This document provides examples of SQL queries using aggregation functions such as SUM, AVG, MIN, MAX, and COUNT. It demonstrates how to use aggregation functions to calculate values across entire tables or groups of rows. It also shows how to use the GROUP BY clause to aggregate values for each unique value in a column, and the HAVING clause to filter groups based on aggregation results. Proper order of operations for aggregation queries is also discussed.

1. Introduction to Databases
Autumn 2022
Egyptian Russian University IS201 IS201: Database Systems 1
Egyptian Russian University
IS201: Database Systems
Dr. Nermeen Kamel

2. SQL – Basics
(More Practice)
© Praphamontripong
IS201
Database Systems

3. Consider sample data of an Orders table. How many days elapsed
between the order date and the ship date for each order?
Orders
Example 1: Translate
and Clean Up
Select the order number,
order date, ship date,
ship date minus order
as DaysElapsed from the
Orders table
Select the order number,
order date, ship date,
ship date minus order as
DaysElapsed from the
Orders table
© Praphamontripong

4. Example 1: Translate and Clean Up
SELECT OrderNumber,
OrderDate, ShipDate,
(ShipDate - OrderDate)
AS DaysElapsed
FROM Orders;
Select the order number,
order date, ship date, ship
date minus order as
DaysElapsed from the Orders
table
Note: rename the column header
© Praphamontripong

5. Consider sample data of an Orders table. Show me a list of orders
made by each customer in descending date order
Orders
Example 2: Translate and Clean Up
Select the customer ID, order
number, order date, ship date, from
the Orders table for each customer
and then sort by customer and
descending order date
Select the customer ID, order
number, order date, ship date, from
the Orders table for each customer
group by customer ID and then sort
by customer order by customer ID
and descending order date desc
© Praphamontripong

6. Example 2: Translate and Clean Up
Select the customer ID, order
number, order date, ship date, from
the Orders table for each customer
group by customer ID and then sort
by customer order by customer ID
and descending order date desc
SELECT CustomerID,
OrderNumber, OrderDate,
ShipDate
FROM Orders
GROUP BY CustomerID,
OrderNumber
ORDER BY CustomerID,
OrderDate DESC;
Note: Several DBMS requires that everything in “ORDER BY” must be in “GROUP BY” and everything in
“GROUP BY” must be in “SELECT.” (MySQL doesn’t seem to enforce these requirements)
© Praphamontripong

7. Find Courses that Minnie is a TA. Also, list all S_id who takes those
courses
SELECT S_id, Course FROM Student_lecture AS S
WHERE S.Teaching_assistant = “Minnie” ;
Student_lecture
S_id Address Course Teaching_assistant
1234 57 Hockanum Blvd Database Systems Minnie
2345 1400 E. Bellows Database Systems Humpty
3456 900 S. Detroit Cloud Computing Dumpty
1234 57 Hockanum Blvd Web Programming Lang. Mickey
5678 2131 Forest Lake Ln. Software Analysis Minnie
Recap 1: SELECT .. FROM .. WHERE
For each row in S:
if (row.Teaching_assistant = ”Minnie”:
output (row.S_id, row.Course)
FROM
Open an iterator
1
WHERE
Filter each row
2
SELECT
output selected attributes
3
© Praphamontripong

8. Recap 2: SELECT .. FROM .. WHERE
Find all S_id who is taking Database Systems and have Humpty as a TA
Student_lecture
S_id Address Course Teaching_assistant
1234 57 Hockanum Blvd Database Systems Minnie
2345 1400 E. Bellows Database Systems Humpty
3456 900 S. Detroit Cloud Computing Dumpty
1234 57 Hockanum Blvd Web Programming Lang. Mickey
5678 2131 Forest Lake Ln. Software Analysis Minnie
SELECT S_id FROM Student_lecture AS S
WHERE S.Course = “Database Systems” AND
S.Teaching_assistant = “Humpty” ;
S_id
2345
© Praphamontripong

9. Recap 3: SELECT .. FROM .. WHERE
hiring
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
List all names of TAs and the number of hours the TAs work per
week, rename the hours as “Hours_per_week”
SELECT Name, two_week_hours/2 AS Hours_per_week
FROM hiring
© Praphamontripong
Name Hours_per_week
Minnie 10
Humpty 12
Dumpty 15
Minnie 6
Mickey 8

10. Recap 4: SELECT .. FROM .. WHERE
List all years
hiring
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
SELECT Year
FROM hiring
Year
4
3
4
3
2
Duplicates
may occur in output
of an operator
© Praphamontripong

11. Recap 5: SELECT .. FROM .. WHERE
hiring
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
List all years the TAs are in. If multiple TAs are in the same year,
list the year only once
SELECT DISTINCT Year
FROM hiring
© Praphamontripong
Year
4
3
2

12. Recap 6: SELECT .. FROM .. WHERE
hiring
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
List all names of TAs and the number of hours the TAs work per
week, rename the hours as “Hours_per_week”, then order the
result set by names and then Hours_per_week
SELECT Name, two_week_hours/2 AS Hours_per_week
FROM hiring
ORDER BY Name, Hours_per_week
© Praphamontripong
Name Hours_per_week
Dumpty 15
Humpty 12
Mickey 8
Minnie 6
Minnie 10

13. SQL – Aggregates
© Praphamontripong
[A. Silberschatz, H. F
. Korth, S. Sudarshan, Database System Concepts, Ch.3.7 and Ch. 5.5]
[C.M. Ricardo, S.D. Urban, Databases Illuminated, Ch. 5.4]

14. Aggregation Functions
© Praphamontripong
Calculate a value across an entire set or across groups of
rows within the set
SQL uses five aggregation operators:
• SUM – produces the sum of a column with numerical values
• AVG – produces the average of a column with numerical values
• MIN – applied to a column with numerical values, produces the
smallest value
• MAX – applied to a column with numerical values, produces the
largest value
• COUNT – produces the number of (not necessarily distinct)
values in a column

15. Example: SUM
Given the loan schema
loan(loan_number, branch_name, amount)
The sum of the amounts of all loans is expressed by
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
SUM(amount)
8700
Total
© Praphamontripong
8700
SELECT SUM(amount)
FROM loan
SELECT SUM(amount) AS Total
FROM loan

16. Example: AVG
Given the loan schema
loan(loan_number, branch_name, amount)
The average of the amounts of all loans is expressed by
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
AVG(amount)
© Praphamontripong
1242.857142857143
SELECT AVG(amount)
FROM loan

17. Example: MIN
Given the loan schema
loan(loan_number, branch_name, amount)
The smallest amount of loans is expressed by
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
min(amount)
© Praphamontripong
500
SELECT MIN(amount)
FROM loan

18. Example: MAX
Given the loan schema
loan(loan_number, branch_name, amount)
The largest amount of loans is expressed by
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
max(amount)
© Praphamontripong
2000
SELECT MAX(amount)
FROM loan

19. Example: COUNT
Given the loan schema
loan(loan_number, branch_name, amount)
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
count(*)
7
Count(loan_number)
7
Count the number of tuples in the loan table
SELECT count(*) SELECT count(loan_number)
FROM loan FROM loan
Count
© Praphamontripong
rows
Count values
of a specified
column

20. Example: COUNT .. DISTINCT
Given the loan schema
loan(loan_number, branch_name, amount)
Count the number of values in the branch_name column
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
count(branch_name)
7
count(distinct branch_name)
© Praphamontripong
5
SELECT count(branch_name)
FROM loan
SELECT count(DISTINCT branch_name)
FROM loan

21. Aggregation: Order of Actions
1. The FROM clause generates the data set
2. The WHERE clause filters the data set generated by the FROM clause
3. The GROUP BY clause aggregates the data set that was filtered by
the WHERE clause (note: GROUP BY does not sort the result set)
4. The HAVING clause filters the data set that was aggregated by the
GROUP BY clause
5. The SELECT clause transforms the filtered aggregated data set
6. The ORDER BY clause sorts the transformed data set
SELECT select_list
FROM table_source
[WHERE search_condition]
[GROUP BY group_by_expression]
[HAVING search_condition]
[ORDER BY order_expression [ASC | DESC] ]
Order matter
© Praphamontripong

22. Grouping Requirement
Several DBMS requires that the columns appear in the SELECT
clause that are not used in an aggregation function must appear
in the GROUP BY clause
SELECT column_A, column_B, some_aggregation_function
FROM table_source
GROUP BY column_A, column_B
SELECT column_A, column_B, some_aggregation_function
FROM table_source
GROUP BY column_B
© Praphamontripong

23. Example: SUM with GROUP BY
© Praphamontripong
Given the loan schema
loan(loan_number, branch_name, amount)
The sum of the amounts of all loans for each branch is
expressed by
loan_number branch_name amount
L-11 Round Hill 900
L-14 Downtown 1500
L-15 Perryridge 1500
L-16 Perryridge 1300
L-17 Downtown 1000
L-23 Redwood 2000
L-93 Mianus 500
branch_name SUM(amount)
Downtown 2500
Mianus 500
Perryridge 2800
Redwood 2000
Round Hill 900
SELECT branch_name, SUM(amount)
FROM loan
GROUP BY branch_name;

24. Grouping, Aggregation, and Null
© Praphamontripong
• The value NULL is ignored in any aggregation
• Not contribute to a sum, average, or count of an attribute
• Cannot be the minimum or maximum in its column
• Null is treated as an ordinary value when forming groups
• Can have a group with NULL attribute(s)
• When performing any aggregation except count over an
empty bag of values, the result is NULL
• The count of an empty bag is 0

25. HAVING
Clauses
© Praphamontripong
• An aggregation in a HAVING clause applies only to the
tuples of the group being tested – filter groups
• Any attributes of relations in the FROM clause may be
aggregated in the HAVING clause
• But only those attributes that are in the GROUP BY list
may appear unaggregated in the having clause
branch_name SUM(amount)
Downtown 2500
Perryridge 2800
The sum of the amounts of all loans for each branch that has
more than one loan is expressed by
SELECT branch_name, sum(amount)
FROM loan
GROUP BY branch_name
HAVING COUNT(branch_name) > 1;

26. Example 1
List the number of customers in each country. Only include
countries with more than 10 customers
count(id) Country
11 France
11 Germany
13 USA
SELECT count(id), country
FROM Customer
GROUP BY country
HAVING COUNT(id) > 10;
© Praphamontripong

27. Example 2
List the number of customers in each country, except USA,
sorted high to low. Only include countries with 9 or more
customers
count(id) Country
11 France
11 Germany
9 Brazil
SELECT COUNT(id), country
Customer
country <> “USA”
FROM
WHERE
GROUP BY country
HAVING
ORDER BY COUNT(id)
COUNT(id) >= 9
DESC;
© Praphamontripong

28. Final Notes about
Aggregation
© Praphamontripong
#1: Keep the GROUP BY clause small and precise
• Several DBMSs require that all non-aggregated columns must
be in the GROUP BY clause
• Excessive columns in GROUP BY can negatively impact the
query’s performance; make the query hard to read,
understand, rewrite
• For queries that need both aggregations and details, do all
aggregations in subqueries first, then join those to the tables to
retrieve the details

29. Final Notes about
Aggregation
© Praphamontripong
#2: COUNT(*) and COUNT(<column_name>) are different
• COUNT(*) – count all rows, including ones with null values
• COUNT(<column_name>) – count only the rows where the
column value is not NULL
• Sometimes, dividing a query into subqueies can be more
efficient than using a GROUP BY.

30. Final Notes about Aggregation
© Praphamontripong
#3: Use DISTINCT to get distinct counts
• COUNT(*) – returns the number of rows in a group, including
NULL value and duplicates
• COUNT(<column_name>) – returns the number of rows where
the column value is not NULL
• COUNT(DISTINCT <column_name>) – returns the number of
rows with unique, non-null values of the column

31. Wrap-Up
© Praphamontripong
• Aggregation functions
• Order of actions matter when applying aggregation
• Aggregation helps make decisions and succinctly convey
information
What’s next?
• SQL – Joins
• Combine techniques (aggregates and joins) to solve
complex questions

32. SQL - Join
© Praphamontripong [A. Silberschatz, H. F
. Korth, S. Sudarshan, Database System Concepts, Ch.4.1]
[ https://www.dofactory.com/sql/join ]

33. Recap 1: Aggregation
© Praphamontripong
ID name dept_name salary
10101 Srinivasan Comp. Sci. 65000.00
12121 Wu Finance 90000.00
15151 Mozart Music 40000.00
22222 Einstein Physics 95000.00
32343 El Said History 60000.00
33456 Gold Physics 87000.00
45565 Kate Comp. Sci. 75000.00
58583 Katz History 62000.00
76543 Singh Finance 80000.00
76766 Brandt Biology 72000.00
83821 Kate Comp. Sci. 92000.00
98345 Kim Elec. Eng. 80000.00
Instructor (refer to alldbs.sql)
Find average salary
for each department
SELECT dept_name, AVG(salary)
FROM instructor
GROUP BY dept_name;

34. Recap 2: Aggregation
© Praphamontripong
ID name dept_name salary
10101 Srinivasan Comp. Sci. 65000.00
12121 Wu Finance 90000.00
15151 Mozart Music 40000.00
22222 Einstein Physics 95000.00
32343 El Said History 60000.00
33456 Gold Physics 87000.00
45565 Kate Comp. Sci. 75000.00
58583 Katz History 62000.00
76543 Singh Finance 80000.00
76766 Brandt Biology 72000.00
83821 Kate Comp. Sci. 92000.00
98345 Kim Elec. Eng. 80000.00
Find average salary
for each department
that is greater than
70000
SELECT dept_name, AVG(salary)
FROM instructor
GROUP BY dept_name
HAVING AVG(salary) > 70000;
Instructor (refer to alldbs.sql)

35. Recap: Primary Keys, Foreign Keys
Hiring (TA_id, Name, Year, Two_week_hours)
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
Key = one or more
attributes that uniquely
identify a row
TA_Assignment (TA_id, Course)
TA_id Course
1234 Database Systems
2345 Database Systems
3456 Cloud Computing
1234 Software Analysis
5678 Web Programming Lang.
Foreign key = one or more
attributes that uniquely
identify a row in another
table
5678 Mickey 2 16
references
Foreign key describes a relationship between tables
© Praphamontripong

36. JOIN
• Combine related tables or sets of data on key values
• All tables (or relations) must have a unique identifier
(primary key)
• Any related tables must contain a copy of the unique
identifier (foreign key) from the related table
left
table
right
table
FULL JOIN
left
table
right
table
LEFT JOIN
left
table
right
table
RIGHT JOIN
left
table
right
table
(INNER) JOIN
left
table
right
table
NATURAL JOIN
(preserve all rows,
uncommon, thus skip)
© Praphamontripong
(most common)
(preserve all rows
in right table)
(preserve all rows
in left table)

37. NATURAL JOIN
left
table
right
table
NATURAL JOIN
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Not include the repeated column
one(p1, colA, colB) two(p2, p1, colX, colY)
SELECT *
FROM one NATURAL JOIN two;
for each row1 in one:
for each row2 in two:
if (row1.p1 = row2.p1):
output (row1.p1, row1.colA,
row2.p2, row2.colX,
row1.colB,
row2.colY)
Join on common
named
attribute(s).
If no match
found, move on.
© Praphamontripong

38. p1 colA colB p2 colX colY
1 A aa 9001 55 ABC
2 B bb 9006 42 STU
3 C cc 9002 77 PQR
3 C cc 9003 95 DEF
5 E ee 9005 50 KLM
7 G gg 9004 62 GHI
7 G gg 9007 83 XYZ
NATURAL JOIN
left
table
right
table
NATURAL JOIN
Not include the repeated column
one(p1, colA, colB) two(p2, p1, colX,
SELECT *
FROM one
NATURAL JOIN two
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
colY)
Note the order of the columns; order of rows does not matter
© Praphamontripong

39. (INNER) JOIN
left
table
right
table
(INNER) JOIN
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Include the repeated column
one(p1, colA, colB) two(p2, p1, colX, colY)
SELECT *
FROM one JOIN two ON one.p1 =
two.p1;
Join on specified
attribute(s).
If no match
found, move on.
© Praphamontripong
for each row1 in one:
for each row2 in two:
if (row1.p1 = row2.p1):
output (row1.p1, row1.colA, row1.colB,
row2.p2, row2.p1, row2.colX, row2.colY)

40. p1 colA colB p2 p1 colX colY
1 A aa 9001 1 55 ABC
2 B bb 9006 2 42 STU
3 C cc 9002 3 77 PQR
3 C cc 9003 3 95 DEF
5 E ee 9005 5 50 KLM
7 G gg 9004 7 62 GHI
7 G gg 9007 7 83 XYZ
(INNER) JOIN
left
table
right
table
(INNER) JOIN
SELECT *
FROM one JOIN two
ON one.p1 = two.p1
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Include the repeated column
one(p1, colA, colB) two(p2, p1, colX, colY)
SELECT *
FROM one INNER JOIN two
ON one.p1 = two.p1
Note the order of the columns; order of rows does not matter
© Praphamontripong

41. Cartesian (Cross)
Product
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
one(p1, colA, colB) two(p2, p1, colX, colY)
SELECT *
FROM one JOIN two;
for each row1 in one:
for each row2 in two:
output (row1.p1, row1.colA, row1.colB,
row2.p2, row2.p1, row2.colX, row2.colY)
SELECT *
FROM one, two;
© Praphamontripong

42. p1 colA colB p2 p1 colX colY
1 A aa 9001 1 55 ABC
1 A aa 9005 5 50 KLM
1 A aa 9004 7 62 GHI
1 A aa 9003 3 95 DEF
1 A aa 9007 7 83 XYZ
1 A aa 9002 3 77 PQR
1 A aa 9006 2 42 STU
2 B bb 9002 3 77 PQR
2 B bb 9006 2 42 STU
2 B bb 9001 1 55 ABC
2 B bb 9005 5 50 KLM
2 B bb 9004 7 62 GHI
2 B bb 9003 3 95 DEF
2 B bb 9007 7 83 XYZ
3 C cc 9003 3 95 DEF
3 C cc 9007 7 83 XYZ
... … … ... … … …
Cartesian (Cross)
Product
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
one(p1, colA, colB)
two(p2, p1, colX, colY)
one × two
Note the order of the columns; order of rows does not matter
© Praphamontripong

43. NATURAL LEFT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
SELECT *
FROM one NATURAL LEFT OUTER JOIN
two;
Join on common
named
attribute(s).
If no match
found, pad with
NULL values
left
table
right
table
LEFT JOIN
Preserve all rows
in left table
© Praphamontripong

44. p1 colA colB p2 colX colY
1 A aa 9001 55 ABC
2 B bb 9006 42 STU
3 C cc 9002 77 PQR
3 C cc 9003 95 DEF
4 D dd NULL NULL NULL
5 E ee 9005 50 KLM
7 G gg 9004 62 GHI
7 G gg 9007 83 XYZ
8 H hh NULL NULL NULL
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Note the order of the columns; order of rows does not matter
NATURAL LEFT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY) left
table
right
table
LEFT JOIN
© Praphamontripong

45. LEFT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
SELECT *
FROM one LEFT OUTER JOIN two
ON one.p1 = two.p1;
Join on specified
attribute(s).
If no match
found, pad with
NULL values
left
table
right
table
LEFT JOIN
Preserve all rows
in left table
© Praphamontripong

46. p1 colA colB p2 p1 colX colY
1 A aa 9001 1 55 ABC
2 B bb 9006 2 42 STU
3 C cc 9002 3 77 PQR
3 C cc 9003 3 95 DEF
4 D dd NULL NULL NULL NULL
5 E ee 9005 5 50 KLM
7 G gg 9004 7 62 GHI
7 G gg 9007 7 83 XYZ
8 H hh NULL NULL NULL NULL
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Note the order of the columns; order of rows does not matter
LEFT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY) left
table
right
table
LEFT JOIN
© Praphamontripong

47. NATURAL RIGHT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
SELECT *
FROM one NATURAL RIGHT OUTER JOIN
two;
Join on common
named
attribute(s).
If no match
found, pad with
NULL values
left
table
right
table
RIGHT JOIN
Preserve all rows
in right table
© Praphamontripong

48. p1 colA colB p2 colX colY
1 A aa 9001 55 ABC
2 B bb 9006 42 STU
3 C cc 9002 77 PQR
3 C cc 9003 95 DEF
5 E ee 9005 50 KLM
7 G gg 9004 62 GHI
7 G gg 9007 83 XYZ
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Note the order of the columns; order of rows does not matter
NATURAL RIGHT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
Since two.p1 is a foreign key
referencing one.p1,
matches are always founded
left
table
right
table
RIGHT JOIN
© Praphamontripong

49. RIGHT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Join on specified
attribute(s).
If no match
found, pad with
NULL values
left
table
right
table
RIGHT JOIN
SELECT *
FROM one RIGHT OUTER JOIN two
ON one.p1 = two.p1;
Preserve all rows
in right table
© Praphamontripong

50. p1 colA colB
1 A aa
2 B bb
3 C cc
4 D dd
5 E ee
7 G gg
8 H hh
p2 p1 colX colY
9001 1 55 ABC
9002 3 77 PQR
9003 3 95 DEF
9004 7 62 GHI
9005 5 50 KLM
9006 2 42 STU
9007 7 83 XYZ
Note the order of the columns; order of rows does not matter
RIGHT OUTER JOIN
Augment a join by the dangling tuples
one(p1, colA, colB) two(p2, p1, colX, colY)
Since two.p1 is a foreign key
referencing one.p1,
matches are always founded
left
table
right
table
RIGHT JOIN
p1 colA colB p2 p1 colX colY
1 A aa 9001 1 55 ABC
2 B bb 9006 2 42 STU
3 C cc 9002 3 77 PQR
3 C cc 9003 3 95 DEF
5 E ee 9005 5 50 KLM
7 G gg 9004 7 62 GHI
7 G gg 9007 7 83 XYZ
© Praphamontripong

51. Self Joins
© Praphamontripong
• Join with itself
• Useful when the table has a FOREIGN KEY which
references its own PRIMARY KEY
• Can be viewed as joining two copies of the same table
• To do self join, use aliases

52. (Challenging) Example: Self Joins
© Praphamontripong
Find names of all TAs who are assigned to Database Systems and
Software Analysis
Hiring(TA_id, Name, Year, Two_week_hours) TA_Assignment(TA_id, Course)
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
TA_id Course
1234 Database Systems
2345 Database Systems
3456 Cloud Computing
1234 Software Analysis
5678 Web Programming Lang.
2345 Software Analysis
How should we write SQL?

53. (Challenging) Example: Self Joins
© Praphamontripong
Find names of all TAs who are assigned to Database Systems and
Software Analysis
Hiring(TA_id, Name, Year, Two_week_hours) TA_Assignment(TA_id, Course)
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
TA_id Course
1234 Database Systems
2345 Database Systems
3456 Cloud Computing
1234 Software Analysis
5678 Web Programming Lang.
2345 Software Analysis
SELECT H.Name
From Hiring AS H, TA_Assignment AS T
WHERE H.TA_id = T.TA_id AND
T.Course = "Database Systems" AND
T.Course = "Software Analysis";
Will this work?
No.
Returns empty set

54. (Challenging) Example: Self Joins
© Praphamontripong
Find names of all TAs who are assigned to Database Systems and
Software Analysis
Hiring(TA_id, Name, Year, Two_week_hours) TA_Assignment(TA_id, Course)
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
TA_id Course
1234 Database Systems
2345 Database Systems
3456 Cloud Computing
1234 Software Analysis
5678 Web Programming Lang.
2345 Software Analysis
SELECT H.Name
From Hiring AS H, TA_Assignment AS T1,
TA_Assignment AS T2
WHERE H.TA_id = T1.TA_id AND
H.TA_id = T2.TA_id AND
T1.Course = "Database Systems" AND
T2.Course = "Software Analysis";
Will this work?
Name
Minnie
Humpty

55. (Challenging) Example: Self Joins
Find names of all TAs who are assigned to Database Systems and
Software Analysis
Hiring(TA_id, Name, Year, Two_week_hours) TA_Assignment(TA_id, Course)
TA_id Name Year Two_week_hours
1234 Minnie 4 20
2345 Humpty 3 24
3456 Dumpty 4 30
3333 Minnie 3 12
5678 Mickey 2 16
TA_id Course
1234 Database Systems
2345 Database Systems
3456 Cloud Computing
1234 Software Analysis
5678 Web Programming Lang.
2345 Software Analysis
SELECT H.Name
From Hiring AS H, TA_Assignment AS T1,
TA_Assignment AS T2
WHERE H.TA_id =
H.TA_id =
T1.Course
T2.Course
T1.TA_id AND
T2.TA_id AND
= "Database Systems" AND
= "Software Analysis";
Let’s consider each
part of the query
All pairs of courses
a TA is assigned
For each pair, check
for courses
© Praphamontripong

56. Wrap-Up
Join combines data across tables
• Nested-loop
• Natural join (most common)
• Inner join (filter Cartesian product)
• Outer joins (preserve non-matching tuples)
• Self join pattern
Different joining techniques can be used to achieve particular goals
left
table
right
table
LEFT JOIN
left
table
right
table
RIGHT JOIN
left
table
right
table
NATURAL JOIN
left
table
right
table
(INNER) JOIN
© Praphamontripong