Chapter 3: SQL

Chapter 3: SQL

اسلاید 1: Chapter 3: SQL

اسلاید 2: Chapter 3: SQLData DefinitionBasic Query StructureSet OperationsAggregate FunctionsNull ValuesNested SubqueriesComplex Queries ViewsModification of the DatabaseJoined Relations**

اسلاید 3: HistoryIBM Sequel language developed as part of System R project at the IBM San Jose Research LaboratoryRenamed Structured Query Language (SQL)ANSI and ISO standard SQL:SQL-86SQL-89SQL-92 SQL:1999 (language name became Y2K compliant!)SQL:2003Commercial systems offer most, if not all, SQL-92 features, plus varying feature sets from later standards and special proprietary features. Not all examples here may work on your particular system.

اسلاید 4: Data Definition LanguageThe schema for each relation.The domain of values associated with each attribute.Integrity constraintsThe set of indices to be maintained for each relations.Security and authorization information for each relation.The physical storage structure of each relation on disk.Allows the specification of not only a set of relations but also information about each relation, including:

اسلاید 5: Domain Types in SQLchar(n). Fixed length character string, with user-specified length n.varchar(n). Variable length character strings, with user-specified maximum length n.int. Integer (a finite subset of the integers that is machine-dependent).smallint. Small integer (a machine-dependent subset of the integer domain type).numeric(p,d). Fixed point number, with user-specified precision of p digits, with n digits to the right of decimal point. real, double precision. Floating point and double-precision floating point numbers, with machine-dependent precision.float(n). Floating point number, with user-specified precision of at least n digits.More are covered in Chapter 4.

اسلاید 6: Create Table ConstructAn SQL relation is defined using the create table command:create table r (A1 D1, A2 D2, ..., An Dn, (integrity-constraint1), ..., (integrity-constraintk))r is the name of the relationeach Ai is an attribute name in the schema of relation rDi is the data type of values in the domain of attribute AiExample:create table branch (branch_namechar(15) not null, branch_citychar(30), assetsinteger)

اسلاید 7: Integrity Constraints in Create Tablenot nullprimary key (A1, ..., An )Example: Declare branch_name as the primary key for branch and ensure that the values of assets are non-negative.create table branch (branch_namechar(15), branch_citychar(30), assetsinteger, primary key (branch_name))primary key declaration on an attribute automatically ensures not null in SQL-92 onwards, needs to be explicitly stated in SQL-89

اسلاید 8: Drop and Alter Table ConstructsThe drop table command deletes all information about the dropped relation from the database.The alter table command is used to add attributes to an existing relation: alter table r add A D where A is the name of the attribute to be added to relation r and D is the domain of A.All tuples in the relation are assigned null as the value for the new attribute. The alter table command can also be used to drop attributes of a relation:alter table r drop A where A is the name of an attribute of relation rDropping of attributes not supported by many databases

اسلاید 9: Basic Query Structure SQL is based on set and relational operations with certain modifications and enhancementsA typical SQL query has the form: select A1, A2, ..., An from r1, r2, ..., rm where P Ai represents an attributeRi represents a relationP is a predicate.This query is equivalent to the relational algebra expression. The result of an SQL query is a relation.

اسلاید 10: The select ClauseThe select clause list the attributes desired in the result of a querycorresponds to the projection operation of the relational algebraExample: find the names of all branches in the loan relation: select branch_name from loanIn the relational algebra, the query would be: branch_name (loan)NOTE: SQL names are case insensitive (i.e., you may use upper- or lower-case letters.) Some people use upper case wherever we use bold font.

اسلاید 11: The select Clause (Cont.)SQL allows duplicates in relations as well as in query results.To force the elimination of duplicates, insert the keyword distinct after select.Find the names of all branches in the loan relations, and remove duplicatesselect distinct branch_name from loan The keyword all specifies that duplicates not be removed. select all branch_name from loan

اسلاید 12: The select Clause (Cont.)An asterisk in the select clause denotes “all attributes”select * from loanThe select clause can contain arithmetic expressions involving the operation, +, –, , and /, and operating on constants or attributes of tuples.The query: select loan_number, branch_name, amount  100 from loanwould return a relation that is the same as the loan relation, except that the value of the attribute amount is multiplied by 100.

اسلاید 13: The where ClauseThe where clause specifies conditions that the result must satisfyCorresponds to the selection predicate of the relational algebra. To find all loan number for loans made at the Perryridge branch with loan amounts greater than $1200.select loan_number from loan where branch_name = ‘ Perryridge’ and amount > 1200Comparison results can be combined using the logical connectives and, or, and not. Comparisons can be applied to results of arithmetic expressions.

اسلاید 14: The where Clause (Cont.)SQL includes a between comparison operatorExample: Find the loan number of those loans with loan amounts between $90,000 and $100,000 (that is,  $90,000 and  $100,000) select loan_number from loan where amount between 90000 and 100000

اسلاید 15: The from ClauseThe from clause lists the relations involved in the queryCorresponds to the Cartesian product operation of the relational algebra.Find the Cartesian product borrower X loanselect  from borrower, loan Find the name, loan number and loan amount of all customers having a loan at the Perryridge branch.select customer_name, borrower.loan_number, amount from borrower, loan where borrower.loan_number = loan.loan_number and branch_name = ‘Perryridge’

اسلاید 16: The Rename OperationThe SQL allows renaming relations and attributes using the as clause:old-name as new-nameFind the name, loan number and loan amount of all customers; rename the column name loan_number as loan_id.select customer_name, borrower.loan_number as loan_id, amount from borrower, loan where borrower.loan_number = loan.loan_number

اسلاید 17: Tuple VariablesTuple variables are defined in the from clause via the use of the as clause.Find the customer names and their loan numbers for all customers having a loan at some branch.select distinct T.branch_name from branch as T, branch as S where T.assets > S.assets and S.branch_city = ‘ Brooklyn’ Find the names of all branches that have greater assets than some branch located in Brooklyn.select customer_name, T.loan_number, S.amount from borrower as T, loan as S where T.loan_number = S.loan_number

اسلاید 18: String OperationsSQL includes a string-matching operator for comparisons on character strings. The operator “like” uses patterns that are described using two special characters:percent (%). The % character matches any substring.underscore (_). The _ character matches any character.Find the names of all customers whose street includes the substring “Main”.select customer_name from customer where customer_street like ‘%Main%’Match the name “Main%”like ‘Main%’ escape ‘’SQL supports a variety of string operations such asconcatenation (using “||”) converting from upper to lower case (and vice versa) finding string length, extracting substrings, etc.

اسلاید 19: Ordering the Display of TuplesList in alphabetic order the names of all customers having a loan in Perryridge branchselect distinct customer_name from borrower, loan where borrower loan_number = loan.loan_number and branch_name = ‘Perryridge’ order by customer_nameWe may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default.Example: order by customer_name desc

اسلاید 20: DuplicatesIn relations with duplicates, SQL can define how many copies of tuples appear in the result.Multiset versions of some of the relational algebra operators – given multiset relations r1 and r2:1.  (r1): If there are c1 copies of tuple t1 in r1, and t1 satisfies selections ,, then there are c1 copies of t1 in  (r1).2. A (r ): For each copy of tuple t1 in r1, there is a copy of tuple A (t1) in A (r1) where A (t1) denotes the projection of the single tuple t1.3. r1 x r2 : If there are c1 copies of tuple t1 in r1 and c2 copies of tuple t2 in r2, there are c1 x c2 copies of the tuple t1. t2 in r1 x r2

اسلاید 21: Duplicates (Cont.)Example: Suppose multiset relations r1 (A, B) and r2 (C) are as follows: r1 = {(1, a) (2,a)} r2 = {(2), (3), (3)}Then B(r1) would be {(a), (a)}, while B(r1) x r2 would be{(a,2), (a,2), (a,3), (a,3), (a,3), (a,3)}SQL duplicate semantics: select A1,, A2, ..., An from r1, r2, ..., rm where Pis equivalent to the multiset version of the expression:

اسلاید 22: Set OperationsThe set operations union, intersect, and except operate on relations and correspond to the relational algebra operations Each of the above operations automatically eliminates duplicates; to retain all duplicates use the corresponding multiset versions union all, intersect all and except all. Suppose a tuple occurs m times in r and n times in s, then, it occurs:m + n times in r union all smin(m,n) times in r intersect all smax(0, m – n) times in r except all s

اسلاید 23: Set OperationsFind all customers who have a loan, an account, or both:(select customer_name from depositor) except (select customer_name from borrower)(select customer_name from depositor) intersect (select customer_name from borrower) Find all customers who have an account but no loan.(select customer_name from depositor) union (select customer_name from borrower) Find all customers who have both a loan and an account.

اسلاید 24: Aggregate FunctionsThese functions operate on the multiset of values of a column of a relation, and return a valueavg: average value min: minimum value max: maximum value sum: sum of values count: number of values

اسلاید 25: Aggregate Functions (Cont.)Find the average account balance at the Perryridge branch. Find the number of depositors in the bank. Find the number of tuples in the customer relation.select avg (balance) from account where branch_name = ‘Perryridge’select count (*) from customerselect count (distinct customer_name) from depositor

اسلاید 26: Aggregate Functions – Group ByFind the number of depositors for each branch.Note: Attributes in select clause outside of aggregate functions must appear in group by listselect branch_name, count (distinct customer_name) from depositor, account where depositor.account_number = account.account_number group by branch_name

اسلاید 27: Aggregate Functions – Having ClauseFind the names of all branches where the average account balance is more than $1,200. Note: predicates in the having clause are applied after the formation of groups whereas predicates in the where clause are applied before forming groupsselect branch_name, avg (balance) from account group by branch_name having avg (balance) > 1200

اسلاید 28: Null ValuesIt is possible for tuples to have a null value, denoted by null, for some of their attributesnull signifies an unknown value or that a value does not exist.The predicate is null can be used to check for null values.Example: Find all loan number which appear in the loan relation with null values for amount.select loan_number from loan where amount is nullThe result of any arithmetic expression involving null is nullExample: 5 + null returns nullHowever, aggregate functions simply ignore nullsMore on next slide

اسلاید 29: Null Values and Three Valued LogicAny comparison with null returns unknownExample: 5 < null or null <> null or null = nullThree-valued logic using the truth value unknown:OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknownAND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknownNOT: (not unknown) = unknown“P is unknown” evaluates to true if predicate P evaluates to unknownResult of where clause predicate is treated as false if it evaluates to unknown

اسلاید 30: Null Values and AggregatesTotal all loan amountsselect sum (amount ) from loanAbove statement ignores null amountsResult is null if there is no non-null amountAll aggregate operations except count(*) ignore tuples with null values on the aggregated attributes.

اسلاید 31: Nested SubqueriesSQL provides a mechanism for the nesting of subqueries.A subquery is a select-from-where expression that is nested within another query.A common use of subqueries is to perform tests for set membership, set comparisons, and set cardinality.

اسلاید 32: Example QueryFind all customers who have both an account and a loan at the bank. Find all customers who have a loan at the bank but do not have an account at the bankselect distinct customer_name from borrower where customer_name not in (select customer_name from depositor )select distinct customer_name from borrower where customer_name in (select customer_name from depositor )

اسلاید 33: Example QueryFind all customers who have both an account and a loan at the Perryridge branch Note: Above query can be written in a much simpler manner. The formulation above is simply to illustrate SQL features.select distinct customer_name from borrower, loan where borrower.loan_number = loan.loan_number and branch_name = ‘Perryridge’ and (branch_name, customer_name ) in (select branch_name, customer_name from depositor, account where depositor.account_number = account.account_number )

اسلاید 34: Set ComparisonFind all branches that have greater assets than some branch located in Brooklyn. Same query using > some clauseselect branch_name from branch where assets > some (select assets from branch where branch_city = ‘Brooklyn’)select distinct T.branch_name from branch as T, branch as S where T.assets > S.assets and S.branch_city = ‘ Brooklyn’

اسلاید 35: Definition of Some ClauseF <comp> some r t r such that (F <comp> t ) Where <comp> can be:     056(5 < some) = true050) = false505(5  some) = true (since 0  5)(read: 5 < some tuple in the relation) (5 < some) = true(5 = some(= some)  inHowever, ( some)  not in

اسلاید 36: Example QueryFind the names of all branches that have greater assets than all branches located in Brooklyn.select branch_name from branch where assets > all (select assets from branch where branch_city = ‘Brooklyn’)

اسلاید 37: Definition of all ClauseF <comp> all r t r (F <comp> t)056(5 < all) = false6104) = true546(5  all) = true (since 5  4 and 5  6)(5 < all) = false(5 = all( all)  not inHowever, (= all)  in

اسلاید 38: Test for Empty RelationsThe exists construct returns the value true if the argument subquery is nonempty.exists r  r  Ønot exists r  r = Ø

اسلاید 39: Example QueryFind all customers who have an account at all branches located in Brooklyn.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 )) Note that X – Y = Ø  X Y Note: Cannot write this query using = all and its variants

اسلاید 40: Test for Absence of Duplicate TuplesThe unique construct tests whether a subquery has any duplicate tuples in its result.Find all customers who have at most 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’ )

اسلاید 41: Example QueryFind all customers who have at least two accounts at the Perryridge branch. select distinct T.customer_namefrom depositor as Twhere not 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’)

اسلاید 42: Derived RelationsSQL allows a subquery expression to be used in the from clauseFind the average account balance of those branches where the average account balance is greater than $1200.select branch_name, avg_balance from (select branch_name, avg (balance) from account group by branch_name ) as branch_avg ( branch_name, avg_balance ) where avg_balance > 1200Note that we do not need to use the having clause, since we compute the temporary (view) relation branch_avg in the from clause, and the attributes of branch_avg can be used directly in the where clause.

اسلاید 43: With ClauseThe with clause provides a way of defining a temporary view whose definition is available only to the query in which the with clause occurs. Find all accounts with the maximum balance with max_balance (value) as select max (balance) from account select account_number from account, max_balance where account.balance = max_balance.value

اسلاید 44: Complex Query using With ClauseFind all branches where the total account deposit is greater than the average of the total account deposits at all branches. 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

اسلاید 45: ViewsIn some cases, it is not desirable for all users to see the entire logical model (that is, all the actual relations stored in the database.)Consider a person who needs to know a customer’s loan number but has no need to see the loan amount. This person should see a relation described, in SQL, by (select customer_name, loan_number from borrower, loan where borrower.loan_number = loan.loan_number )A view provides a mechanism to hide certain data from the view of certain users. Any relation that is not of the conceptual model but is made visible to a user as a “virtual relation” is called a view.

اسلاید 46: View DefinitionA view is defined using the create view statement which has the formcreate view v as < query expression >where <query expression> is any legal SQL expression. The view name is represented by v.Once a view is defined, the view name can be used to refer to the virtual relation that the view generates.View definition is not the same as creating a new relation by evaluating the query expression Rather, a view definition causes the saving of an expression; the expression is substituted into queries using the view.

اسلاید 47: Example QueriesA view consisting of branches and their customers Find all customers of the Perryridge branchcreate 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 )select customer_name from all_customer where branch_name = ‘Perryridge’

اسلاید 48: Views Defined Using Other ViewsOne view may be used in the expression defining another view A view relation v1 is said to depend directly on a view relation v2 if v2 is used in the expression defining v1A view relation v1 is said to depend on view relation v2 if either v1 depends directly to v2 or there is a path of dependencies from v1 to v2 A view relation v is said to be recursive if it depends on itself.

اسلاید 49: View ExpansionA way to define the meaning of views defined in terms of other views.Let view v1 be defined by an expression e1 that may itself contain uses of view relations.View expansion of an expression repeats the following replacement step:repeat Find any view relation vi in e1 Replace the view relation vi by the expression defining vi until no more view relations are present in e1As long as the view definitions are not recursive, this loop will terminate

اسلاید 50: Modification of the Database – DeletionDelete all account tuples at the Perryridge branchdelete from account where branch_name = ‘Perryridge’Delete all accounts at every branch located in the city ‘Needham’.delete from account where branch_name in (select branch_name from branch where branch_city = ‘Needham’)

اسلاید 51: Example QueryDelete the record of all accounts with balances below the average at the bank. delete from account where balance < (select avg (balance ) from account )Problem: as we delete tuples from deposit, the average balance changesSolution used in SQL: 1. First, compute avg balance and find all tuples to delete 2. Next, delete all tuples found above (without recomputing avg or retesting the tuples)

اسلاید 52: Modification of the Database – InsertionAdd a new tuple to accountinsert into account values (‘A-9732’, ‘Perryridge’,1200) or equivalently insert into account (branch_name, balance, account_number) values (‘Perryridge’, 1200, ‘A-9732’)Add a new tuple to account with balance set to nullinsert into account values (‘A-777’,‘Perryridge’, null )

اسلاید 53: Modification of the Database – InsertionProvide as a gift for all loan customers of the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account insert into account select loan_number, branch_name, 200 from loan where branch_name = ‘Perryridge’ insert into depositor select customer_name, loan_number from loan, borrower where branch_name = ‘ Perryridge’ and loan.account_number = borrower.account_numberThe select from where statement is evaluated fully before any of its results are inserted into the relation (otherwise queries like insert into table1 select * from table1 would cause problems)

اسلاید 54: Modification of the Database – UpdatesIncrease all accounts with balances over $10,000 by 6%, all other accounts receive 5%.Write two update statements:update account set balance = balance  1.06 where balance > 10000update account set balance = balance  1.05 where balance  10000The order is importantCan be done better using the case statement (next slide)

اسلاید 55: Case Statement for Conditional UpdatesSame query as before: Increase all accounts with balances over $10,000 by 6%, all other accounts receive 5%. update account set balance = case when balance <= 10000 then balance *1.05 else balance * 1.06 end

اسلاید 56: Update of a ViewCreate a view of all loan data in the loan relation, hiding the amount attributecreate view branch_loan as select branch_name, loan_number from loanAdd a new tuple to branch_loaninsert into branch_loan values (‘Perryridge’, ‘L-307’)This insertion must be represented by the insertion of the tuple(‘L-307’, ‘Perryridge’, null )into the loan relation

اسلاید 57: Updates Through Views (Cont.)Some updates through views are impossible to translate into updates on the database relationscreate view v as select branch_name from account insert into v values (‘L-99’, ‘ Downtown’, ‘23’)Others cannot be translated uniquelyinsert into all_customer values (‘ Perryridge’, ‘John’)Have to choose loan or account, and create a new loan/account number!Most SQL implementations allow updates only on simple views (without aggregates) defined on a single relation

اسلاید 58: Joined Relations**Join operations take two relations and return as a result another relation.These additional operations are typically used as subquery expressions in the from clauseJoin condition – defines which tuples in the two relations match, and what attributes are present in the result of the join.Join type – defines how tuples in each relation that do not match any tuple in the other relation (based on the join condition) are treated.

اسلاید 59: Joined Relations – Datasets for ExamplesRelation loanRelation borrowerNote: borrower information missing for L-260 and loan information missing for L-155

اسلاید 60: Joined Relations – Examples loan inner join borrower on loan.loan_number = borrower.loan_numberloan left outer join borrower on loan.loan_number = borrower.loan_number

اسلاید 61: Joined Relations – Examplesloan natural inner join borrowerloan natural right outer join borrower

اسلاید 62: Joined Relations – Examplesloan full outer join borrower using (loan_number)Find all customers who have either an account or a loan (but not both) at the bank.select customer_name from (depositor natural full outer join borrower ) where account_number is null or loan_number is null

اسلاید 63: End of Chapter 3

اسلاید 64: Figure 3.1: Database Schemabranch (branch_name, branch_city, assets)customer (customer_name, customer_street, customer_city)loan (loan_number, branch_name, amount)borrower (customer_name, loan_number)account (account_number, branch_name, balance)depositor (customer_name, account_number)

اسلاید 65: Figure 3.3: Tuples inserted into loan and borrower

اسلاید 66: Figure 3.4: The loan and borrower relatio