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1、Appendix C:Other Relational Languages,Appendix C:Other Relational Languages,Query-by-Example(QBE)AccessDatalog,Query-by-Example(QBE),Basic StructureQueries on One RelationQueries on Several RelationsThe Condition BoxThe Result RelationOrdering the Display of TuplesAggregate Operations Modification o
2、f the Database,QBE Basic Structure,A graphical query language which is based(roughly)on the domain relational calculusTwo dimensional syntax system creates templates of relations that are requested by usersQueries are expressed“by example”,QBE Skeleton Tables for the Bank Example,QBE Skeleton Tables
3、(Cont.),Queries on One Relation,Find all loan numbers at the Perryridge branch.,_x is a variable(optional;can be omitted in above query)P.means print(display)duplicates are removed by default To retain duplicates use P.ALL,P._x,Queries on One Relation(Cont.),Display full details of all loans,P._x,P.
4、_y,P._z,Method 1:,Method 2:Shorthand notation,Queries on One Relation(Cont.),Find names of all branches that are not located in Brooklyn,Find the loan number of all loans with a loan amount of more than$700,Queries on One Relation(Cont.),Find the loan numbers of all loans made jointly to Smith and J
5、ones.,Find all customers who live in the same city as Jones.,Queries on Several Relations,Find the names of all customers who have a loan from the Perryridge branch.,Queries on Several Relations(Cont.),Find the names of all customers who have both an account and a loan at the bank.,Negation in QBE,F
6、ind the names of all customers who have an account at the bank,but do not have a loan from the bank.,means“there does not exist”,Negation in QBE(Cont.),Find all customers who have at least two accounts.,means“not equal to”,The Condition Box,Allows the expression of constraints on domain variables th
7、at are either inconvenient or impossible to express within the skeleton tables.Complex conditions can be used in condition boxesExample:Find the loan numbers of all loans made to Smith,to Jones,or to both jointly,Condition Box(Cont.),QBE supports an interesting syntax for expressing alternative valu
8、es.,Condition Box(Cont.),Find all account numbers with a balance greater than$1,300 and less than$1,500.,Find all account numbers with a balance greater than$1,300 and less than$2,000 but not exactly$1,500.,Condition Box(Cont.),Find all branches that have assets greater than those of at least one br
9、anch located in Brooklyn.,The Result Relation,Find the customer_name,account_number,and balance for all customers who have an account at the Perryridge branch.We need to:Join depositor and account.Project customer_name,account_number and balance.To accomplish this we:Create a skeleton table,called r
10、esult,with attributes customer_name,account_number,and balance.Write the query.,The Result Relation(Cont.),The resulting query is:,Ordering the Display of Tuples,AO=ascending order;DO=descending order.Example:list in ascending alphabetical order all customers who have an account at the bank.When sor
11、ting on multiple attributes,the sorting order is specified by including with each sort operator(AO or DO)an integer surrounded by parentheses.Example:List all account numbers at the Perryridge branch in ascending alphabetic order with their respective account balances in descending order.,Aggregate
12、Operations,The aggregate operators are AVG,MAX,MIN,SUM,and CNTThe above operators must be postfixed with“ALL”(e.g.,SUM.ALL.or AVG.ALL._x)to ensure that duplicates are not eliminated.Example:Find the total balance of all the accounts maintained at the Perryridge branch.,Aggregate Operations(Cont.),UN
13、Q is used to specify that we want to eliminate duplicates.Find the total number of customers having an account at the bank.,Query Examples,Find the average balance at each branch.,The“G”in“P.G”is analogous to SQLs group by construct.The“ALL”in the“P.AVG.ALL”entry in the balance column ensures that a
14、ll balances are considered.To find the average account balance at only those branches where the average account balance is more than$1,200,we simply add the condition box:,Query Example,Find all customers who have an account at all branches located in Brooklyn.Approach:for each customer,find the num
15、ber of branches in Brooklyn at which they have accounts,and compare with total number of branches in Brooklyn.QBE does not provide subquery functionality,so both above tasks have to be combined in a single query.Can be done for this query,but there are queries that require subqueries and cannot alwa
16、ys be expressed in QBE.,In the query on the next pagew specifies the number of distinct branches in Brooklyn.Note:The variable _w is not connected to other variables in the query.z specifies the number of distinct branches in Brooklyn at which customer x has an account.,Query Example(Cont.),Modifica
17、tion of the Database Deletion,Deletion of tuples from a relation is expressed by use of a D.command.In the case where we delete information in only some of the columns,null values,specified by,are inserted.Delete customer SmithDelete the branch_city value of the branch whose name is“Perryridge”.,Del
18、etion Query Examples,Delete all loans with a loan amount greater than$1300 and less than$1500.For consistency,we have to delete information from loan and borrower tables,Deletion Query Examples(Cont.),Delete all accounts at branches located in Brooklyn.,Modification of the Database Insertion,Inserti
19、on is done by placing the I.operator in the query expression.Insert the fact that account A-9732 at the Perryridge branch has a balance of$700.,Modification of the Database Insertion(Cont.),Provide as a gift for all loan customers of the Perryridge branch,a new$200 savings account for every loan acc
20、ount they have,with the loan number serving as the account number for the new savings account.,Modification of the Database Updates,Use the U.operator to change a value in a tuple without changing all values in the tuple.QBE does not allow users to update the primary key fields.Update the asset valu
21、e of the Perryridge branch to$10,000,000.Increase all balances by 5 percent.,Microsoft Access QBE,Microsoft Access supports a variant of QBE called Graphical Query By Example(GQBE).GQBE differs from QBE in the following waysAttributes of relations are listed vertically,one below the other,instead of
22、 horizontally.Instead of using variables,lines(links)between attributes are used to specify that their values should be the same.Links are added automatically on the basis of attribute name,and the user can then add or delete links.By default,a link specifies an inner join,but can be modified to spe
23、cify outer joins.Conditions,values to be printed,as well as group by attributes are all specified together in a box called the design grid.,An Example Query in Microsoft Access QBE,Example query:Find the customer_name,account_number and balance for all accounts at the Perryridge branch.,An Aggregati
24、on Query in Access QBE,Find the name,street and city of all customers who have more than one account at the bank.,Aggregation in Access QBE,The row labeled Total specifies:which attributes are group by attributes.which attributes are to be aggregated upon(and the aggregate function).For attributes t
25、hat are neither group by nor aggregated,we can still specify conditions by selecting where in the Total row and listing the conditions below.As in SQL,if group by is used,only group by attributes and aggregate results can be output.,Datalog,Basic Structure Syntax of Datalog RulesSemantics of Nonrecu
26、rsive DatalogSafety Relational Operations in DatalogRecursion in DatalogThe Power of Recursion,Basic Structure,Prolog-like logic-based language that allows recursive queries;based on first-order logic.A Datalog program consists of a set of rules that define views.Example:define a view relation v1 co
27、ntaining account numbers and balances for accounts at the Perryridge branch with a balance of over$700.v1(A,B):account(A,“Perryridge”,B),B 700.Retrieve the balance of account number“A-217”in the view relation v1.?v1(“A-217”,B).To find account number and balance of all accounts in v1 that have a bala
28、nce greater than 800?v1(A,B),B 800,Example Queries,Each rule defines a set of tuples that a view relation must contain.E.g.,v1(A,B):account(A,“Perryridge”,B),B 700 is read as:for all A,B if(A,“Perryridge”,B)account and B 700 then(A,B)v1The set of tuples in a view relation is then defined as the unio
29、n of all the sets of tuples defined by the rules for the view relation.Example:interest_rate(A,5):account(A,N,B),B=10000,Negation in Datalog,Define a view relation c that contains the names of all customers who have a deposit but no loan at the bank:c(N):depositor(N,A),not is_borrower(N).is_borrower
30、(N):borrower(N,L).NOTE:using not borrower(N,L)in the first rule results in a different meaning,namely there is some loan L for which N is not a borrower.To prevent such confusion,we require all variables in negated“predicate”to also be present in non-negated predicates.,Named Attribute Notation,Data
31、log rules use a positional notation that is convenient for relations with a small number of attributes.It is easy to extend Datalog to support named attributes.E.g.,v1 can be defined using named attributes as v1(account_number A,balance B):account(account_number A,branch_name“Perryridge”,balance B),
32、B 700.,Formal Syntax and Semantics of Datalog,We formally define the syntax and semantics(meaning)of Datalog programs,in the following steps:We define the syntax of predicates,and then the syntax of rules.We define the semantics of individual rules.We define the semantics of non-recursive programs,b
33、ased on a layering of rules.It is possible to write rules that can generate an infinite number of tuples in the view relation.To prevent this,we define what rules are“safe”.Non-recursive programs containing only safe rules can only generate a finite number of answers.It is possible to write recursiv
34、e programs whose meaning is unclear.We define what recursive programs are acceptable,and define their meaning.,Syntax of Datalog Rules,A positive literal has the formp(t1,t2.,tn)p is the name of a relation with n attributeseach ti is either a constant or variableA negative literal has the form not p
35、(t1,t2.,tn)Comparison operations are treated as positive predicates E.g.,X Y is treated as a predicate(X,Y)“”is conceptually an(infinite)relation that contains all pairs of values such that the first value is greater than the second valueArithmetic operations are also treated as predicatesE.g.,A=B+C
36、 is treated as+(B,C,A),where the relation“+”contains all triples such that the third value is thesum of the first two,Syntax of Datalog Rules(Cont.),Rules are built out of literals and have the form:p(t1,t2,.,tn):L1,L2,.,Lm.head bodyeach Li is a literalhead the literal p(t1,t2,.,tn)body the rest of
37、the literalsA fact is a rule with an empty body,written in the form:p(v1,v2,.,vn).indicates tuple(v1,v2,.,vn)is in relation pA Datalog program is a set of rules.,Semantics of a Rule,A ground instantiation of a rule(or simply instantiation)is the result of replacing each variable in the rule by some
38、constant.E.g.,Rule defining v1 v1(A,B):account(A,“Perryridge”,B),B 700.An instantiation above rule:v1(“A-217”,750):account(“A-217”,“Perryridge”,750),750 700.The body of rule instantiation R is satisfied in a set of facts(database instance)l if1.For each positive literal qi(vi,1,.,vi,ni)in the body o
39、f R,l contains the fact qi(vi,1,.,vi,ni).2.For each negative literal not qj(vj,1,.,vj,nj)in the body of R,l does not contain the fact qj(vj,1,.,vj,nj).,Semantics of a Rule(Cont.),We define the set of facts that can be inferred from a given set of facts l using rule R as:infer(R,l)=p(t1,.,tn)|there i
40、s a ground instantiation R of R where p(t1,.,tn)is the head of R,and the body of R is satisfied in l Given an set of rules=R1,R2,.,Rn,we defineinfer(,l)=infer(R1,l)infer(R2,l).infer(Rn,l),Layering of Rules,Define the interest on each account in Perryridgeinterest(A,l):perryridge_account(A,B),interes
41、t_rate(A,R),l=B*R/100.perryridge_account(A,B):account(A,“Perryridge”,B).interest_rate(A,5):account(N,A,B),B=10000.Layering of the view relations,Layering Rules(Cont.),A relation is a layer 1 if all relations used in the bodies of rules defining it are stored in the database.A relation is a layer 2 i
42、f all relations used in the bodies of rules defining it are either stored in the database,or are in layer 1.A relation p is in layer i+1 if it is not in layers 1,2,.,iall relations used in the bodies of rules defining a p are either stored in the database,or are in layers 1,2,.,i,Formally:,Semantics
43、 of a Program,Define I0=set of facts stored in the database.Recursively define li+1=li infer(i+1,li)The set of facts in the view relations defined by the program(also called the semantics of the program)is given by the set of facts ln corresponding to the highest layer n.,Let the layers in a given p
44、rogram be 1,2,.,n.Let i denote theset of all rules defining view relations in layer i.,Note:Can instead define semantics using view expansion likein relational algebra,but above definition is better for handlingextensions such as recursion.,Safety,It is possible to write rules that generate an infin
45、ite number of answers.gt(X,Y):X Ynot_in_loan(B,L):not loan(B,L)To avoid this possibility Datalog rules must satisfy the following conditions.Every variable that appears in the head of the rule also appears in a non-arithmetic positive literal in the body of the rule.This condition can be weakened in
46、 special cases based on the semantics of arithmetic predicates,for example to permit the rulep(A):-q(B),A=B+1Every variable appearing in a negative literal in the body of the rule also appears in some positive literal in the body of the rule.,Relational Operations in Datalog,Project out attribute ac
47、count_name from account.query(A):account(A,N,B).Cartesian product of relations r1 and r2.query(X1,X2,.,Xn,Y1,Y1,Y2,.,Ym):r1(X1,X2,.,Xn),r2(Y1,Y2,.,Ym).Union of relations r1 and r2.query(X1,X2,.,Xn):r1(X1,X2,.,Xn),query(X1,X2,.,Xn):r2(X1,X2,.,Xn),Set difference of r1 and r2.query(X1,X2,.,Xn):r1(X1,X2
48、,.,Xn),not r2(X1,X2,.,Xn),Recursion in Datalog,Suppose we are given a relation manager(X,Y)containing pairs of names X,Y such that Y is a manager of X(or equivalently,X is a direct employee of Y).Each manager may have direct employees,as well as indirect employees.Indirect employees of a manager,say
49、 Jones,are employees of people who are direct employees of Jones,or recursively,employees of people who are indirect employees of Jones.Suppose we wish to find all(direct and indirect)employees of manager Jones.We can write a recursive Datalog program.empl_jones(X):-manager(X,Jones).empl_jones(X):-m
50、anager(X,Y),empl_jones(Y).,Semantics of Recursion in Datalog,Assumption(for now):program contains no negative literalsThe view relations of a recursive program containing a set of rules are defined to contain exactly the set of facts l computed by the iterative procedure Datalog-Fixpointprocedure Da
