1. 1 Database Systems SQL: Advanced Queries

2. 2 Union, Intersection, and Except (2)   Find the names of those people who are either a graduate student or an instructor or both in CS department. select Name from Student where DeptName='CS' union select Name from Instructor where DeptName='CS'   union removes duplicate rows.   union all keeps duplicate rows.

3. 3 Union, Intersection, and Except (3)   Find the names of those who are both a graduate student and an instructor in CS department. select Name from Student where DeptName='CS‘ intersect select Name from Instructor where DeptName='CS‘   Find the names of those who are an instructor but not a graduate student in CS department. select Name from Instructor where DeptName='CS' except select Name from Student where DeptName='CS'

4. 4 Union, Intersection, and Except (5)   make a list of all project numbers for projects that involve an employee whose last name is “Smith”, either as a worker or as a manager of the department that controls the project. (SELECT Distinct Pnumber FROM PROJECT, Works-on, Employee WHERE Pnumber = Pno AND ESSN = SSN AND Lname = “Smith”) UNION (SELECT PNumber FROM Project, Department, Employee WHERE Dnum=Dnumber AND MGRSSN=SSN AND Lname=“Smith”)

5. 5 Comparisons Involving NULL   SQL allows queries that check if a value is NULL (missing or unknown or not applicable)   SQL uses IS or IS NOT to compare NULLs because it considers each NULL value distinct from other NULL values, so equality comparison is not appropriate.   Find all employees who have no managers. select * from Employees where Manager-SSN is null;   Find all employees who have a manager. select * from Employees where Manager-SSN is not null;

6. 6 Null Values and Three Valued Logic   The result of any arithmetic expression involving null is null  E.g. 5 + null returns null   Any comparison with null returns unknown (null)  E.g. 5 null or null = null   Three-valued logic using the truth value unknown:  OR(unknown or true) = true (unknown or false) = unknown (unknown or unknown)= unknown  AND(true and unknown)= unknown (false and unknown) = false (unknown and unknown) = unknown  NOT(not unknown)= unknown   Result of where clause predicate is treated as false if it evaluates to unknown (null)

7. 7 Null Values and Three Valued Logic   Eaxmple: Consider table Employees with the following data: EmpIdName Salary Bonus 123Smith 25000 5000 234John 28000 null select Name, Salary+Bonus Total from Employee   Result is: NameTotal Smith 30000 John null

8. 8 NVL function   Use NVL function to convert null values   Format: nvl(exp1, exp2)  exp1 is a source value that may be null  exp2 is the target value for converting null   Ex: select name, salary + nvl(bonus, 0) Total from Employee   Result is: NameTotal Smith 30000 John 28000

9. 9 Using IN   Retrieve the social security numbers of all employees who work on project number 1, 2, or 3; SELECT DISTINCT ESSN FROM Works-on WHERE PNO IN (1, 2, 3);   Find all students who are 20, 22, or 24 years old. select * from Students where age in (20, 22, 24);   not in is the opposite of in.

10. 10 Nested Queries   A complete SELECT query, called a nested query, can be specified within the WHERE-clause of another query, called the outer query   Many of the previous queries can be specified in an alternative form using nesting   Retrieve the name and address of all employees who work for the 'Research' department. SELECTFNAME, LNAME, ADDRESS FROM EMPLOYEE WHEREDNO IN (SELECT DNUMBER FROM DEPARTMENT WHERE DNAME='Research');

11. 11 Nested Queries   Make a list of all project numbers for projects that involve an employee whose last name is “Smith”, either as a worker or as a manager of the department that controls the project. select distinct Pnumber from Project wherePnumber IN (select pnumber from project, department, employee where dnum=dnumber and MGRSSN=SSN and Lname=‘Smith’) OR Pnumber IN (select pno from works-on, employee where ESSN=SSN and Lname = ‘Smith’);

12. 12 Nested Queries   Find the names of all students who take at least one course offered by the CS department. Select Name from student s, enrollment e where s.SSN=e.SSN and e.cno in (select cno from course where dept_name = 'CS')   The previous query is equivalent to: (1) select Name from student s, snrollment e, course c where s.ssn = e.ssn and e.cno=c.cno and c.dept_name = 'CS’; (2) select Name from Student where SSN in (select SSN from Enrollment where cno in (select cno from Course where Dept_Name='CS'));

13. 13 Nested Queries   The previous query is a nested query. The query outside is an outer query and the query nested under the outer query is an inner query. Multiple nestings are allowed.   The semantics of the above query is to evaluate the inner query first and evaluate the outer query last.   Many nested queries have equivalent non-nested versions.

14. 14 Nested Queries   Example: Consider the relations: Employee (SSN, Name, Age), Dependent (Name, Sex, ESSN)   Find the names of all employees who has a dependent with the same name. select Name from Employee where Name in (select Name from Dependent where ESSN = SSN)

15. 15 Nested Queries   in accepts multi-column list.   Find all enrollments that have 25 years old students taking CS courses. select * from enrollment where (SSN, Crs_no) in (select s.SSN, c.crs_no from student s, course c where s.age=25 and c.dept_name='CS');   Another way to write the query: select * from enrollment where SSN in (select SSN from student where Age = 25) and crs_no in (select crs_no from course where dept_name='CS');

16. 16 Using ANY   There’s an alternate way of seeing if a value is IN some table   The evaluation of ANY expressions is not always intuitive   Examples:  4 = ANY (1,2,3,4)  True, because it’s equal to the 4.  4 <> ANY (1,2,3,4)  True, because it’s not equal to the 1.

17. 17 Using ANY   Find the names of those students who are 18 or younger and whose GPA is higher than the GPA of any students who are 25 or older. select Name from Student where age any (select GPA from Student where Age >= 25);   Other set comparison operators: >any, =any, =any, <>any

18. 18 Using ALL   Similarly, we can use ALL to see how a value compares to all values in a table   Other set comparison operators: >all, =all, =all, <>all   Again, the evaluation of ALL expressions is not always intuitive   Examples:  4=ALL(1,2,3,4)  No, because it’s not equal to the 1.  4<>ALL(1,2,3,4)  Yes, because it’s not equal to the 1.  4=ALL(4,4,4,4)  Yes, because all of the values are 4.

19. 19 Using ALL   Retrieve the names of employees whose salary is greater than the salary of all the employees in Department 5 SELECT Lname, Fname FROM Employee where salary > ALL (select salary from employee where DNO=5);

20. 20 Using ANY and ALL   =any is equivalent to in   <>any is not equivalent to not in.   <>all is equivalent to not in. Let x = a and S = {a, b}. Then x <> any S is true ( x <> b ) but x not in S is false. x <> all S is also false ( x <> b but x=a ).

21. 21 Correlated Nested Queries   If a condition in the WHERE-clause of a nested query references an attribute of a relation declared in the outer query, the two queries are said to be correlated.   The result of a correlated nested query is different for each tuple (or combination of tuples) of the relation(s) the outer query   Retrieve the name of each employee who has a dependent with the same first name and same sex as the employee. SELECT FNAME, LNAME FROMEMPLOYEE E WHERESSN IN (SELECT ESSN FROM DEPENDENT WHERE ESSN = SSN AND FNAME = DNAME AND E.Sex = Sex);

22. 22 Correlated Nested Queries   In the previous query, the nested query has a different result for each tuple in the outer query   A query written with nested SELECT... FROM... WHERE... blocks and using the = or IN comparison operators can always be expressed as a single block query. SELECT FNAME, LNAME FROM EMPLOYEE E, DEPENDENT D WHERE SSN=ESSN AND FNAME=DNAME AND E.Sex = D.Sex;

23. 23 The EXISTS Function   EXISTS is used to check whether the result of a correlated nested query is empty (contains no tuples) or not.   We can formulate the previous query in an alternative form that uses EXISTS as below: SELECTFNAME, LNAME FROMEMPLOYEE E WHEREEXISTS (SELECT * FROMDEPENDENT WHERESSN=ESSN AND FNAME=DNAME AND E.Sex = Sex);

24. 24 The EXISTS Function   Retrieve the names of employees who have no dependents. SELECT FNAME, LNAME FROMEMPLOYEE WHERE NOT EXISTS (SELECT * FROM DEPENDENT WHERE SSN=ESSN);   The correlated nested query retrieves all DEPENDENT tuples related to an EMPLOYEE tuple. If none exist, the EMPLOYEE tuple is selected   EXISTS is necessary for the expressive power of SQL

25. 25 The EXISTS Function   Find all students who take at least one course. select * from Student s where exists (select * from Enrollment where SSN = s.SSN);   The previous query is equivalent to: select distinct s.* from Student s, Enrollment e where s.SSN = e.SSN; select * from Student where SSN in (select SSN from Enrollment);

26. 26 The EXISTS Function   exists () is true if the set () is not empty.   exists () is false if the set () is empty.   not exists () is true if the set () is empty.   not exists () is false if the set () is not empty.

27. 27 The EXISTS Function   Find all students who do not take CS532. select * from Student s where not exists (select * from Enrollment where SSN=s.SSN and cno= 'CS532');   This query is equivalent to: select * from Student where SSN not in (select SSN from Enrollment where cno = 'CS532');  SQL function UNIQUE(Q)  Returns TRUE if there are no duplicate tuples in the result of query Q

28. 28 Joining Tables   Find the titles of all courses offered by departments located in building EB. Department(Name, Location)Dept (DName, Location) Course(CNo, Title, DName) selectTitle fromCourse join Department on Dname=Name whereLocation = 'EB’; or select Title fromCourse natural join Department As D(DName, Location, Chairman) whereLocation = 'EB' or selectTitle fromCourse natural join Dept whereLocation = 'EB'

29. 29 Joining Tables SELECTFNAME, LNAME, ADDRESS FROM EMPLOYEE, DEPARTMENT WHEREDNAME='Research' AND DNUMBER=DNO; could be written as: SELECTFNAME, LNAME, ADDRESS FROM (EMPLOYEE JOIN DEPARTMENT ON DNUMBER=DNO) WHEREDNAME='Research’; or as: SELECTFNAME, LNAME, ADDRESS FROM (EMPLOYEE NATURAL JOIN DEPARTMENT AS DEPT(DNAME, DNO, MSSN, MSDATE) WHEREDNAME='Research’

30. 30   Retrieve pnumber, dnum, lname, bdate for employees working on projects located at “Stafford”. SELECT PNUMBER, DNUM,LNAME, BDATE FROM((PROJECT JOIN DEPARTMENT ON DNUM=DNUMBER) JOIN EMPLOYEE ON MGRSSN=SSN ) WHERE PLOCATION='Stafford’; Joining Tables

31. 31 Joining Tables   Employee(SSN, Name, Position, Pno) Project(Pno, Title, Budget)   Find who works for which project. select SSN, Name, Title from Employee e join Project p on e.Pno = p.Pno;   Identify also those who do not work for any project. select SSN, Name, Title from Employee e left outer join Project p on e.Pno = p.Pno;

32. 32 Joining Tables   Identify also those projects no one works for. select SSN, Name, Title from Employee e right outer join on e.Pno = p.Pno; Project p on e.Pno = p.Pno;   Identify even those who do not work for any project and those projects no one works for: select SSN, Name, Title from Employees natural full join Projects;

33. 33 Aggregate Functions   SQL supports five aggregate functions: Name Argum. type Result type Description avg numeric numeric average count any numeric count min char or num. same as arg minimum max char or num. same as arg maximum sum numeric numeric sum

34. 34 Aggregate Functions   Find the maximum salary, the minimum salary, and the average salary among all employees. SELECT MAX(salary), MIN(salary), AVG(salary) FROMEMPLOYEE;   Find the maximum salary, the minimum salary, and the average salary among employees who work for the 'Research' department. SELECT MAX(salary), MIN(salary), AVG(salary) FROMEMPLOYEE, DEPARTMENT WHEREDNO=DNUMBER AND DNAME='Research‘;

35. 35 Aggregate Functions   Retrieve the total number of employees in the company SELECT COUNT (*) FROMEMPLOYEE;   Retrieve the number of employees in the 'Research' department SELECT COUNT (*) FROMEMPLOYEE, DEPARTMENT WHEREDNO=DNUMBER AND DNAME='Research’;   Find the number of courses offered. select count(distinct Course_no) from Enrollment   Note that the above is different from select distinct count(Course_no) from Enrollment   Find the average, the minimum and the maximum GPAs among all students' GPAs. select avg(GPA), min(GPA), max(GPA) from Student

36. 36 Aggregate Functions   Find the SSNs and names of all students who take 5 or more courses. select SSN, Name from Students s where 5 <= (select count(*) from enrollment where ssn = s.ssn);   Count the number of distinct salary values in the database. SELECT COUNT(DISTINCT Salary) FROM Employee;   Retrieve the names of all employees who have two or more dependents. SELECT Lname, Fname FROM Employee where (select count(*) from dependent where ssn= essn) >= 2;

37. 37 Aggregate Functions   Find the number of students who are 17 years old. select count(*) from Student where Age = 17   The above last query is equivalent to select count(SSN) from Students where Age = 17   but may not be equivalent to select count(Name) from Students where Age=17   Aggregate functions, except count(*), ignore null values.

38. 38 Group By & Having Clause   In many cases, we want to apply the aggregate functions to subgroups of tuples in a relation   Each subgroup of tuples consists of the set of tuples that have the same value for the grouping attribute(s)   The function is applied to each subgroup independently   SQL has a GROUP BY-clause for specifying the grouping attributes, which must also appear in the SELECT-clause

39. 39 Grouping   For each department, retrieve the department number, the number of employees in the department, and their average salary. SELECT DNO, COUNT (*), AVG (SALARY) FROMEMPLOYEE GROUP BYDNO;   the EMPLOYEE tuples are divided into groups--each group having the same value for the grouping attribute DNO   The COUNT and AVG functions are applied to each such group of tuples separately   The SELECT-clause includes only the grouping attribute and the functions to be applied on each group of tuples   A join condition can be used in conjunction with grouping

40. 40 Grouping   For each project, retrieve the project number, project name, and the number of employees who work on that project. SELECT PNUMBER, PNAME, COUNT (*) FROMPROJECT, WORKS_ON WHEREPNUMBER=PNO GROUP BYPNUMBER, PNAME;   In this case, the grouping and functions are applied after the joining of the two relations

41. 41 Grouping   Find the average GPA for students of different age groups. select Age, avg(GPA) from Students group by Age;   One tuple will be generated for each distinct value of age. STUDENTS RESULT SSN Name Age GPA Age avg(GPA) 123456789 John 19 3.6 19 3.7 234567891 Tom 20 3.2 20 3.5 345678912 Tom 19 3.8 21 2.8 345678912 Bill 21 2.8 345678912 Mary 20 3.8

42. 42 Grouping   When group by is used, each attribute in the select clause must have a single atomic value for each group of common group by values.   Each attribute in the select clause should be either a grouping attribute or an attribute on which a set function is applied.   Every grouping attribute should be listed in the select clause.   The following is an illegal query: select Age, SSN, avg(GPA) from Students group by Age

43. 43 Grouping   Find the SSN, name and the number of credit hours each student still needs to graduate.  Course(CNo, Title, Dept_Name, CreditHour)  Enrollment(SSN, CNo, Grade, Semester)   Assume that each student needs 120 credit hours to graduate. select ssn, name, 120-sum(credithour) CreditNeeded from Student s, Enrollment e, Course c where s.ssn=e.ssn and e.CNo=c.CNo group by s.ssn, name

44. 44 The Having Clause   Sometimes we want to retrieve the values of these functions for only those groups that satisfy certain conditions   Having is used in conjunction with group by in the cases where the groups must satisfy a condition to appear in the result.   The HAVING-clause is used for specifying a selection condition on groups (rather than on individual tuples)   Conditions on aggregate functions are specified in the having clause.

45. 45 The Having Clause   For each project on which more than two employees work, retrieve the project number, project name, and the number of employees who work on that project. SELECT PNUMBER, PNAME, COUNT (*) FROM PROJECT, WORKS_ON WHERE PNUMBER=PNO GROUP BY PNUMBER, PNAME HAVING COUNT (*) > 2;   Find the number of students of each age group and output only those groups having more than 50 members. select Age, count(*) from Students group by Age having count(*) > 50;

46. 46 The Having Clause   For each department having more than five employees, retrieve the department name and the number of employees making more than $40000. SELECT Dname, COUNT(*) FROM Department, Employee WHERE Dnumber = DNO and Salary > 40000 AND DNO IN (Select DNO FROM Employee Group by DNO Having COUNT(*)>5) GROUP BY Dname;

47. 47 Summary of SQL Queries   A query in SQL can consist of up to six clauses, but only the first two, SELECT and FROM, are mandatory.   The clauses are specified in the following order: SELECT FROM [WHERE ] [GROUP BY ] [HAVING ] [ORDER BY ]

48. 48 Evaluation of a six-clause query 1. Form all combinations of tuples from the relations in the from clause (Cartesian product). 2. Among all the tuples generated in step 1, find those that satisfy the conditions in the where clause. 3. Group the remaining tuples based on the grouping attributes. 4. Among all the tuples generated in step 3, find those that satisfy the conditions in the having clause. 5. Using the order by clause to order the tuples produced in step 4. 6. Project on the desired attribute values as specified in the select clause.

49. 49 SQL: Example 1   Find the average GPAs of students of different age groups. Only students younger than 35 are considered and only groups with the average GPAs higher than 3.2 are listed. The listing should be in ascending age values. select Age, avg(GPA) from Student where Age < 35 group byAge havingavg(GPA) > 3.2 order byAge

50. 50 SQL: Example 2   Consider two tables:  Agents(aid, aname, city, percent)  Orders(ordno, month, cid, aid, pid, qty, dollars)   Find the aid and total commission of each agent. An agent’s total commission is computed based on his/her commission rate and total sale (in dollars). select a.aid, a.percent*sum(dollars) as commission from agents a, orders o where a.aid = o.aid group by a.aid, a.percent;

51. 51 The Drop Command   DROP SCHEMA COMPANY CASCADE;  the schema and all its elements (tables, views, domain definitions … etc.) are also deleted.   DROP SCHEMA COMPNAY RESTRICT;  the schema will be deleted if it has no elements. If it has any element the drop schema statement will not be executed.   DROP TABLE DEPENDENT CASCADE;  The table dependent and all other elements that reference the dependent table are also deleted (e.g. foreign keys and views).   DROP TABLE DEPENDENT RESTRICT;  The table is deleted only if it is not referenced in any constraint (foreign key definitions in other tables, views).

52. 52 ALTER TABLE   Add attributes to base relations:  ALTER TABLE COMPANY.EMPLOYEE ADD column JOB VARCHAR(12);   Drop a column:  ALTER TABLE COMPANY.EMPLOYEE DROP column ADDRESS CASCADE;  all views and constraints that reference this column are automatically deleted.  ALTER TABLE COMPANY.EMPLOYEE DROP column ADDRESS RESTRICT;  the address attribute won’t be deleted if it is being referenced by other schema elements.

53. 53 ALTER TABLE   Drop the default value of an attribute:  ALTER TABLE COMPNAY.DEPARTMENT alter column MGRSSN DROP DEFAULT;   Change the default value of an attribute:  ALTER TABLE COMPANY.DEPARTMENT ALTER column MGRSSN SET DEFAULT “333445555”;  Drop constraints  ALTER TABLE COMPANY.EMPLOYEE DROP CONSTRAINT EMPSUPERFK CASCADE;  Add constraints  ALTER TABLE COMPANY.EMPLOYEE ADD CONSTRAINT ……;

54. 54 ALTER TABLE   alter table Student add column Address varchar2(40);   alter table Student add constraint unique(Address);   not null cannot be specified for newly added attributes.   DDL statements ( create, alter, drop, truncate, etc ), once executed, are irreversible! NO UNDO!!!

55. 55 Union, Intersection, and Except (1)  SQL supports three set operations:  union, intersect, minus  Union compatibility is required.  Consider the following two relations:  Students(SSN, Name, GPA, GRE, DeptName)  Instructors(SSN, Name, Office, DeptName)