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©Silberschatz, Korth and Sudarshan3.1Database System Concepts Banking Example  branch (branch-name, branch-city, assets)  customer (customer-name, customer-street,

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Presentation on theme: "©Silberschatz, Korth and Sudarshan3.1Database System Concepts Banking Example  branch (branch-name, branch-city, assets)  customer (customer-name, customer-street,"— Presentation transcript:

1 ©Silberschatz, Korth and Sudarshan3.1Database System Concepts Banking Example  branch (branch-name, branch-city, assets)  customer (customer-name, customer-street, customer-city)  account (account-number, branch-name, balance)  loan (loan-number, branch-name, amount)  depositor (customer-name, account-number)  borrower (customer-name, loan-number)

2 ©Silberschatz, Korth and Sudarshan3.2Database System Concepts Example Queries  Find the loan-number, for loans of over $1200  Find the loan number for each loan of an amount greater than $1200 Notice that a relation on schema [loan-number] is implicitly defined by the query {t.loan_number | t  loan   s  loan (t[loan-number] = s[loan- number]  s [amount]  1200)} {t.loan-number | t  loan  t [amount]  1200}

3 ©Silberschatz, Korth and Sudarshan3.3Database System Concepts Example Queries  Find the names of all customers having a loan, or an account { t.customer_name | t  customer   s  borrower( t[customer-name] = s[customer-name])   u  depositor( t[customer-name] = u[customer-name])  Find the names of all customers who have a loan and an account at the bank {t.customer_name | t  customer   s  borrower( t[customer-name] = s[customer-name])   u  depositor( t[customer-name] = u[customer-name])

4 ©Silberschatz, Korth and Sudarshan3.4Database System Concepts Example Queries  Find the names of all customers having a loan at the Perryridge branch { t.customer_name | t  customer   s  borrower( t[customer-name] = s[customer-name]   u  loan(u[branch-name] = “Perryridge”  u[loan-number] = s[loan-number]))  not  v  depositor (v[customer-name] = t[customer_name]) }  Find the names of all customers who have a loan at the Perryridge branch, but no account at any branch of the bank {t.customer_name | t  customer   s  borrower(t[customer-name] = s[customer-name]   u  loan(u[branch-name] = “Perryridge”  u[loan-number] = s[loan-number]))}

5 ©Silberschatz, Korth and Sudarshan3.5Database System Concepts Example Queries  Find the names of all customers having a loan from the Perryridge branch, and the cities they live in { t.customer_name,t.customer_city | t  customer   s  loan(s[branch-name] = “Perryridge”   u  borrower (u[loan-number] = s[loan-number]  t [customer-name] = u[customer-name])   v  customer (u[customer-name] = v[customer-name]  t[customer-city] = v[customer-city])))} Note, v is redundant!

6 ©Silberschatz, Korth and Sudarshan3.6Database System Concepts Example Queries  Find the names of all customers who have an account at all branches located in Brooklyn: {t.customer_name | t  customer   b  branch(b[branch-city] = “Brooklyn”   u  account ( b[branch-name] = u[branch-name]   s  depositor ( t[customer-name] = s[customer-name]  s[account-number] = u[account-number] )) )} Also {t.customer_name | t  customer   b  branch(b[branch-city] != “Brooklyn” V  u  account ( b[branch-name] = u[branch-name]   s  depositor ( t[customer-name] = s[customer-name]  s[account-number] = u[account-number] )) )}

7 ©Silberschatz, Korth and Sudarshan3.7Database System Concepts Example Queries  Find the names of all customers who have an account at all branches located in Brooklyn: {t.customer_name | t  customer  not (  b) b  branch(b[branch-city] = “Brooklyn”  not (  u  account ( b[branch-name] = u[branch-name]   s  depositor ( t[customer-name] = s[customer-name]  s[account-number] = u[account-number] )) ) )} For each output customer, there does not exist a branch in Brooklyn such that There does not exist an account for that customer in that branch

8 ©Silberschatz, Korth and Sudarshan3.8Database System Concepts Transforming the Universal and Existential Quantifiers ( x) (P(x))  not ( x)(not (P(x))) ( x) (P(x) and Q(x))  not ( x) (not (P(x)) or not (Q(x))) ( x) (P(x) or Q(x))  not ( x) (not (P(x)) and not (Q(x))) ( x) (P(x)) or Q(x))  not ( x) (not (P(x)) and not (Q(x))) ( x) (P(x) and Q(x))  not ( x) (not (P(x)) or not (Q(x))) A => B  not(A) or B Notice also that the following is true, where the => symbol stands for implies: ( x) (P(x)) => ( x) (P(x)) Not ( x) (P(x)) => not ( x) (P(x))

9 ©Silberschatz, Korth and Sudarshan3.9Database System Concepts Safety of Expressions  It is possible to write tuple calculus expressions that generate infinite relations.  For example, {t |  t  r} results in an infinite relation if the domain of any attribute of relation r is finite  To guard against the problem, we restrict the set of allowable expressions to safe expressions.  An expression {t | P(t)} in the tuple relational calculus is safe if all t values which cause P to be true, are taken from dom (p), where dom (P) is the cartezian product of the domains of all relations appearing in P. E.g. { t | t[A]=5  true } is not safe --- it defines an infinite set with attribute values that do not appear in any relation or tuples or constants in P.

10 ©Silberschatz, Korth and Sudarshan3.10Database System Concepts Safety of Expressions {  x 1, x 2, …, x n  | P(x 1, x 2, …, x n )} is safe if all of the following hold: 1.All values that appear in tuples of the expression are values from dom(P) The values appear either in P or in a tuple of a relation mentioned in P. 2.For every “there exists” subformula of the form  x (P 1 (x)) The subformula is true if and only if there is a value of x in dom(P 1 ) such that P 1 (x) is true. 3. For every “for all” subformula of the form  x (P 1 (x)), the sub formula is true if and only if P 1 (x) is true for all values x from dom (P 1 ).

11 ©Silberschatz, Korth and Sudarshan3.11Database System Concepts

12 ©Silberschatz, Korth and Sudarshan3.12Database System Concepts TRC – Additional Examples  QUERY 1 Retrieve the name and address of all employees who work in the ‘Research’ department. Q1 : {t.FNAME, t.LNAME, t.ADDRESS | EMPLOYEE(t) and ( d) (DEPARTMENT (d) and d.DNAME = ‘Research’ and d.DNUMBER=t.DNO) }  QUERY 2 for every project located in ‘Stafford’, list the project number, the controlling department number, and the department manager’s last name, birthday, and address. Q2 : {p.PNUMBER, p.DNUM, m.LNAME, m.BDATE, m.ADDRESS | PROJECT(p) and EMPLOYEE (m) and p.PLOCATION=‘Stafford’ and (( d)(DEPARTMENT(d) and p.DNUM=d.DNUMBER an d.MGRSSN=m.SSN)) }  QUERY 3 find the name of each employee who works on a project controlled by department number 5. Q3 : {e.LNAME, e.FNAME | EMPLOYEE(e) and ( ( x) ( w) (PROJECT(x) and WORKS_ON(w) and x.DNUM=5 and w.ESSN=e.SSN and x.PNUMBER=w.PNO) ) }

13 ©Silberschatz, Korth and Sudarshan3.13Database System Concepts  QUERY 3 Find the name of employees who work on all the projects controlled by department number 5. One way of specifying this query is by using the universal quantifier as shown. Q3: {e.LNAME, e.FNAME | EMPLOYEE(e) and (( x)(not(PROJECT(x))or not(x.DNUM=5) Or( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO) ) ) ) } QUERY 4 Make a list of project numbers for projects that involve an employee whose last name is ‘Smith’, either as a worker or as manager of the controlling department for the project. Q4 : {p.PNUMBER | PROJECT(p) and ( ( e)( w)(EMPLOYEE(e) and WORKS_ON(w) and w.PNO=p.PNUMBER and e.LNAME=‘Smith’ and e.SSN=w.ESSN) ) Or( ( m)( d)(EMPLOYEE(m) and DEPARTMENT(d) and p.DNUM=d.DNUMBER and d.MGRSSN=m.SSN and m.LNAME=‘Smith’) ) ) } TRC – Additional Examples

14 ©Silberschatz, Korth and Sudarshan3.14Database System Concepts TRC – Additional Examples cont.  QUERY 6 find the names of employees who have no dependents. Q6 : {e.FNAME, e.LNAME EMPLOYEE(e) and (not ( d)(DEPENDENT(d) and e.SSN=d.ESSN))} Using the general transformation rule, we we can rephrase Q6 as follows: Q6A : {e.FNAME, e.LNAME EMPLOYEE(e) and (( d)(not (DEPENDENT(d)) or not (e.SSN=d.ESSN)))}  QUERY 7 List the names of managers who have at least one dependent. Q7 : {e.FNAME, e.LNAME EMPLOYEE(e)and (( d)( p) (DEPARTMENT(d) and DEPENDENT(p) and e.SSN=d.MGRSSN and p.ESSN=e.SSN))}

15 ©Silberschatz, Korth and Sudarshan3.15Database System Concepts Domain Relational Calculus  A nonprocedural query language equivalent in power to the tuple relational calculus  Each query is an expression of the form: {  x 1, x 2, …, x n  | P(x 1, x 2, …, x n )} x 1, x 2, …, x n represent domain variables P represents a formula similar to that of the predicate calculus NOTE: The attribute Xi represents is by its location!

16 ©Silberschatz, Korth and Sudarshan3.16Database System Concepts Example Queries  Find the loan-number, branch-name, and amount for loans of over $1200 {  c, a  |  l (  c, l   borrower   b(  l, b, a   loan  b = “Perryridge”))} or {  c, a  |  l (  c, l   borrower   l, “Perryridge”, a   loan)}  Find the names of all customers who have a loan from the Perryridge branch and the loan amount: {  c  |  l, b, a (  c, l   borrower   l, b, a   loan  a > 1200)}  Find the names of all customers who have a loan of over $1200 {  l, b, a  |  l, b, a   loan  a > 1200}

17 ©Silberschatz, Korth and Sudarshan3.17Database System Concepts Example Queries  Find the names of all customers having a loan, an account, or both at the Perryridge branch: {  c  |  s, n (  c, s, n   customer)   x,y,z(  x, y, z   branch  y = “Brooklyn”)   a,b(  a,x,b   account   c,a   depositor)}  Find the names of all customers who have an account at all branches located in Brooklyn: {  c  |  l ({  c, l   borrower   b,a(  l, b, a   loan  b = “Perryridge”))   a(  c, a   depositor   b,n(  a, b, n   account  b = “Perryridge”))}

18 ©Silberschatz, Korth and Sudarshan3.18Database System Concepts  QUERY 1 Retrieve the name and address of all employees who work for the ‘Research’ department. Q1 : {qsv l ( z) ( l) ( m) (EMPLOYEE(qrstuvwxyz) and DEPARTMENT(lmno) and l=‘Research’ and m=z)} note implicit existenial notation  QUERY 2 For every project located in ‘stafford’, list the project number, the controlling department number, and the department manager’s last name, birthdate and address. Q2 : {iksuv l ( j) ( m) ( n) ( t)(PROJECT(hijk)and EMPLOYEE(qrstuvwxyz) and DEPARTMENT(lmno) and k=m and n=t and j=‘stanfford’)}  QUERY 6 find the names of employees who have no dependents. Q6 : {qs l ( t) (EMPLOYEE(qrstuvwxyz) and (( l)(not(DEPENDENT(lmnop)) or not (t=l))))} Query 6 can be restated using universal quantifiers instead of the existensial quantifiers, as shown in Q6A: Q6A : {qs l ( t) (EMPLOYEE(qrstuvwxyz) and (( l) (not(DEPENDENT(lmnop))or not (t=l))))} DRC Examples

19 ©Silberschatz, Korth and Sudarshan3.19Database System Concepts Four ways of specifying the query Q0 in QBE

20 ©Silberschatz, Korth and Sudarshan3.20Database System Concepts The notion of Relational Complete  Theorem: The Relational algebra (without functions), the Tuple relational calculus, and the Domain relational calculus are equivalent

21 ©Silberschatz, Korth and Sudarshan3.21Database System Concepts Views  In some cases, it is not desirable for all users to see the entire logical model (i.e., 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 the relational algebra, by  customer-name, loan-number (borrower loan)  Any relation that is not of the conceptual model but is made visible to a user as a “virtual relation” is called a view.

22 ©Silberschatz, Korth and Sudarshan3.22Database System Concepts View Definition  A view is defined using the create view statement which has the form create view v as where is any legal relational algebra query 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.

23 ©Silberschatz, Korth and Sudarshan3.23Database System Concepts View Examples  Consider the view (named all-customer) consisting of branches and their customers.  We can find all customers of the Perryridge branch by writing: create view all-customer as  branch-name, customer-name (depositor account)   branch-name, customer-name (borrower loan)  branch-name (  branch-name = “Perryridge” (all-customer))

24 ©Silberschatz, Korth and Sudarshan3.24Database System Concepts Updates Through View  Database modifications expressed as views must be translated to modifications of the actual relations in the database.  Consider the person who needs to see all loan data in the loan relation except amount. The view given to the person, branch- loan, is defined as: create view branch-loan as  branch-name, loan-number (loan)  Since we allow a view name to appear wherever a relation name is allowed, the person may write: branch-loan  branch-loan  {(“Perryridge”, L-37)}

25 ©Silberschatz, Korth and Sudarshan3.25Database System Concepts Updates Through Views (Cont.)  The previous insertion must be represented by an insertion into the actual relation loan from which the view branch-loan is constructed.  An insertion into loan requires a value for amount. The insertion can be dealt with by either. rejecting the insertion and returning an error message to the user. inserting a tuple (“L-37”, “Perryridge”, null) into the loan relation  Some updates through views are impossible to translate into database relation updates create view v as  branch-name = “Perryridge” (account)) v  v  (L-99, Downtown, 23)  Others cannot be translated uniquely all-customer  all-customer  {(“Perryridge”, “John”)}  Have to choose loan or account, and create a new loan/account number!

26 ©Silberschatz, Korth and Sudarshan3.26Database System Concepts Data Dictionary Storage  Information about relations  names of relations  names and types of attributes of each relation  names and definitions of views  integrity constraints  User and accounting information, including passwords  Statistical and descriptive data  number of tuples in each relation  Physical file organization information  How relation is stored (sequential/hash/…)  Physical location of relation  operating system file name or  disk addresses of blocks containing records of the relation  Information about indices (Chapter 12) Data dictionary (also called system catalog) stores metadata: that is, data about data, such as

27 ©Silberschatz, Korth and Sudarshan3.27Database System Concepts Data Dictionary Storage (Cont.)  Catalog structure: can use either specialized data structures designed for efficient access a set of relations, with existing system features used to ensure efficient access The latter alternative is usually preferred  A possible catalog representation:  Relation-metadata = (relation-name, number-of-attributes, storage-organization, location) Attribute-metadata = (attribute-name, relation-name, domain-type, position, length) User-metadata = (user-name, encrypted-password, group) Index-metadata = (index-name, relation-name, index-type, index-attributes) View-metadata = (view-name, definition)

28 ©Silberschatz, Korth and Sudarshan3.28Database System Concepts System Catalogs  For each index: structure (e.g., B+ tree) and search key fields  For each relation: name, file name, file structure (e.g., Heap file) attribute name and type, for each attribute index name, for each index integrity constraints  For each view: view name and definition  Plus statistics, authorization, buffer pool size, etc. * Catalogs are themselves stored as relations !

29 ©Silberschatz, Korth and Sudarshan3.29Database System Concepts Attr_Cat(attr_name, rel_name, type, position)

30 End of Chapter 3


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