1. Schedule Today: Next After that Subqueries, Grouping and Aggregation.
Read Sections
Next
Modifications, Schemas, Views.
Read Sections
After that
Constraints.
Read Sections ,
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2. Union, Intersection, Difference
“(subquery) UNION (subquery)” produces the union of the two relations.
Similarly for INTERSECT, and EXCEPT = set difference.
But: in Oracle set difference is MINUS, not EXCEPT.
Example
Find the drinkers and beers such that the drinker likes the beer and frequents a bar that serves it.
Likes(drinker, beer)
Sells(bar, beer, price)
Frequents(drinker, bar)
(SELECT * FROM Likes)
INTERSECT
(SELECT drinker, beer
FROM Sells, Frequents
WHERE Frequents.bar = Sells.bar);
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3. Forcing Set Semantics SELECT DISTINCT price FROM Sells;
Default for select-from-where is bag; default for union, intersection, and difference is set.
Force set semantics with DISTINCT after SELECT.
Find the different prices charged for beers.
Sells(bar, beer, price)
SELECT DISTINCT price
FROM Sells;
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4. Forcing Bag Semantics Force bag semantics with ALL
In the case of set operations, duplicates are automatically eliminated. To retain duplicates, use union all, intersect all, and except all.
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5. J. Holliday
More on Bags
There may be duplicate tuples in the results of SQL queries. To determine the number of duplicate tuples in the result.
1. If there are c1 copies of a tuple t1 in r and t1 satisfies the predicate P, then there will be c1 copies of t1 in
select * from r where P
If there are c1 copies of a tuple t1 in r, then there will be c1 copies of select A from t1
in select A from r
If there are c1 copies of a tuple t1 in r and c2 copies of t2 in s, then there will be c1* c2 copies of the tuple t1.t2 in select * from r, s
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6. Bag vs Set Bag Semantics Set Semantics SQL query relational algebra
SELECT DISTINCT ...
(query)UNION(query)
(query)INTERSECT(query)
(query)UNION ALL(query) (query)EXCEPT(query)
(query)INTERSECT ALL(query)
(query)EXCEPT ALL(query)
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7. How many tuples? A B 1 a 2 C 2 3 select * r= from s where C = 3
select B
from r
select B s=
from r, s
select B select A
from r, s from r,s
where C = 3 where A=C A B 1 a 2 C 2 3
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8. Forcing Set Semantics with DISTINCTB 1 a 2
select distinct B r=
from r
select distinct B
from r, s s=
select distinct C
from s C 2 3
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9. How many tuples? a 1 2 B A 3 2 C (C=3)(s) r= B(r) B(rs) s=
B(rs) s=
 B (C=3)(rs)
A (C=3)(rs) a 1 2 B A 3 2 C
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10. What is the Result? r = r2 = a 1 2 B A a 1 b B A (Select A from r2)
UNION
(Select A from r)
(Select * from r2)
UNION ALL
(Select * from r)
(Select * from r2)
INTERSECT
(Select * from r)
(Select * from r2)
UNION
(Select * from r)
(Select * from r2)
EXCEPT ALL
(Select * from r)
(Select * from r2)
EXCEPT
(Select * from r)
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11. What is the Result? r = r2 = a 1 2 B A (Select * from r2) INTERSECT
UNION ALL
(Select * from r)
(Select * from r2)
INTERSECT ALL
(Select * from r)
(Select * from r2)
EXCEPT ALL
(Select * from r)
(Select * from r2)
EXCEPT
(Select * from r)
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12. Exercise: Use set operations to-
Find all customers who have both a loan and an account.
Find all customers who have an account but no loan.
Depositor = (customer-name, account#)
Borrower = (customer-name, loan#)
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13. Answers (select customer-name from Depositor) intersect
(select customer-name from Borrower)
except
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14. Join-Based Expressions
Natural join does a cross product and requires the common attributes to be equal.
Can be used to relate 2 select-from-where statements or 2 tables or to define a relation in the FROM-clause.
(Select A,B FROM R) NATURAL JOIN
(Select B,C FROM S)
R NATURAL JOIN S
Select A,B,C FROM R NATURAL JOIN S
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15. Join Semantics Two types of join:
Natural Join - do cross product and choose rows where common attributes are equal (equi-join), delete duplicate attribute.
R NATURAL JOIN S
Cross Join - do cross product, and choose rows as specified by ON.
R JOIN S ON condition
e.g., condition: R.B=S.B
If no condition, just cross product
R CROSS JOIN S
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16. Join Exercise R = S = What is R natural join S ? (no second B) A B C a
101 35 b
103 24
104 65 D B
Smith
101
Jones
102
Johnson
103 A B C D a
101 35
Smith b
103 24
Johnson
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17. Outer Join Expressions
Natural join and Cross Join...On gives us a way to choose tuples by matching attributes from tables. Outer join gives us a way to report on the tuples that don’t match up.
Optional LEFT, RIGHT, or FULL (default) before OUTER.
LEFT = pad (with NULL) dangling tuples of R only; RIGHT = pad dangling tuples of S only.
To do an outer join, do a cross or natural join then add in the tuples from the left or right table (as specified) that didn’t match up and pad them with Null.
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18. Outer Join Exercise R = S = What is R natural left outer join S? A B C
101 35 b
103 24
104 65 D B
Smith
101
Jones
102
Johnson
103 A B C D a
101 35
Smith b
103 24
Johnson
104 65
--Null--
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19. Outer Join Exercise R = S = What is R natural right outer join S? A B
101 35 b
103 24
104 65 D B
Smith
101
Jones
102
Johnson
103
Jones
102
--null--
Johnson 24
103 b
Smith 35
101 a D C B A
-null-
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20. Full Outer Join Exercise
R = S =
What is R natural full outer join S? A B C a
101 35 b
103 24
104 65 D B
Smith
101
Jones
102
Johnson
103 A B C D a
101 35
Smith b
103 24
Johnson
104 65
--null--
102
Jones
--null--
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21. Outer Join Exercise R = S = R left outer join S on R.B=S.B A B C a 10135 b
103 24
104 65 D B
Smith
101
Jones
102
Johnson
103 A B C D a
101 35
Smith b
103 24
Johnson
104 65
--Null--
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22. Outer Join Exercise
Find all customers who have an account but no loan.
Depositor=(customer-name,account#)
Borrower=(customer-name, loan#)
Select customer-name
From Depositor LEFT OUTER NATURAL JOIN Borrower
Where loan# is NULL
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23. Aggregations
Sum, avg, min, max, and count apply to attributes (columns). Also, count(*) applies to tuples.
Use these in lists following SELECT. (also having)
Output has only one tuple
Sells(bar,beer,price)
Find the average price of Bud.
SELECT AVG(price)
FROM Sells
WHERE beer = 'Bud';
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24. Eliminating Duplicates Before Aggregation
Find the number of different prices at which Bud is sold. Sells(bar, beer, price)
SELECT COUNT(DISTINCT price)
FROM Sells
WHERE beer = 'Bud';
DISTINCT may be used in any aggregation, but typically only makes sense with COUNT.
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25. Example Using AVG
Find the average account balance at the Oakland branch.
Account = (branch-name,account#,balance)
select avg(balance)
from Account
where branch-name = "Oakland“
What does the result look like?
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26. Example Using AVG
Find the average account balance at the Oakland branch of accounts over $3000.
Account = (branch-name, account#, balance)
select avg(balance)
from Account
where branch-name =
"Oakland“ and balance>3000
B-name
Acc#
Balance
Oakland
101
2000
102
7000
103
5000
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27. Grouping Example:Find the average sales price for each beer.
Follow select-from-where by GROUP BY and a list of attributes.
The relation that is the result of the FROM and WHERE clauses is grouped according to the values of these attributes, and aggregations take place only within a group.
Example:Find the average sales price for each beer.
Sells(bar, beer, price)
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer;
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28. Avg sales price for each beer
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer;
Joe’s
Bud
2.00
Kelly’s
3.00
Miller
O’Shea
5.00
Bar
Beer
Price
Joe’s
Bud
2.00
Miller
3.00
Kelly’s
O’Shea
5.00
Bud
2.00
3.00
Miller
5.00
Bud
2.50
Miller
4.00
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29. Example
Find, for each drinker, the average price of Bud at the bars they frequent.
Sells(bar, beer, price)
Frequents(drinker, bar)
SELECT drinker, AVG(price)
FROM Frequents, Sells
WHERE beer = 'Bud' AND
Frequents.bar = Sells.bar
GROUP BY drinker;
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30. When Grouping…
Note: grouping occurs after the cross product in FROM and the condition in the WHERE is performed.
When rows (tuples) are grouped, one line of output is produced for each group.
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31. Restriction on SELECT Lists With Aggregation
If any aggregation is used, then each element of a SELECT clause must either be aggregated or appear in a group-by clause (if a where clause is present).
The following might seem a tempting way to find the bar that sells Bud the cheapest:
Sells(bar, beer, price)
SELECT bar, MIN(price)
FROM Sells
WHERE beer = 'Bud';
But it is illegal in most SQL implementations.
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32. HAVING Clauses
HAVING clauses are selections on groups, just as WHERE clauses are selections on tuples.
Condition can use the tuple variables or relations in the FROM and their attributes, just like the WHERE can.
But the tuple variables range only over the group.
And the attribute better make sense within a group; i.e., be one of the grouping attributes.
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33. Example
Find the average price of those beers that are either served in at least 3 bars or manufactured by Anheuser-Busch.
Beers(name, manf)
Sells(bar, beer, price)
SELECT beer, AVG(price)
FROM Sells
GROUP BY beer
HAVING COUNT(*) >= 3 OR
beer IN (
SELECT name
FROM Beers
WHERE manf = 'Anheuser-Busch' );
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34. Before we go on..
How do we find the bar that sells Bud the cheapest? Sells(bar, beer, price)
Select bar
From Sells
Where beer = ‘Bud’ and price =
(select min(price) from Sells
where beer = ‘Bud’ )
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35. Alternatively SELECT bar FROM Sells WHERE beer = ‘Bud’ and
price <= ALL(
SELECT price
WHERE beer = ‘Bud’);
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36. Database Modifications
So far, we have looked at queries that ask about the current state of the database (instance).
We use similar syntax to make changes to the database.
Modification = insert + delete + update.
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37. DB Insert Insertion of a Tuple Example
INSERT INTO relation VALUES (list of values).
Inserts the tuple = list of values, associating values with attributes in the order the attributes were declared.
Forget the order? List the attributes as arguments of the relation.
Example
Likes(drinker, beer)
Insert the fact that Sally likes Bud.
INSERT INTO Likes(drinker, beer)
VALUES('Sally', 'Bud');
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38. Insertion of the Result of a Query
INSERT INTO relation (subquery).
Example
Create a (unary) table of all Sally's buddies, i.e., the people who frequent bars that Sally also frequents.
Frequents(drinker, bar)
CREATE TABLE Buddies(
name char(30) );
INSERT INTO Buddies
(SELECT DISTINCT d2.drinker
FROM Frequents d1, Frequents d2
WHERE d1.drinker = 'Sally' AND
d2.drinker <> 'Sally' AND
d1.bar = d2.bar
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39. Deletion DELETE FROM relation WHERE condition.
Deletes all tuples satisfying the condition from the named relation.
Sally no longer likes Bud.
Likes(drinker, beer)
DELETE FROM Likes
WHERE drinker = 'Sally' AND
beer = 'Bud';
Make the Likes relation empty.
DELETE FROM Likes;
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40. Example
Delete all beers for which there is another beer by the same manufacturer. Beers(name,manf)
DELETE FROM Beers b
WHERE EXISTS
(SELECT name
FROM Beers
WHERE manf = b.manf AND
name <> b.name );
Note alias for relation from which deletion occurs.
Subquery evaluated once for each row of b
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41. Semantics is tricky. If A. B
Semantics is tricky. If A.B. makes Bud and BudLite (only), does deletion of Bud make BudLite not satisfy the condition?
SQL semantics: all conditions in modifications must be evaluated by the system before any changes due to that command occur.
In Bud/Budlite example, we would first identify both beers as targets, and then delete both.
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42. More on Delete
Oracle 8i does not allow complex conditions in the where clause.
You can only delete from one table at a time.
Delete all accounts at every branch located in Fremont. 
delete from Account
where branch-name in
(select branch-name
from Branch
where branch-city="Fremont" )
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43. Updates
UPDATE relation SET list of assignments WHERE condition
Example
Drinker Fred's phone number is now
Drinkers(name, addr, phone)
UPDATE Drinkers
SET phone = ' '
WHERE name = 'Fred';
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44. Example - Update Make $4 the maximum price for beer.
Updates many tuples at once.
Sells(bar, beer, price)
UPDATE Sells
SET price = 4.00
WHERE price > 4.00;
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45. Review/Quiz select branch-name, avg(balance) from Account
where account# > 5000
group by branch-name
having avg(balance) > 2000
B-name
Acc#
Balance
Main
105
1000
5005
3000
Oak
203
3100
5033
1900
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46. Answer
If both a where clause and a having command are present, the where clause is done first.
B-name
AVG(Balance)
Main
3000
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