1. CPSC-608 Database Systems
Fall 2018
Instructor: Jianer Chen
Office: HRBB 315C
Phone:
Notes 7

2. SQL: Structured Query language
a very-high-level language. * say “what to do” rather than “how to do it.” * avoid a lot of data-manipulation details needed in procedural languages like C or Java.
Database management system figures out the “best” way to execute queries * called “query optimization”
For both data definition and data manipulation.

3. Aggregations
SUM, AVG, COUNT, MIN, and MAX can be applied to a column in a SELECT clause to produce that aggregation on the column.

4. Aggregations
SUM, AVG, COUNT, MIN, and MAX can be applied to a column in a SELECT clause to produce that aggregation on the column.
Also, COUNT(*) counts the number of tuples.

5. Example: Aggregation
From Sells(bar, beer, price), find the average price of Bud:
SELECT AVG(price)
FROM Sells
WHERE beer = ’Bud’;

6. Eliminating Duplicates in Aggregation
Sells(bar, beer, price)
Eliminating Duplicates in Aggregation

7. Eliminating Duplicates in Aggregation
Sells(bar, beer, price)
Eliminating Duplicates in Aggregation
Use DISTINCT inside an aggregation.

8. Eliminating Duplicates in Aggregation
Sells(bar, beer, price)
Eliminating Duplicates in Aggregation
Use DISTINCT inside an aggregation.
Example: find the number of different prices charged for Bud:

9. Eliminating Duplicates in Aggregation
Sells(bar, beer, price)
Eliminating Duplicates in Aggregation
Use DISTINCT inside an aggregation.
Example: find the number of different prices charged for Bud:
SELECT COUNT(DISTINCT price)
FROM Sells
WHERE beer = ’Bud’;

10. NULL is Ignored in Aggregation
NULL never contributes to a sum, average, or count, and can never be the minimum or maximum of a column.

11. NULL is Ignored in Aggregation
NULL never contributes to a sum, average, or count, and can never be the minimum or maximum of a column.
But if there are no non-NULL values in a column, then the result of the aggregation is NULL.

12. Example: Effect of NULL’s
Sells(bar, beer, price)
Example: Effect of NULL’s
SELECT count(*)
FROM Sells
WHERE beer = ’Bud’;
SELECT count(price)

13. Example: Effect of NULL’s
Sells(bar, beer, price)
Example: Effect of NULL’s
SELECT count(*)
FROM Sells
WHERE beer = ’Bud’;
SELECT count(price)
The number of bars
that sell Bud.

14. Example: Effect of NULL’s
Sells(bar, beer, price)
Example: Effect of NULL’s
SELECT count(*)
FROM Sells
WHERE beer = ’Bud’;
SELECT count(price)
The number of bars
that sell Bud.
The number of bars
that sell Bud at a
known price.

15. Grouping
We may follow a SELECT-FROM- WHERE expression by GROUP BY and a list of attributes.

16. Grouping
We may follow a SELECT-FROM- WHERE expression by GROUP BY and a list of attributes.
The relation that results from the SELECT-FROM-WHERE is grouped according to the values of all those attributes, and any aggregation is applied only within each group.

17. Example: Grouping
From Sells(bar, beer, price), find the average price for each beer:

18. Example: Grouping
From Sells(bar, beer, price), find the average price for each beer:
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer;

19. Example: Grouping
From Sells(bar, beer, price), find the average price for each beer:
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer;
Output one tuple
for each group

20. Sells(bar, beer, price)
Frequents(drinker, bar)
Example: Grouping
From Sells and Frequents, find for each drinker the average price of Bud at the bars they frequent:

21. Sells(bar, beer, price)
Frequents(drinker, bar)
Example: Grouping
From Sells and Frequents, find for each drinker the average price of Bud at the bars they frequent:
SELECT drinker, AVG(price)
FROM Frequents, Sells
WHERE beer = ’Bud’ AND
Frequents.bar = Sells.bar
GROUP BY drinker;

22. Sells(bar, beer, price)
Frequents(drinker, bar)
Example: Grouping
From Sells and Frequents, find for each drinker the average price of Bud at the bars they frequent:
compute
drinker-bar-
price for Bud
tuples first,
then group
by drinker.
SELECT drinker, AVG(price)
FROM Frequents, Sells
WHERE beer = ’Bud’ AND
Frequents.bar = Sells.bar
GROUP BY drinker;

23. Restriction on SELECT Lists With Aggregation
If any aggregation is used, then each element of the SELECT list must be either:
Aggregated, or
An attribute on the GROUP BY list.

24. Sells(bar, beer, price)
Illegal Query Example
You might think you could find the bar that sells Bud the cheapest by:
SELECT bar, MIN(price)
FROM Sells
WHERE beer = ’Bud’;
But this query is illegal in SQL.

25. HAVING Clauses HAVING <condition> may follow a GROUP BY clause.
If so, the condition applies to each group, and groups not satisfying the condition are eliminated.

26. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.

27. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer

28. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer

29. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(bar) >= 3 beer IN (SELECT name
FROM Beers
WHERE manf = ’Pete’’s’);
at least 3 bars appear in the beer group

30. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(bar) >= 3 OR
beer IN (SELECT name
FROM Beers
WHERE manf = ’Pete’’s’);
at least 3 bars appear in the beer group

31. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(bar) >= 3 OR
beer IN (SELECT name
FROM Beers
WHERE manf = ’Pete’’s’);
at least 3 bars appear in the beer group

32. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(bar) >= 3 OR
beer IN (SELECT name
FROM Beers
WHERE manf = ’Pete’’s’);
at least 3 bars appear in the beer group
beers made by Pete’s

33. Sells(bar, beer, price)
Beers(name, manf)
Example. From Sells and Beers, find the average price of those beers that are either served in at least three bars or are manufactured by Pete’s.
group tuples (bar, beer, price) in Sells in terms of beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(bar) >= 3 OR
beer IN (SELECT name
FROM Beers
WHERE manf = ’Pete’’s’);
at least 3 bars appear in the beer group
the beer is made by Pete’s
beers made by Pete’s

34. Requirements on HAVING Conditions
These conditions may refer to any relation or tuple-variable in the FROM clause.

35. Requirements on HAVING Conditions
These conditions may refer to any relation or tuple-variable in the FROM clause.
They may refer to attributes of those relations, as long as the attribute makes sense within a group; i.e., it is either:
A grouping attribute, or
Aggregated.

36. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING

37. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING
step 1, input

38. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING
step 1, input
step 2, pick the proper tuples

39. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING
step 1, input
step 2, pick the proper tuples
step 3, group the picked tuples

40. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING
step 1, input
step 2, pick the proper tuples
step 3, group the picked tuples
step 4, pick the proper groups

41. Requirements on HAVING Conditions
It is easier to understand this from an implementation viewpoint:
SELECT
FROM
WHERE
GROUP BY
HAVING
step 5, compute the output
step 1, input
step 2, pick the proper tuples
step 3, group the picked tuples
step 4, pick the proper groups

42. Database Modifications

43. Database Modifications
A modification command does not return a result (as a query does), but changes the database in some way.

44. Database Modifications
A modification command does not return a result (as a query does), but changes the database in some way.
Three kinds of modifications:
Insert a tuple or tuples.
Delete a tuple or tuples.
Update the value(s) of an existing tuple or tuples.

45. Insertion

46. Insertion To insert a single tuple:
INSERT INTO <relation> VALUES (<list of values>);

47. Insertion To insert a single tuple:
Likes(drinker, beer)
Insertion
To insert a single tuple:
INSERT INTO <relation> VALUES (<list of values>);
Example: add to Likes(drinker, beer) the fact that Sally likes Bud.
INSERT INTO Likes VALUES(’Sally’, ’Bud’);

48. Insertion To insert a single tuple:
Likes(drinker, beer)
Insertion
To insert a single tuple:
INSERT INTO <relation> VALUES (<list of values>);
Example: add to Likes(drinker, beer) the fact that Sally likes Bud.
INSERT INTO Likes VALUES(’Sally’, ’Bud’);
We may add a list of attributes to <relation>.

49. Insertion To insert a single tuple:
Likes(drinker, beer)
Insertion
To insert a single tuple:
INSERT INTO <relation> VALUES (<list of values>);
Example: add to Likes(drinker, beer) the fact that Sally likes Bud.
INSERT INTO Likes VALUES(’Sally’, ’Bud’);
We may add a list of attributes to <relation>. Two reasons for doing so:
Forget the order of attributes for the relation.
Don’t have values for all attributes, and want the system to fill in missing ones with default values.

50. Insertion To insert a single tuple:
Likes(drinker, beer)
Insertion
To insert a single tuple:
INSERT INTO <relation> VALUES (<list of values>);
Example: add to Likes(drinker, beer) the fact that Sally likes Bud.
INSERT INTO Likes VALUES(’Sally’, ’Bud’);
We may add a list of attributes to <relation>. Two reasons for doing so:
Forget the order of attributes for the relation.
Don’t have values for all attributes, and want the system to fill in missing ones with default values.
So another solution for the above example:
INSERT INTO Likes(beer, drinker) VALUES(‘Bud’, ’Sally’);

51. Inserting Many Tuples
We may insert the entire result of a query into a relation, using the form:
INSERT INTO <relation>
(<subquery>);

52. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.

53. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);

54. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);
(Sally, Joe’s, Tom, Joe’s)
(Sally, Sue’s, Jeff, Sue’s)
(Sally, Sue’s, Mary, Sue’s) ……

55. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
2. collect the drinkers
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);
(Sally, Joe’s, Tom, Joe’s)
(Sally, Sue’s, Jeff, Sue’s)
(Sally, Sue’s, Mary, Sue’s) ……

56. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
2. collect the drinkers
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);
Tom
Jeff
Mary …

57. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
2. collect the drinkers
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);
Tom
Jeff
Mary …
3. add the drinkers to PotBuddies

58. Frequents(drinker, bar)
Example. Using Frequents, enter into the new relation PotBuddies(name) all of Sally’s “potential buddies,” i.e., those drinkers who frequent at least one bar that Sally also frequents.
1. find all potential buddies of Sally by pairing Sally with those who frequent the bars Sally frequents.
2. collect the drinkers
PotBuddies
D2.drinker
Tom
Jeff
Mary …
INSERT INTO PotBuddies
(SELECT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = ’Sally’ AND
d2.drinker <> ’Sally’ AND
d1.bar = d2.bar);
3. add the drinkers to PotBuddies

59. Deletion

60. Deletion To delete tuples satisfying a condition from some relation:
DELETE FROM <relation>
WHERE <condition>;

61. Deletion To delete tuples satisfying a condition from some relation:
Likes(drinker, beer)
Deletion
To delete tuples satisfying a condition from some relation:
DELETE FROM <relation>
WHERE <condition>;
Example. Delete from Likes the fact that Sally likes Bud:

62. Deletion To delete tuples satisfying a condition from some relation:
Likes(drinker, beer)
Deletion
To delete tuples satisfying a condition from some relation:
DELETE FROM <relation>
WHERE <condition>;
Example. Delete from Likes the fact that Sally likes Bud:
DELETE FROM Likes
WHERE drinker = ’Sally’ AND beer = ’Bud’;

63. Deletion To delete tuples satisfying a condition from some relation:
Likes(drinker, beer)
Deletion
To delete tuples satisfying a condition from some relation:
DELETE FROM <relation>
WHERE <condition>;
Example. Delete from Likes the fact that Sally likes Bud:
DELETE FROM Likes
WHERE drinker = ’Sally’ AND beer = ’Bud’;
To make the relation Likes empty:
Note that no WHERE clause is needed

64. Example: Delete Many Tuples
Delete from Beers(name, manf) all beers for which there is another beer by the same manufacturer.

65. Example: Delete Many Tuples
Delete from Beers(name, manf) all beers for which there is another beer by the same manufacturer.
DELETE FROM Beers b
WHERE EXISTS (
SELECT name FROM Beers
WHERE manf = b.manf AND
name <> b.name);

66. Example: Delete Many Tuples
Delete from Beers(name, manf) all beers for which there is another beer by the same manufacturer.
DELETE FROM Beers b
WHERE EXISTS (
SELECT name FROM Beers
WHERE manf = b.manf AND
name <> b.name);
Beers with the same manufacturer and a different name from the name of the beer represented by tuple b.

67. Semantics of Deletion

68. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.

69. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.

70. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud.

71. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud.

72. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion ?
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)

73. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion ?
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)
Now, when b is the tuple for Bud Lite, do we delete that tuple too? ?

74. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)
Now, when b is the tuple for Bud Lite, do we delete that tuple too?
Answer: we do delete Bud Lite as well.

75. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)
Now, when b is the tuple for Bud Lite, do we delete that tuple too?
Answer: we do delete Bud Lite as well.
Reason: Deletion proceeds in two stages:

76. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)
Now, when b is the tuple for Bud Lite, do we delete that tuple too?
Answer: we do delete Bud Lite as well.
Reason: Deletion proceeds in two stages:
-- Mark all tuples for which WHERE condition holds.

77. name
manf
Bud
Anheuser-Busch
Bud Lite
Semantics of Deletion
Suppose Anheuser-Busch makes only Bud and Bud Lite.
If we come to the tuple b for Bud first.
The subquery is nonempty, because of the Bud Lite tuple, so we delete Bud. (?)
Now, when b is the tuple for Bud Lite, do we delete that tuple too?
Answer: we do delete Bud Lite as well.
Reason: Deletion proceeds in two stages:
-- Mark all tuples for which WHERE condition holds.
-- Delete the marked tuples.

78. Updates

79. Updates To change certain attributes in certain tuples of a relation:
UPDATE <relation>
SET <list of attribute assignments>
WHERE <condition on tuples>;

80. Updates To change certain attributes in certain tuples of a relation:
Drinkers(name, addr, phone)
Updates
To change certain attributes in certain tuples of a relation:
UPDATE <relation>
SET <list of attribute assignments>
WHERE <condition on tuples>;
Change Fred’s phone number to :
UPDATE Drinkers
SET phone = ’ ’
WHERE name = ’Fred’;

81. Updates To change certain attributes in certain tuples of a relation:
Drinkers(name, addr, phone)
Sells(bar, beer, price)
Updates
To change certain attributes in certain tuples of a relation:
UPDATE <relation>
SET <list of attribute assignments>
WHERE <condition on tuples>;
Change Fred’s phone number to :
UPDATE Drinkers
SET phone = ’ ’
WHERE name = ’Fred’;
Make $4 the maximum price for beer
(make updates for several tuples):
UPDATE Sells
SET price = 4.00
WHERE price > 4.00;