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Κεφάλαιο 4

Κεφάλαιο 4. Η Γλώσσα SQL και Σχεσιακός Λογισμός. Relational Calculus: Introduction. The Relational Calculus (RC) is a non-procedural, formal language based on first-order predicate calculus

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Κεφάλαιο 4

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  1. Κεφάλαιο 4 Η Γλώσσα SQL και Σχεσιακός Λογισμός

  2. Relational Calculus: Introduction • TheRelational Calculus (RC) is a non-procedural, formal language based onfirst-order predicate calculus • Queries in RC specifyWHATis to be retrieved (declarative) while the system takes care ofHOW • Most commercial relational languages are based on relational calculus (QUEL, QBE, and SQL). Such languages emphasize ease and convenience of use. • In terms of expressiveness, RELATIONAL ALGEBRAandRELATIONAL CALCULUSare identical.

  3. Relational Calculus: Introduction (2) • The Relational Calculus uses the notion of VARIABLE • There are two flavors of the language: • tuple calculus variables refer to tuples from a specified relation e.g., t is a tuple variable referring to a tuple of r(R) • domain calculus variables refer to individual values from a specified domain e.g., d is a domain variable referring to the value which a tuple in r(R) has for attribute A (from domain D)

  4. Tuple Calculus • A TUPLE CALCULUS EXPRESSION defines a new relation in terms of existing (base) relations. • An expression is constructed from the following elements: (1) tuple variables(e.g., t, v, w, t1, t2, t3, ... tn), which are defined to range over a specified relation instance r(R) Tuple variables may be restricted, where, t.A, with A an attribute of R, denotes the A-component of the value of t Example: t.Name (2)conditionsof the form x operator y, where -- x, y are restricted tuple variables or constant values -- operator Î { =, ¹, <, >, £, ³ } Examples: t.Name = ‘mary’, t.City ¹ v.City

  5. Tuple Calculus Expressions (3)Well-Formed Formulas (WFFs), defined as: t a condition is a WFF t if f is a WFF, so is (f) and Ø(f) (where Ø is the logical NOT) t if f and g are WFFs, so are (f & g) and (f | g) (where &, |, are the logical AND, OR respectively) t if f is a WFF with free variable t, so are $t(f), "t(f) (where $, "are the existential and universal quantifiers) A tuple variable t is said to be bound in a formula f if it is one of the quantified variables in f. Otherwise, t is free in f. Example: t is free in f1 : (t.City = ‘london’) and is bound in f2 : ("t ) (t.DNumber = v.Dno), while v is free in f2

  6. Tuple Calculus Expressions • Examples of WFFs • t.City = v.City • (t.City = v.City) • Ø (t.City = v.City) • (t.City ¹ v.City) • (t.City = v.City) & (w.City = london) • $t (t.City = v.City) • "t (t.City = athens) • if f is a WFF, then a tuple calculus expression is any expression of the form: { ti.Aj | ti Î r(Rk) AND f } where Rk are relation schemes and Aj are attributes in Rk

  7. Calculus Query Examples • Recall the sailors-boats database SAILORS (Sid, SName, Rating) BOATS (Bid, BName, Color) RESERVE (Sid, Bid, Date) • QUERY1: Find the names of sailors whose id is greater than 10 and have rating ‘a’ RANGE of t is SAILORS t.SName where& (t.Sid > 10) & (t.Rating = ‘a’) This is an equivalent form of the pure RC query: { t.SName | t Î r(SAILORS)& ((t.Sid > 10) & (t.Rating = ‘a’)) }

  8. Calculus Query Examples (2) • QUERY2: Find the names and ratings of sailors who have reserved boat number 3 RANGE of t is SAILORS RANGE of v is RESERVE t.SName, t.Rating where$v ( (t.Sid = v.Sid) & (v.Bid = 3) ) • QUERY3: Find the names sailors who have not reserved boat number 3 RANGE of t is SAILORS RANGE of v is RESERVE t.SName whereØ ( $v ( (t.Sid = v.Sid) & (v.Bid = 3) ) )

  9. SQL - Introduction • SQL (Structured Query Language) has become the “standard” in query languages. It was first used in IBM’s prototype , called SYSTEM-R, developed at San Jose in the mid-seventies. SQL has gone over many evaluations. • There are 4 basic commands: • select (not to be confused with SELECTION in algebra) • insert • update • delete • The result of any query on relations is again a relation

  10. SQL - Informal Definition • Assume the EMPLOYEE relation and the following query: “Find the names of employees who earn more than 30000” This is expressed in SQL as: select e.Name from EMPLOYEE e where (e.Salary > 30000) • e is a tuple variable defined to range over the relation EMPLOYEE (in the from clause) • e.Name, as a restricted tuple variable, specifies the value of e in attribute Name, and is the target list (specifies in the select clause the projections of columns) • (e.Salary > 30000) is the qualification (specifies in the where clause all selections and joins)

  11. SQL -- Formal Definition • A selection clause is a comparison between a restricted tuple variable x and a constant c of the form : x operator c where operator Î { =, ¹, <, >, £, ³ } Example:t.Name = ‘mary’, t.Salary > 30000 • A join clause is a comparison between two restricted tuple variables x and y of the form x operator y where x, y belong to different relations and operator Î { =, ¹, <, >, £, ³ } Example:t.Name = v.EName

  12. SQL -- Formal Definition (2) • A qualification is a Boolean combination (i.e., with logical and, or, not) of selection and joinclauses.) Example: (t.Name = v.EName) and (t.Salary > 30000) • QUALIFICATION SEMANTICS A qualification Q describes the subset of the Cartesian product of the ranges of its tuple variables that satisfy Q Example: Consider relations: EMPLOYEE(SSN, DNumber), and DEPARTMENT(Dno, mgrSSN), with e and d tuple variables Q = (e.DNumber = d.Dno) and (d.mgrSSN = 9876) Semantics of Q: The set of e, d pairs that satisfy Q

  13. SQL Qualification Semantics EMPLOYEE DEPARTMENT . = X Cartesian Product Finally, after establishing the subset of EMPLOYEE X DEPARTMENT that satisfies the qualification Q, we get two tuples: Ü Ü

  14. SQL -- Complete Format select [ distinct ] target_list from tuple_variable_list [ where qualification ] [ group by grouping_attributes ] [ having group_condition ] [ order by target_list_subset ] • A query is evaluated by first applying the WHERE clause, then GROUP-BY and HAVING (all optional), and finally the SELECT clause (target list) - ordering the resulting tuples if required in the ORDER BY clause (also optional).

  15. SQL: Sailor Examples • Consider again the Sailors-Boats database: SAILORS (Sid, SName, Rating) BOATS (Bid, BName, Color) RESERVE (Sid, Bid, Date) • SQUERY1:Find the names of sailors who have reserved boat number 2 select s.SName from SAILORS s, RESERVE r where s.Sid = r.Sid and r.Bid = 2

  16. SQL Sailor Examples (2) • SQUERY2: Find the names of sailors who have reserved a red boat select s.SName from SAILORS s, BOATS b, RESERVE r where s.Sid = r.Sid and r.Bid = b.Bid and b.Color = “red” • SQUERY3: Find the colors of the boats reserved by eleni select b.Color from SAILORS s, BOATS b, RESERVE r where s.Sid = r.Sid and r.Bid =b.Bid and s.SName= “eleni”

  17. SQL Sailor Examples (3) • SQUERY4: Find the names of the sailors who have reserved at least one boat select s.SName from SAILORS s, RESERVE r where s.Sid = r.Sid • SQUERY5: Find the names of sailors who have reserved a red or a green boat select s.SName from SAILORS s, BOATS b, RESERVE r where s.Sid = r.Sid and r.Bid = b.Bid and (b.Color = “red” or b.Color = “green”)

  18. SQL Sailor Examples (4) • SQUERY6: Find the names of sailors who have reserved both a red and a green boat select s.SName from SAILORS s, BOATS b1, RESERVE r1, BOATS b2, RESERVE r2 where s.Sid = r1.Sid and r1.Bid = b1.Bid and b1.Color = “red” and s.Sid = r2.Sid and r2.Bid = b2.Bid and b2.Color = “green”

  19. SQL Elaboration: Examples from the COMPANY database EMPLOYEE ( SSN, Name, BirthDate, Address, Sex, Salary, SupSSN, DNumber) DEPARTMENT ( DNumber, DName, MgrSSN, MgrStartDate) PROJECT ( PNumber, PName, Location, DNumber) DEPT_LOCATION ( DNumber, DLocation) WORKS_ON ( SSN, PNumber, HoursPW) DEPENDENT ( SSN, DependName, Sex, BirthDate, Relationship)

  20. SQL: Target List Examples (1) • TARGETLISTS: Each item in a target list can be as general as: attribute_name = expression where expression is any arithmetic or string expression over restricted tuple variables and constants (also built-ins and aggregates.) • CQUERY1: List, increased by 10000 the salary of employees who have worked on two different projects more than 25 hours select e.Name, Salary = e.Salary + 10000 from EMPLOYEE e, WORKS_ON w1, WORKS_ON w2 where e.SSN = w1.SSN and e.SSN = w2.SSN and w1.HoursPW > 25 and w2.HoursPW > 25 and w1.PNumber != w2.PNumber

  21. SQL: Target List Examples (2) • TARGET Lists may also contain the keyword DISTINCT • Since SQL does not treat relations as sets, duplicate tuples may appear, therefore DISTINCT is used to eliminate the duplicates • CQUERY2: Show all distinct (different values) salaries that employees earn select distinct e.Salary from EMPLOYEE e • The above query also shows that the WHERE clause is optional (with missing WHERE, all tuples qualify)

  22. SQL: Target List Examples (3) • The TARGET List may contain the wild character: “*” When a * is used, it implies that all attributes from the relation(s) should be retrieved • CQUERY3: Show all employees in department number 4 select * from EMPLOYEE e where e.Dnumber = 4

  23. SQL: Use of Tuple Variables • Relation names can be used instead of tuple variables • Tuple variables can be implicit if the system can figure out which relation each attribute belongs to. • CQUERY4: For every project located in Athens, list the project name, the controlling department number and the department’s manager name select PName, DEPARTMENT.DNumber, Name from EMPLOYEE, DEPARTMENT, PROJECT where PROJECT.DNumber = DEPARTMENT.DNumber and MgrSSN = SSN and Location = “athens”

  24. SQL Qualification Examples -1 • QUALIFICATIONS: Each item in a qualification can be as general as: expression = expression • CQUERY5: Find the names of employees whose salary is more than double the salary of some other employee (also show their name) select Name1 = e1.Name, Name2 = e1.Name from EMPLOYEE e1, EMPLOYEE e2 where 2 * e1.Salary < e2.Salary

  25. SQL Qualification Examples -2 • SQL provides direct support of several SET operations, like: • union • minus • intersect • CQUERY6: Find the names of employees who work in department number 4 and earn at most 40000 (select Name from EMPLOYEE where DNumber = 4) minus (select Name from EMPLOYEE where Salary > 40000 )

  26. SQL Qualification Examples -3 • CQUERY7: List all project names for projects that involve an employee whose name is “jenny” as a worker or as a manager of the department that controls the project (select PName from EMPLOYEE, PROJECT, DEPARTMENT where EMPLOYEE.DNumber=DEPARTMENT.DNumber and MgrSSN = SSN and Name = “jenny” ) union (select PName from EMPLOYEE, PROJECT, WORKS_ON where PROJECT.PNumber = WORKS_ON.PNumber and WORKS_ON.SSN=EMPLOYEE.SSN and Name = “jenny” )

  27. SQL Qualification Examples-4 • NESTING OF SQL QUERIES: A complete SELECT query (called the nested query) can be specified in the qualification of another query (called the outer query) • CQUERY8: List all employees that work in the research department select Name from EMPLOYEE where DNumber in (select DNumber from DEPARTMENT where DName = “research” )

  28. SQL Qualification Examples-5 • CORRELATED NESTED Queries: If a condition in the qualification of a nested query references an attribute in the outer query, the two are said to be CORRELATED The result of the nested query is different for each tuple of the relation(s) in the outer query • CQUERY9: List the name of each employee who has a dependent with the same sex as the employee select Name from EMPLOYEE e where e.SSN in (select d.SSN from DEPENDENT d where d.SSN = e.SSN and d.Sex = e.Sex )

  29. SQL Qualification Examples-6 • Any query that uses the IN comparison operator (tests for set membership) can always be expressed as a single block query (flat query) • CQUERY9a: List the name of each employee who has a dependent with the same sex as the employee select Name from EMPLOYEE e, DEPENDENT d where e.SSN = d.SSN and d.Sex = e.Sex )

  30. SQL Qualification Examples-7 • Similar connectives to INare: t not in (tests for set non-membership) t OP any (OP relationship with some tuple in a set) t OP all (OP relationship with all tuples in a set) where OP Î { =, ¹, <, >, £, ³ } • CQUERY10: List all employees that earn more than everybody in the research department select Name from EMPLOYEE where Salary > all (select Salary from EMPLOYEE e, DEPARTMENT d where e.DNumber = d.DNumber and DName = “research” )

  31. SQL Qualification Examples-8 • SQL also provides SET COMPARATORS: • contains, not contains (a set (not) being a superset) • exists, not exists (a set (not) being empty) • CQUERY11: List all employees who work on all projects controlled by department number 4 select Name from EMPLOYEE e where (select w.PNumber from WORKS_ON where w.SSN = e.SSN) contains (select PNumber from PROJECT where DNumber = 4)

  32. SQL Qualification Examples-9 • CQUERY9b: List the name of each employee who has a dependent with the same sex as the employee select Name from EMPLOYEE e where exists (select * from DEPENDENT d where d.SSN=e.SSN and d.Sex = e.Sex) • CQUERY12: List the employees with no dependents select Name from EMPLOYEE e where not exists (select * from DEPENDENT d where d.SSN=e.SSN )

  33. SQL Aggregates and Groups-1 • SQL supports FIVE AGGREGATE FUNCTIONS (can be applied to any attribute X of a relation): t count ( [DISTINCT] X) :number of unique values in X t sum ( [DISTINCT] X) :sum of unique values in X t avg ( [DISTINCT] X) :average of unique values in X t max (X) :maximum value in X t min (X) :minimum value in X • Aggregates return a single value

  34. SQL Aggregates and Groups-2 • Some SQL implementations do not allow more than one value in the target list • CQUERY13: List the maximum salary, the minimum salary, and the average salary among all employees select max(Salary), min(Salary), avg(Salary from EMPLOYEE • QUERY14: Find the number of employees select count(*) from EMPLOYEE

  35. SQL Aggregates and Groups-3 • QualifiedAggregates: The set on which aggregates apply can be restricted by the where-clause • CQUERY15: Find the average salary of employees in department with number 4 select avg(Salary) from EMPLOYEE where DNumber = 4

  36. SQL Aggregates and Groups-4 • AggregateFunctions: Aggregates or groups of tuples are computed using the GROUP BY clause • CQUERY16: In each department, find the minimum age of employees who earn more than 40000 select DNumber, max(BirthDate) from EMPLOYEE where Salary > 40000 group by DNumber • Note that the grouping attributes MUST ALSO appear in the select clause

  37. SQL Aggregates and Groups-5 • CQUERY17: For each project, retrieve the project number, project name, and the number of employees that work on that project selectp.PNumber, p.PName, count(*) from PROJECT p, WORKS_ON w where p.PNumber = w.PNumber group by PNumber, PName • In the above query, the grouping and functions are applied afterjoining the relations PROJECT and WORKS_ON.

  38. SQL Aggregates and Groups-6 • HAVING CLAUSE: Qualifications that have to be satisfied by each group formed by the group by- clause are put in a HAVING clause • CQUERY18: Find the average salary of employees born after 1950 for each department with more than 10 such employees select DNumber, avg(Salary) from EMPLOYEE where BirthDate > “1.1.51” group by DNumber having count(*) > 10

  39. SQL Aggregates and Groups-6 • CQUERY19: Find the average salary of employees born after 1950 for each department with more than 10 employees select e1.DNumber, avg(e1.Salary) from EMPLOYEE e1 where e1.BirthDate > “1.1.51” group by e1.DNumber having 10 < any (select count(e2.SSN) from EMPLOYEE e2 where e2.DNumber = e1.DNumber )

  40. SQL Updates (1) • INSERT command insert into relation_name select-statement or insert into relation_name values (value_list) • CUPDATE1: Insert a new department insert into DEPARTMENT values (6, “inventory”, 9879, “30.5.45”)

  41. SQL Updates (2) • Suppose we have a relation DEPT_INFO as in: DEPT_INFO ( DeptName, NoOfEmpl, TotalSalary) • We can insert tuples in this relation with (CUPDATE2): insert into DEPT_INFO select d.DName, count(*), sum(e.Salary) from DEPARTMENT d, EMPLOYEE e where d.DNumber=e.DNumber group by d.DName

  42. SQL Updates (3) • DELETE command delete from relation_name where-qualification SEMANTICS: -- Execute the corresponding SELECT command and then remove the resulting tuples from relation_name • CUPDATE3: delete from EMPLOYEE where DNumber in (select DNumber from DEPARTMENT where DName = “admin”)

  43. SQL Updates (4) • UPDATE command update relation_name set target_list where qualification SEMANTICS: -- Execute the two corresponding SELECT commands, then remove the old tuples from relation, then insert the new ones • CUPDATE4: update EMPLOYEE set Salary = Salary * 1.14 where DNumber in (select DNumber from DEPARTMENT where DName = “admin”)

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