time

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CREATING TABLES

CREATE TABLE Student( StdNo VARCHAR(50), StdFirstName VARCHAR(50), StdLastName VARCHAR(50), StdState CHAR(2), StdGPA DECIMAL(3,2), StdEnrollDate DATE, StdProbation BOOLEAN)

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REPRESENTING RELATIONSHIPS

Represented relationships by having the same column in two tables e.g., Part(pid, pname, color) & Catalog(sid, pid,

cost) Connect 1 row from table A to one or more

rows from table B put primary key column from A into table B primary key of A is a foreign key in B

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REPRESENTING RELATIONSHIPS

Customer#

Name Address

C1 Phil 112 Forest

C2 Emily 56 Lawn

C3 Tom 84 Lynn

C4 Erica 96 5th Ave

Customer

Order#

<Customer#>

Total

O1 C1 56.94

O2 C2 1.59

O3 C3 65.39

O4 C4 153.58

O5 C1 14.87

Order

101

INTEGRITY CONSTRAINTS

Must tell DBMS about rules governing data which columns should be unique which columns are primary keys which columns are foreign keys which columns should never be null

Use SQL CONSTRAINT key word table constraints govern keys and uniqueness row constraints govern nulls should provide a name for each constraint

CONSTRAINT ConstraintName ConstraintType

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KEY CONSTRAINTS

PRIMARY KEY (columns) list of columns that define the primary key for a

table no two rows can contain same data in these

columns columns can be a comma separated list goes at the end of table definition

UNIQUE (columns) list of columns that must be unique no two rows can contain same data in these

columns candidate key goes at the end of table definition

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KEY CONSTRAINTS

FOREIGN KEY (column) REFERENCES table column contains either null OR a primary key from table goes at the end of table definition

NOT NULL a column cannot contain a null value mandatory relationships require this goes right after data type of column

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KEY CONSTRAINTS

CREATE TABLE Order( OrderNo INTEGER, CustomerNo INTEGER

CONSTRAINT CustomerRequired NOT NULL,

OrderTotal DECIMAL(10,2),

CONSTRAINT PKOrder PRIMARY KEY (OrderNo), CONSTRAINT FKCustNo FOREIGN KEY (CustumerNo) REFERENCES Customer,)

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MAINTAINING INTEGRITY CONSTRAINTS

Recall we need to take special care when a FK is modified/deleted in its own table two separate cases: update and delete e.g., CustomerNo is deleted in Customer table what to do with CustomerNo in Order table

4 possible actions restrict: do not allow if reference exists cascade: perform same action for referencing

rows nullify: replace foreign key with null default: replace foreign key with default

106


Update ON UPDATE clause goes at end of FOREIGN KEY constraint

Delete ON DELETE clause goes at end of FOREIGN KEY constraint

4 options restrict is default, no need to make a clause CASCADE SET NULL SET DEFAULT

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Default values DEFAULT DefaultValue clause goes at the end of the column definition

CREATE TABLE Customer( CustomerName VARCHAR(50) DEFAULT “Phil”)

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MAINTAINING KEY CONSTRAINTSCREATE TABLE Order( OrderNo INTEGER, CustomerNo INTEGER

CONSTRAINT CustomerRequired NOT NULL, OrderTotal DECIMAL(10,2),

CONSTRAINT PKOrder PRIMARY KEY (OrderNo), CONSTRAINT FKCustNo FOREIGN KEY (CustumerNo) REFERENCES Customer

ON UPDATE CASCADEON DELETE CASCADE,

)

QUESTIONS?

109

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Basic SQL

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