Chapter 4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
Basic Structure
Set Operations
Aggregate Functions
Null Values
Nested Subqueries
Views
Complex Queries
Recursion in SQL
Modification of the Database
Joined Relations
Data-Definition Language
Embedded SQL
Exercises
SQL
4.1 Basic Structure
The Basic Structure of an SQL expression consists of
three clauses: select, from and where:
① The select clause corresponds to the projection
operation of the relational algebra. It is used to list the
attributes desired in the result of a query.
② The from clause corresponds to the Cartesianproduct operation of the relational algebra. It lists the
relations to be scanned in the evaluation of the
expression.
③ The where clause corresponds to the selection
predicate of the relational algebra. It consists of a
predicate involving attributes of the relations that appear
in the from clause.
4.1 Basic Structure
A typical SQL query has the form
select A1,A2,……An
from r1,r2……rm
where p
Each Ai represents an attribute and each ri a relation.
P is a predicate. The query is equivalent to the
relational-algebra expression.
∏ A1,A2,……An (p(r1r2……  rm))
4.1.1 The Select Clause
SQL allows duplicates in relations as well as in the
results of SQL expressions.
Examples:
① Find the names of all branches in the loan relation
Loan-schema=( branch-name, loan-number ,amount)
select branch-name
from
loan
loan-number branch-name amount
loan
branch-name
L-11
Perryridge
900
Perryridge
L-14
Perryridge
1500
Perryridge
L-15
Mianus
2000
Mianus
4.1.1 The Select Clause
② If you want to force the elimination of duplicates, you
can insert the keyword distinct after select.
select distinct branch-name
from loan
loan-number branch-name amount
loan
branch-name
L-11
Perryridge
900
Perryridge
L-14
Perryridge
1500
Mianus
L-15
Mianus
2000
4.1.1 The Select Clause
③ The asterisk symbol “” can be used to denote “all
attributes”.
Loan-schema=(branch-name, loan-number ,amount)
select 
from loan
select branch-name,loan-number,amount
from loan
loan
loan-number branch-name amount
loan-number branch-name amount
L-11
Perryridge
900
L-11
Perryridge
900
L-14
Perryridge
1500
L-14
Perryridge
1500
L-15
Mianus
2000
L-15
Mianus
2000
4.1.1 The Select Clause
④ The select clause can also contain arithmetic
expressions involving the operators,+, -, ,and /, and
operating on constants or attributes of tuples.
select branch-name,loan-number,amount 10
from loan
loan
loan
loan-number branch-name amount
branch-name loan-number amount
L-11
Perryridge
900
Perryridge
L-11
9000
L-14
Perryridge
1500
Perryridge
L-14
15000
L-15
Mianus
2000
Mianus
L-15
20000
4.1.2
The Where Clause
SQL uses the logical connectives and, or, and not. The
operands of the logical connectives can be expressions
involving the comparison operators <, <=, >, >=, =, and
<>.
Examples:
① Find all loan numbers for loans made at the Perryride
branch with loan amounts greater that $1200.
Loan-schema=( branch-name, loan-number ,amount)
select loan-number
from loan
where branch-name=“Perryridge” and amount>1200
4.1.2
The Where Clause
Between:
② Find the loan number of those loans with loan
amounts between $90,000 and $100,000.
Loan-schema=( branch-name, loan-number ,amount)
select loan-number
from loan
where amount between 90000 and 100000
where amount >= 90000 and amount <= 100000
Similarly, we can use the not between comparison
operator.
4.1.3 The from Clause
The from clause by itself defines a Cartesian product
of the relations in the clause.
Examples:
① For all customers who have a loan form the bank, find
their names and loan number and loan amount.
Borrower-schema=(customer-name, loan-number)
Loan-schema=( branch-name, loan-number ,amount)
select distinct customer-name, borrower.loan-number,amount
from borrower, loan
where borrower.loan-number=loan.loan-number
4.1.3 The from Clause
borrower
customer-name
loan
loan-number
loan-number branch-name amount
Hayes
L-10
L-11
Perryridge
900
Johnson
L-14
L-14
Perryridge
1500
Smith
L-15
L-15
Mianus
2000
where borrower.loan-number=loan.loan-number
customer-name
borrower.loannumber
Johnson
L-14
L-14
Perryridge
1500
Smith
L-15
L-15
Mianus
2000
loan.loan-number branch-name
amount
select distinct customer-name, borrower.loan-number,amount
4.1.3 The from Clause
② Find the names and loan numbers of all customers
who have a loan at the Perryridge branch.
Borrower-schema=(customer-name, loan-number)
Loan-schema=( branch-name, loan-number ,amount)
select distinct customer-name, borrower.loan-number
from borrower, loan
where borrower.loan-number=loan.loan-number and
branch-name=“Perryridge”
4.1.4
The Rename Operation
Problems:
①Two relations in the from clause may have attributes
with the same name, in which case a attribute name is
duplicated in the result.
Example: loan-number amount
loan-number amount
loan
loan
d
②If we used an arithmetic expression in the select clause
the resultant attribute does not have a name.
Example: select branch-name, amount 100
namount
③ Even if a attribute name can be derived from the base
relations as in the preceding example, we may want to
change the attribute name in the result.
4.1.4
The Rename Operation
SQL provides a way of renaming the attributes of a
result relation.
Example:
① If we want the attribute name loan-number to be
replaced with the name loan-id. Then we can write.
select distinct customer-name, borrower.loan-number
as loan-id
from borrower, loan
where borrower.loan-number=loan.loan-number and
branch-name=“Perryridge”
4.1.5 Tuple Variables
Tuple variables are defined in the from clause via the
use of the as clause. Tuple variables are most useful for
comparing two tuples in the same relation.
Example:
① Find the names of all branches that have assets greater than at
least one branch located in Brooklyn.
Branch-schema=(branch-name, branch-city,assets)
Brighton 5100
5000
Redwood 6800
Rye
4800
…
>
6200
4900
assets
Brooklyn
4.1.5 Tuple Variables
branch-name branch-city assets
branch-name branch-city assets
Brooklyn 5000
Pownal
Redwood
Rye
4900
Brighton
Brooklyn
6200
Mianus
Palo
5100
branch
T
>
Pownal
Brooklyn
5000
Redwood
Rye
4900
Brighton
Brooklyn
6200
Mianus
Palo
5100
branch
S
select distinct T.branch-name
from branch as T, branch as S
where T.assets＞S.assets and S.branch-city=“Brooklyn”
4.1.6 String Operations
like:
⑴ Percent(%): The % character matches any substring.
⑵ Underscore( _ ): The _ character matches any
character.
Examples:
① “Perry%” matches any string beginning with “Perry”.
② “_ _ _” matches any string of exactly three characters.
4.1.6 String Operations
③ Find the names of all customers whose street address
includes the substring “main”
Customer-schema=(customer-name,customer-street,customer-city)
select customer-name
from customer
where customer-street like “%main%”
customer-name customer-street customer-city
Pittsfield
Adams
Spmain
customer-name
Adams
Brooks
main
Brooklyn
Brooks
Curry
Nmainh
Rye
Curry
Hayes
Park
Stamford
4.1.6 String Operations
⑶ The escape character is used immediately before a
special pattern character to indicate that the special
pattern character is to be treated like a normal
character.
Examples: using a backslash (\) as the escape character
① like “ab\%cd%” escape “\” matches all string
beginning with “ab%cd”
② like “ab\\cd%” escape “\”matches all strings beginning
with “ab\cd”.
4.1.7 Ordering the Display of
Tuples
Order by clause: desc asc
Examples:
① To list in alphabetic order all customers who have a loan at the
Perryridge branch.
Borrower-schema=(customer-name, loan-number)
Loan-schema=( branch-name, loan-number ,amount)
select distinct customer-name
from borrower, loan
where borrower.loan-number=loan.loan-number and
branch-name=“Perryridge”
order by customer-name
(asc)
4.1.7 Ordering the Display of
Tuples
Ordering can be performed on multiple attributes
② List the entire loan relation in descending order of
amount. If several loans have the same amount, we
order than in ascending order by loan number.
Loan-schema=( branch-name, loan-number ,amount)
select ﹡ from loan
order by amount desc, loan-number asc
loan-number branch-name amount
loan-number branch-name amount
L-11
Perryridge
900
L-14asc
L-15
Perryridge
1500
1500
L-14
Mianus
Mianus
1500
L-15
Perryridge
1500
L-11
Perryridge
900
desc
4.2 Set Operations
union, intersect, except： ∪, ∩, －
1. The Union Operation
Example:
① Find all customers having a loan, an account or both at the bank.
customers who
have a loan
customers who
have an account
(select customer-name
from depositor)
union
(select customer-name
from borrower)
Union operation automatically eliminates duplicates. If we want to
retain all duplicates, we must write union all in place of union.
4.2 Set Operations
2. The Intersect Operation
Example:
① Find all customers who have both a loan and an account at the
bank
(select customer-name
from depositor)
intersect
(select customer-name
from borrower)
If we want to retain all duplicates, we must write
intersect all in place of intersect.
4.2 Set Operations
3. The Except Operation
Example:
① Find all customers who have an account but no loan at the bank
(select customer-name
from depositor)
except
(select customer-name
from borrower)
If we want to retain all duplicates, we must write except
all in place of except.
4.3 Aggregate Functions
1. Aggregate functions are functions that take a
collection of values as input and return a single value.
SQL offers five built-in aggregate functions.
Average: avg
Minimum: min
Maximum: max
Total: sum
Count: count
Example: ① Find the average account balance at the
Perryridge branch.
account-number branch-name balance
Perryridge
L-16
500
select avg(balance)
L-93
from account
L-98
where branch-name=“Perryridge”
Redwood
700
Perryridge
900
700
4.3 Aggregate Functions
2. group by clause.
Examples:
①Find the average account balance at each branch.
account-number branch-name balance
Perryridge
L-16
600
L-18
Redwood
700
L-21
Perryridge
800
L-25
Redwood
900
L-33
Perryridge
700
branch-name ebalance
Perryridge
700
Redwood
800
select branch-name, avg(balance) as ebalance
from account group by branch-name
4.3 Aggregate Functions
② Find the number of depositor for each branch.
Depositor-schema=(customer-name, account-number)
Account-schema=(branch-name, account-number, balance)
select branch-name, count(distinct customer-name)
from depositor,account
where depositor.account-number=account.accountnumber
group by branch-name
4.3 Aggregate Functions
3. having clause
Example: ① we might be interested in only those
branches where the average account balances is more
than $750.
account-number branch-name balance
Perryridge
L-16
600
branch-name ebalance
Perryridge
700
Redwood
800
L-18
Redwood
700
L-21
Perryridge
800
ebalance> $750
L-25
Redwood
900
branch-name ebalance
L-33
Perryridge
700
Redwood
800
4.3 Aggregate Functions
branch-name ebalance ebalance> $750 branch-name ebalance
Perryridge
700
Redwood
800
Redwood
800
select branch-name, avg(balance)
from account
group by branch-name
having avg(balance)>750
4.3 Aggregate Functions
4. At times, we wish to treat the entire relation as a
single group, in such cases, we do not use a group by
clause.
Example: ① Find the average balance for all account
select avg(balance) customer-name customer-street customer-city
Pittsfield
Adams
Spmain
from account
Brooks
main
5. count (﹡)
Curry
Nmainh
Example:① Find the
Hayes
Park
number of tuples in the customer relation
select count(﹡) from customer
4
Brooklyn
Rye
Stamford
4.3 Aggregate Functions
6. A where clause and a having clause appear in the
same query
Example:① Find the average balance for each customer who
lives in Harrison and has at least three accounts.
Account-schema=(branch-name, account-number, balance)
Depositor-schema=(customer-name, account-number)
Customer -schema=(customer-name, customer-street,customer-city)
① innerjion depositor, account, customer
② customer-city=“Harrison”
(where)
③ group by depositor.customer-name
④ count(distinct depositor.account-number)>=3
(having)
4.3 Aggregate Functions
① Find the average balance
for each customer who lives
in Harrison. (where)
② Find the average balance
for each customer who has at
least three accounts. (having)
select depositor.customer-name, avg(balance)
from depositor, account, customer
where depositor.account-number=account.account-number and
depositor.customer-name=customer.customer-name and
customer-city=“Harrison”
group by depositor.customer-name
having count(distinct depositor.account-number)>=3
4.4 Null Valus
SQL allows the use of null values to indicate absence of
information about the value of an attribute.
We use the special keyword null in a predicate to test for a null
value.
Example:
① Find all loan numbers that appear in the loan relation with null
value for amount.
select loan-number
select sum(amount)
from loan
from loan
where amount is null
4.5 Nested Subqueries
A subquery is a select-from-where expression that is nested
within another query.
1. Set Membership
in and not in
Examples:
① Find all customers who have both a loan and an account at the
bank. Borrower-schema=(customer-name, loan-number)
Depositor-schema=(customer-name, account-number)
select distinct customer-name
from borrower
where customer-name in (select customer-name
from depositor)
4.5 Nested Subqueries
customer-name account-number
customer-name
loan-number
Hayes
A-102
Hayes
L-16
Johnson
A-101
Curry
depositor
L-93
borrower
① select customer-name
customer-name
Hayes
from depositor
② select distinct customer-name
Curry
from borrower
in
Hayes
where customer-name in
③ select customer-name
Hayes
Johnson
not in
4.5 Nested Subqueries
② Find all customers who have both an account and a loan at the
perryridge branch.
select distinct customer-name
from borrower, loan
where borrower.loan-number=loan.loan-number and
branch-name=“perryridge” and
(branch-name,customer-name) in (“perryridge” ,customer-name)
(select branch-name,customer-name
from depositor, account
where depositor.account-number=account.account-number)
4.5 Nested Subqueries
(select branch-name,customer-name
from depositor, account
where depositor.account-number=account.account-number)
account-number branch-name balance
customer-name account-number
L-16
Perryridge
500
Hayes
L-16
L-93
Redwood
700
Curry
L-93
L-98
Mianus
900
Smith
L-95
account
depositor
branch-name customer-name
Perryridge
Hayes
Redwood
Curry
4.5 Nested Subqueries
loan-number branch-name amount
customer-name loan-number
L-11
Perryridge
900
Hayes
L-11
L-14
Redwood
1500
Smith
L-14
L-15
Mianus
800
Curry
borrower
L-16
loan
(branch-name,customer-name)
(Perryridge, Hayes)
Hayes have a loan at the Perryridge branch
in
Hayes and Curry branch-name customer-name
have a account
Perryridge
Hayes
Redwood
Curry
Hayes have both an account and a loan at the perryridge branch
4.5 Nested Subqueries
③ Find all customer who do have a loan at the bank but
do not have an account at the bank.
select distinct customer-name
from borrower
where customer-name not in
(select customer-name
from depositor)
4.5 Nested Subqueries
④ Select the names of customers who have a loan at
the bank, and whose names are neither “smith” nor
“jones”
select distinct customer-name
from borrower
where customer-name not in (“smith”, “jones”)
4.5 Nested Subqueries
2. Test for Empty Relations exist and not exist
Examples:
① Find all customers who have both an account and a loan at the bank
Borrower-schema=(customer-name, loan-number)
Depositor-schema=(customer-name, account-number)
select distinct customer-name
from borrower
where exists (select *
from depositor
Where depositor.customername=borrower.customer-name)
4.5 Nested Subqueries
customer-name
loan-number
customer-name account-number
Hayes
A-102
Hayes
L-16
Johnson
A-101
Curry
depositor
L-93
borrower
① borrower(Hayes,A-102) (Johnson,A-101)
customer-name
② (select *
from depositor
Where depositor.customer-name=‘Hayes’)
③ where exists(‘Hayes’, ‘L-16’) True
④ select customer-name form (Hayes,A-102)
Hayes
4.5 Nested Subqueries
② Find all customers who have an account at all the branches
located in brooklyn.
Brighton
Brighton
contain
Downtown
Smith brighton
downtown
Downtown
B
except
curry brighton
all the branches
in brooklyn
relation A contains relation B
not exists(B except A)
brighton
A
downtown
Redwood
…
smith
Null
not exists
smith
true
4.5 Nested Subqueries
Depositor-schema=(customer-name, account-number)
Account-schema=(branch-name, account-number, balance)
Branch-schema=(branch-name, branch-city,assets)
not exists
Brighton
Downtown
( (select branch-name
B
Brighton
A
Downtown
smith
from branch
B
where branch-city=“brooklyn”)
except
(select customer-name ,branch-name
from depositor as T, account as R
where T.account-number=R.account-number) )
A
4.5 Nested Subqueries
Depositor-schema=(customer-name, account-number)
Account-schema=(branch-name, account-number, balance)
select distinct S.customer-name
from depositor as S
Where not exists ((select branch-name
from branch
where branch-city=“brooklyn”)
except
(select R.branch-name
from depositor as T, account as R
where T.account-number=R.account-number and
S.customer-name=T.customer-name))
4.5 Nested Subqueries
③ find the names of all students who have not chosen
all courses.(who choose all courses?)
S-schema=(sno,sname,ssex,sage) PK =sno
C-schema=(cno,cname,teacher) PK=cno
SC-schema=(sno,cno,score) PK =(sno,cno)
Smith c-11
C-10
C-11
all the courses
number
not contain
curry c-11
c-10
…
Smith
4.5 Nested Subqueries
S-schema=(sno,sname,ssex,sage)
C-schema=(cno,cname,teacher)
SC-schema=(sno,cno,score)
PK =sno
PK=cno
PK =(sno,cno)
select sname
from s
where exists
false
(select *
true
from c
false
where not exists
true
(select *
from sc
where sc.cno=c.cno and sc.sno=s.sno))
4.5 Nested Subqueries
④ find the names of all students who have chosen the
courses including one course which the NO.1 student
has chosen at lease.
S-schema=(sno,sname,ssex,sage) PK =sno
C-schema=(cno,cname,teacher) PK=cno
SC-schema=(sno,cno,score) PK =(sno,cno)
Smith c-11
C-10
C-11
NO.1
contain
Curry c-10
c-11
Smith
Curry
…
4.5 Nested Subqueries
S-schema=(sno,sname,ssex,sage) PK =sno
C-schema=(cno,cname,teacher) PK=cno
SC-schema=(sno,cno,score) PK =(sno,cno)
sx sno cno score sy sno cno score
select sname
from s
No.1 C-10 90
No.1 C-10 90
true
where exists
No.1 C-11 89
No.1 C-11 89
false
(select *
No.2 C-10 78
No.2 C-10 78
from sc as sx
No.2 C-11 86
No.2 C-11 86
where sno=“NO.1”
No.3 C-11 89
true
and exists (select * No.3 C-11 89
from sc as sy
false
where sy.sno=s.sno and sx.cno=sy.cno))
4.5 Nested Subqueries
⑤ Find the names of all students who choose the courses
include all courses which NO.1 students has chosen at
least.
sx sno cno score sy sno cno score
select sname
No.1 C-10 90
No.1 C-10 90
from s
true
No.1 C-11 89
No.1 C-11 89
where not exists false
(select *
No.2 C-10 78
No.2 C-10 78
from sc as sx
No.2 C-11 86
No.2 C-11 86
where sno=“NO.1”
No.3 C-11 89
No.3 C-11 89
false
and not exists (select *
true
from sc as sy
where sy.sno=s.sno and sx.cno=sy.cno))
4.5 Nested Subqueries
⑥ Find the names of all students who have chosen the
courses not including all courses which NO.1 students
has chosen.
exists not exists
4.5 Nested Subqueries
3. Set comparison
＞some ==“greater than at least one” ( any )
Examples:
① Find the names of all branches that have assets
greater than those of at least one branch located in
brooklyn.
Branch-schema=(branch-name, branch-city,assets)
select distinct T.branch-name
from branch as T, branch as S
where T.assets＞S.assets and S.branch-city=“Brooklyn”
4.5 Nested Subqueries
1350 (Bighton)
1500
select branch-name
1550 (Mianus)
1600
from branch
1900
where assets > some(select assets
?
>
……. (……….)
assets
(brooklyn)
>some
>min()
=some
in
from branch
where branch-city=“brooklyn”)
② Find the names of all branches that have assets
greater than that of each branch in brooklyn.
>some >all
<>all
not in
4.5 Nested Subqueries
③ Find the branch that has the highest average balance
Account-schema=(account-number,branch-name,balance)
1350 (Bighton)
avg(balance)
1900 (Mianus)
……. (……….)
select branch-name
from account
group by branch-name
1500
>=
?
1600
>=all
max()
1900
avg(balance)
all branchs
having avg(balance) ＞=all (select avg(balance) from account
group by branch-name)
4.5 Nested Subqueries
4. Test for the Absence of Duplicate Tuples
unique construct
Examples:
① Find all customers who have only one account at the perryridge
branch.
select T.customer-name
from depositor as T
where unique(select R.customer-name
from account,depositor as R
where T.customer-name=R.customer-name and
R.account-number=account.account-number and
account.branch-name=“perryridge”)
4.5 Nested Subqueries
② Find all customers who have at least two accounts at
the perryridge branch.
unique not unique
4.6 views
The form of the create view command is:
create view v as <query expression>
Examples:
① consider the view consisting of branch names and the names of
customers who have either an account or a loan at that branch.
Assume that we want this view to be called all-customer.
create view all-customer as
(select branch-name, customer-name
from depositor, account
where depositor.account-number=account.account-number)
union
(select branch-name, customer-name
from borrower, loan
where borrower.loan-number=loan.loan-number)
4.6 views
② The attribute names of a view can be specified
explicitly as follows:
create view branch-total-loan(branch-name,total-loan) as
select branch-name, sum(amount)
from loan
group by branch-name
③ Using the view all-customer, we can find all customers
of the perryridge branch.
select customer-name
from all-customer
where branch-name=“perryridge”
4.7 Complex Queries
Derived Relations: Result relation should be given a
name, and the attributes can be renames.
Examples: ① Find the average account balance of those branches
where the average account balance is greater than$1200
Account-schema=(branch-name, account-number, balance)
select branch-name, avg-balance
from (select branch-name, avg(balance)
from account
group by branch-name)
as result(branch-name, avg-balance)
where avg-balance ＞ 1200
4.7 Complex Queries
② Find the branch name which has the largest sum of
account balance in the bank.
select max(tot-balance)
from (select branch-name, sum(balance)
from account
group by branch-name) as branch-total(branchname, tot-balance)
4.7 Complex Queries
The with clause
Examples: ① Find the account number which has the
largest account balance in the bank.
with max-balance(value) as
select max(balance)
from account
select account-number
from account,max-balance
where account.balance=max-balance.value
② Find all branches where the total account deposits
less than the average of the total account deposits at all
branches.
4.7 Complex Queries
with branch-total(branch-name,value) as
select branch-name,sum(balance)
from account
group by branch-name
with branch-total-avg(value) as
select avg(value)
from branch-total
select branch-name
from branch-total,branch-total-avg
where branch-total.value>=branch-total-avg.value
4.8 Recursion in SQL
With recursive clause:
Example: ① suppose the relation manager has a
attributes emp and mgr. We can find every pair(X,Y)such
that X is directly or indirectly managed by Y.
with recursive empl(emp,mgr) as (
mgr
emp
select emp, mgr
Alon Jak
from manager
union
Jak
Kat
select emp,empl.mgr
Met
Fin
from manager,empl
Fin
Lat
where manager.mgr=empl.emp)
select *
Lat
Gut
from empl
4.9 Modification of the Database
1. Deletion
delete from r
where p
(relation)
(predicate)
Notice: We can delete only whole tuples; we can’t delete values
on only particular attributes.
Examples:
① delete all smith’s account records
delete from depositor
where customer-name=“smith”
② delete all loans with loan amounts between $1300 and $1500
delete from loan
where amount between 1300 and 1500
4.9 Modification of the Database
③ delete all accounts at every branch located in perryridge.
delete from account
where branch-name in (select branch-name
from branch
where branch-city=“perryridge”)
④ delete the records of all accounts with balances below the
average at the bank.
delete from account
where balance (select avg(balance)
from account)
4.9 Modification of the Database
2. Insertion
To insert data into a relation, we either specify a tuple to be
inserted or write a query whose result is a set of tuples to be inserted.
Examples:
① Suppose that we wish to insert the fact that there is an account
A-9732 at the perryridge branch and that is has a balance of $1200
insert into account
values(“perryridge”, “A-9732”,1200)
or
insert into account(account-number,branch-name,balance)
values(“A-9732”, “perryridge”,1200)
4.9 Modification of the Database
② suppose that we want to provide as a gift for all loan
customers of the perryride branch, a new $200 savings account
for each loan account they have.let the loan number serve as the
account number for the savings account.
insert into account
select branch-name, loan-number, 200
from loan
where branch-name=“perryridge”
We also need to add tuples to the depositor relation.
4.9 Modification of the Database
Insert into depositor
select customer-name, loan-number
from borrower, loan
where borrower.loan-number=loan.number and
branch-name=“perryridge”
Problems:
① insert into account
select *
from accont
② insert into account
values(null, “A-401”, 1200)
select account-number
from account
where branch-name=“perryridge”
4.9 Modification of the Database
3. Updates
In certain situations, we may wish to change a value in a tuple
without changing all values in the tuple.
Examples:
① account with balance over $10,000 receive 6 percent interest,
whereas all other receive a percent.
update account
set balance=balance*1.06
where balance>10000
update account
set balance=balance*1.05
where balance<=10000
4.9 Modification of the Database
② pay 5 percent interest on accounts whose balance is greater
than average.
update account
set balance=balance*1.05
where balance> (select avg(balance)
form account)
Case construct:
update account
set balance= case
when balance<=10000 then balance*1.05
esle balance*1.06
end
4.9 Modification of the Database
4. Update of a view
Example:
create view branch-loan as
select branch-name, loan-number
from loan
insert into branch-loan
values(“perryridge”, “L-307”)
* A modification is permitted through a view only if the view in
question is defined in terms of one relation of the actual relational
database---that is, of the logical-level database.
4.10 Joined Relations
SQL provide various mechanisms for joining relations,
including condition joins, natural joins and various forms of
outer joins. (from clause)
1. Inner join
Example:
branchname
loannumber
amount
customer Loan-name
number
Downtown
L-170
3000
Jones
L-170
Redwood
L-230
4000
Smith
L-230
Perryridge
L-260
loan
1700
Hayes
L-155
borrower
4.10 Joined Relations
1. Inner join
loan inner join borrower on loan.loan-number=borrower.loan-number
branchname
loanamount customer
number
-name
Loannumber
Downtown
L-170
3000
Jones
L-170
Redwood
L-230
4000
Smith
L-230
 the attributes of the result consist of the attributes of the lefthand-side relation followed by the attributes of the right-hand-side
relation.
Using as
loan inner join borrower on loan.loan-number=borrower.loan-number
as lb ( branch, loan-number, amount, cust, cust-loan-num)
4.10 Joined Relations
2. Left outer join
loan left outer join borrower on loan.loan-number=borrower.loan-number
Steps:
① compute the inner join
② for every tuple t in the left-hand-side relation loan that did not
match any tuple in the right-hand-side relation borrower in the inner
join, a tuple r is added to the result of the join as follows. The
attributes of tuple r that are derived from the left-hand-side relation
are filled with null values.
branchname
loannumber
amount
customer
-name
Loannumber
Downtown
L-170
3000
Jones
L-170
Redwood
L-230
4000
Smith
L-230
Perryridge
L-260
1700
null
null
4.10 Joined Relations
3. Natural join
loan natural inner join borrower
branchname
loanamount customer
number
-name
Downtown
L-170
3000
Jones
Redwood
L-230
4000
Smith
Notices:
The attributes loan-number appears only once in the result of the
natural join.
4.10 Joined Relations
Join type and conditions
① join condition: define which tuples in the two relations match,
and what attributes are present in the result of the join. (natural;
on<predicate>; using<A1,A2……An>)
② join type: define how tuples in each relation that do not match
any tuple in the other relation are treated. (inner join, left outer join,
right outer join, full outer join)
Examples:
① loan natural right outer join borrower
branch-name
loannumber
amount
customer
-name
Downtown
L-170
3000
Jones
Redwood
L-230
4000
Smith
null
L-155
null
Hayes
4.10 Joined Relations
② loan full outer join borrower using (loan-number)
branchname
loannumber
amount
customer
-name
Downtown
L-170
3000
Jones
Redwood
L-230
4000
Smith
Perryridge
L-260
1700
null
L-155
null
Hayes
null
4.10 Joined Relations
③ find all customers who have an account but no loan at the bank.
select d-CN
from (depositor left join borrower
on depositor.customer-name=borrower.customer-name
as db1(d-CN, account-number, b-CN, loan-number)
where b-CN is null
④ find all customers who have either an account or a loan at the
bank.
select customer-name
from (depositor natural full outer join borrower)
where account-number is null or loan-number is null
4.11 Data-Definition Language
The set of relations in a database must be specified to
the system by means of a data definition language(DDL)
Specification function:
① the schema for each relation
② the domain of values associated with each attribute
③ the integrity constraints
④ the set of indices to be maintained for each relation
⑤ the security and authorization information for each relation
⑥ the physical storage structure of each relation on disk
4.11 Data-Definition Language
Domain Types in SQL
Domain types:
char(n) varchar(n) int
smallint
numeric(p,d)
real
time
interval
float(n)
date
 a particular case where it is essential to prohibit null
values is in the primary key of a relation schema.
create domain clause
create domain person-name char(20)
4.11 Data-Definition Language
Schema Definition in SQL
create table command:
create table r (A1D1, A2D2……AnDn,
<integrity-constrant1>,
……
<integrity-constrantk>)
The allowed integrity constraints include:
primary key (Aj1,Aj2……Ajm) nonnull unique
and
check(p)
4.11 Data-Definition Language
check clause:
Examples:
① create table student
(name char(15) not null
student-id char(10) not null
degree-level char(15) not null
primary key (student-id)
check(degree-level in(“bachelors”, “masters”,
“doctorate”)))
① check (branch-name in (select branch-name
from branch))
4.11 Data-Definition Language


Remove a relation:
drop table r
( delete from r )
Add attributes to an existing relation:
alter table r add AD

Drop attributes from a relation:
alter table r drop A
4.12 Embedded SQL
Most SQL products allow SQL statements to be executed both
directly and as part of an application program can typically be
written in a variety of host language.
The fundamental principle underlying embedded SQL, which we
refer to as the dual mode principle, is that any SQL statement that
can be used interactively can also be used in an application
program.
Points:
① Embedded SQL statement are prefixed by EXEC SQL, to
distinguish them from statement of the host language, and are
terminated by a special terminator symbol.
4.12 Embedded SQL
② An executable SQL statement can appear wherever an executable
host statement can appear. Embedded SQL includes some
statements that are purely declarative, not executable.
③ SQL statements can include references to host variable; such
references must include a colon prefix to distinguish them from SQL
column names.
④ All host variable referenced in SQL statements must be declared .
Within an embedded SQL declare section, which is delimited by the
BEGIN and EDN DECLARE SECTION statement.
4.12 Embedded SQL
⑤ Every program containing embedded SQL statements must
include a host variable called SQLSTATE.
⑥ Host variable must have a data type appropriate to the uses to
which they are put.
 Host variables and SQL columns can have the same name.
Problem:
Operations retrieve may rows, not just one, and host languages
are generally not equipped to handle the retrieved of more than one
row at a time.
4.12 Embedded SQL
Solution: cursors
A cursor is a special kind of SQL object that applies to embedded
SQL only. It consists essentially of a kind of pointer that can be
used to run through a collection of rows, pointing to each of the
rows in turn and thereby providing address ability to those rows
one at a time.
4.12 Embedded SQL
Example:
EXEC SQL DECLARE C SURSOR FOR
select customer-name, customer-city
from depositor, customer, account
where depositor.customer-name=customer.customer-name
and account.account-number=depositor.account-number
and account.balance>:amount
END-EXEC
4.12 Embedded SQL
EXEC SQL OPEN C
Do for all depositor rows accessible via c
END-EXEC
EXEC SQL FETCH C INTO :cn, :cc
END-EXEC
EXEC SQL CLOSE C END-EXEC
4.12 Embedded SQL
① The variable c in the preceding expression is called a
cursor for the query. The statement “DECLARE C
CURSOR…” defines a cursor called c.
② DECLARECURSOR is a purely declarative statement,
not executable. The expression is evaluated when the
cursor is opened.
③ The statement “FETCH C INTO…” is then used to
retrieve rows one at a time from the resulting set,
assigning retrieved values to host variables in
accordance with the specifications of the INTO clause in
that statement.
4.12 Embedded SQL
④ Since there will be many rows in the result set, the FEICH with
normally appear within a loop; the loop will be repeated so long as
there are more rows still to come in that result set. On exit from the
loop, cursor c is closed.
⑤ There executable statement are provided to operate on cursors:
OPEN, FETCH and CLOSE
ⅰEXEC SQL OPEN<cursor name>
ⅱEXEC SQL FETCH<cursor name>
INTO <host variable reference commalist>
ⅲEXEC SQL CLOSE<cursor name>
4.12 Embedded SQL
Example:
using the suppliers-parts-projects database, write a program
with embedded SQL statement to list all supplier rows, in supplier
number order. Each supplier row should be immediately followed in
the listing by all project rows for projects supplied by that supplier,
in project number order.
Note that there might be some suppliers who supply no projects at
all.
S={SNO, SNAME, STATUS, CITY}
J={JNO, JNAME, CITY} SPJ={JNO,SNO,PNO,QUENTITY}
4.12 Embedded SQL
First we define two cursors, CS and CJ.
# define no_more_tuples!(strcmp(SQLSTATE,”02000”))
void pinfo(){
EXEC SQL BEGIN DECLARE SECTION;
int SNO, JNO;
char SNAME[20], SCITY[50], STATUS[12], JNAME[20];
char JCITY[50], SQLSTATE[6];
EXEC SQL END DECLARE SECTION;
EXEC SQL DECLARE CS SURSOR FOR
SELECT SNO, SNAME, STATUS, CITY
FROM S
ORDER BY SNO;
4.12 Embedded SQL
EXEC SQL DECLARE CJ SURSOR FOR
SELECT JNO, JNAME, CITY
FROM J
WHERE JNO IN
(SELECT SPJ.JNO
FROM SPJ
WHERE SPJ.SNO=:SNO)
ORDER BY JNO;
EXEC SQL OPEN CS;
while(1){
EXEC SQL TETCH CS INTO :SNO, :SNAME, :STATUS,:SCITY;
if(no_more_tuples) break;
printf(“%d,%c,%c,%c”,SNO,SNAME,STATUS,SCITY);
4.12 Embedded SQL
EXEC SQL OPEN CJ;
while(1){
EXEC SQL TETCH CJ INTO JNO,JNAME,JCITY;
if(no_more_tuples) break;
printf(“%d,%c,%c”,JNO,JNAME,JCITY);
}
EXEC SQL CLOSE CJ;
}
EXEC SQL CLOSE CS;
}
4.12 Embedded SQL
Dynamic SQL
The two principal dynamic statements are PREPARE
and EXECUTE.
Example:
DCL SQLSOURCE CHAR VARYING(65000)
SQLSOURE=“DELETE FROM loan
WHERE amount<300”;
EXEC SQL PREPARE SQLPREPPED FROM: SQLSOURCE;
EXEC SQL ECECUTE SQLPREPPED;
Exercises:
① Relational Model:
Relation S(S#, SNAME, CITY) KEY=S#
Relation P(P#, PNAME, COLOR, WEIGHT) KEY=P#
Relation J(J#, JNAME, CITY) KEY=J#
Relation SPJ(S#, P#, J#, QTY) KEY=S# P# J#
Consider the relational model, for each of the following queries,
given an expression in the relational algebra, the tuple-relationalcalculus, the domain-relational-calculus and the SQL:
a. Find the suppliers’ numbers s# of all suppliers who offer
the parts to both projects J1 and J2.
b. Find the suppliers’ numbers s# of all suppliers who offer the
parts to any projects of either ShangHai or BeiJing.
Exercises:
c. Find the suppliers’ numbers s# of all suppliers of ShangHai who
offer the parts to the projects of their own places.
d. Find the suppliers’ numbers s# of all suppliers of BeiJing who
never offer the red parts.
e. Find the project numbers J# which use the parts offered by S1 at
least.
f. Find the names of all suppliers who offer all parts.
g. Find the suppliers’ numbers s# of all suppliers who offer the
all parts that S1 offered.