1. ©Silberschatz, Korth and Sudarshan3.1Database System Concepts - 6 th Edition Introduction to SQL Introduction to SQL Data Definition Basic Query Structure Additional Basic Operations Set Operations Null Values Aggregate Functions Nested Subqueries Modification of the Database

2. ©Silberschatz, Korth and Sudarshan3.2Database System Concepts - 6 th Edition Domain Types in SQL char(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.

3. ©Silberschatz, Korth and Sudarshan3.3Database System Concepts - 6 th Edition Create Table Construct An SQL relation is defined using the create table command: create table r (A 1 D 1, A 2 D 2,..., A n D n, (integrity-constraint 1 ),..., (integrity-constraint k )) r is the name of the relation each A i is an attribute name in the schema of relation r D i is the data type of values in the domain of attribute A i Example: create table instructor ( ID char(5), name varchar(20) not null, dept_name varchar(20), salary numeric(8,2)) insert into instructor values (‘10211’, ’Smith’, ’Biology’, 66000); insert into instructor values (‘10211’, null, ’Biology’, 66000);

4. ©Silberschatz, Korth and Sudarshan3.4Database System Concepts - 6 th Edition Integrity Constraints in Create Table not null primary key (A 1,..., A n ) foreign key (A m,..., A n ) references r Example: Declare branch_name as the primary key for branch create table instructor ( ID char(5), name varchar(20) not null, dept_name varchar(20), salary numeric(8,2), primary key (ID), foreign key (dept_name) references department) primary key declaration on an attribute automatically ensures not null

5. ©Silberschatz, Korth and Sudarshan3.5Database System Concepts - 6 th Edition And a Few More Relation Definitions create table student ( ID varchar(5) primary key, name varchar(20) not null, dept_name varchar(20), tot_cred numeric(3,0), foreign key (dept_name) references department) ); create table takes ( ID varchar(5) primary key, course_id varchar(8), sec_id varchar(8), semester varchar(6), year numeric(4,0), grade varchar(2), foreign key (ID) references student, foreign key (course_id, sec_id, semester, year) references section );

6. ©Silberschatz, Korth and Sudarshan3.6Database System Concepts - 6 th Edition And more still create table course ( course_id varchar(8) primary key, title varchar(50), dept_name varchar(20), credits numeric(2,0), foreign key (dept_name) references department) );

7. ©Silberschatz, Korth and Sudarshan3.7Database System Concepts - 6 th Edition Drop and Alter Table Constructs drop table alter table 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. alter table r drop A  where A is the name of an attribute of relation r  Dropping of attributes not supported by many databases.

8. ©Silberschatz, Korth and Sudarshan3.8Database System Concepts - 6 th Edition Basic Query Structure A typical SQL query has the form: select A 1, A 2,..., A n from r 1, r 2,..., r m where P A i represents an attribute R i represents a relation P is a predicate. The result of an SQL query is a relation.

9. ©Silberschatz, Korth and Sudarshan3.9Database System Concepts - 6 th Edition The select Clause The select clause list the attributes desired in the result of a query corresponds to the projection operation of the relational algebra Example: find the names of all instructors: select name from instructor NOTE: SQL names are case insensitive (i.e., you may use upper- or lower-case letters.) E.g., Name ≡ NAME ≡ name Some people use upper case wherever we use bold font.

10. ©Silberschatz, Korth and Sudarshan3.10Database System Concepts - 6 th Edition 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 departments with instructor, and remove duplicates select distinct dept_name from instructor The keyword all specifies that duplicates not be removed. select all dept_name from instructor

11. ©Silberschatz, Korth and Sudarshan3.11Database System Concepts - 6 th Edition The select Clause (Cont.) An asterisk in the select clause denotes “all attributes” select * from instructor The select clause can contain arithmetic expressions involving the operation, +, –, , and /, and operating on constants or attributes of tuples. The query: select ID, name, salary/12 from instructor would return a relation that is the same as the instructor relation, except that the value of the attribute salary is divided by 12.

12. ©Silberschatz, Korth and Sudarshan3.12Database System Concepts - 6 th Edition The where Clause The where clause specifies conditions that the result must satisfy Corresponds to the selection predicate of the relational algebra. To find all instructors in Comp. Sci. dept with salary > 80000 select name from instructor where dept_name = ‘Comp. Sci.' and salary > 80000 Comparison results can be combined using the logical connectives and, or, and not. Comparisons can be applied to results of arithmetic expressions.

13. ©Silberschatz, Korth and Sudarshan3.13Database System Concepts - 6 th Edition The from Clause The from clause lists the relations involved in the query Corresponds to the Cartesian product operation of the relational algebra. Find the Cartesian product instructor X teaches select  from instructor, teaches generates every possible instructor – teaches pair, with all attributes from both relations. Cartesian product not very useful directly, but useful combined with where-clause condition (selection operation in relational algebra).

14. ©Silberschatz, Korth and Sudarshan3.14Database System Concepts - 6 th Edition Joins For all instructors who have taught courses, find their names and the course ID of the courses they taught. select name, course_id from instructor, teaches where instructor.ID = teaches.ID Find the course ID, semester, year and title of each course offered by the Comp. Sci. department select section.course_id, semester, year, title from section, course where section.course_id = course.course_id and dept_name = ‘Comp. Sci.'

15. ©Silberschatz, Korth and Sudarshan3.15Database System Concepts - 6 th Edition Natural Join Natural join matches tuples with the same values for all common attributes, and retains only one copy of each common column select * from instructor natural join teaches;

16. ©Silberschatz, Korth and Sudarshan3.16Database System Concepts - 6 th Edition Natural Join (Cont.) Danger in natural join: beware of unrelated attributes with same name which get equated incorrectly List the names of instructors along with the the titles of courses that they teach Incorrect version (equates course.dept_name with instructor.dept_name) select name, title from instructor natural join teaches natural join course; Correct version select name, title from instructor natural join teaches, course where teaches.course_id= course.course_id; Another correct version select name, title from (instructor natural join teaches) join course using(course_id);

17. ©Silberschatz, Korth and Sudarshan3.17Database System Concepts - 6 th Edition The Rename Operation The SQL allows renaming relations and attributes using the as clause: old-name as new-name E.g., select ID, name, salary/12 as monthly_salary from instructor Find the names of all instructors who have a higher salary than some instructor in ‘Comp. Sci’. select distinct T. name from instructor as T, instructor as S where T.salary > S.salary and S.dept_name = ‘Comp. Sci.’ Keyword as is optional and may be omitted instructor as T ≡ instructor T

18. ©Silberschatz, Korth and Sudarshan3.18Database System Concepts - 6 th Edition String Operations SQL 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 instructors whose name includes the substring “dar”. select name from instructor where name like ' %dar% ' Match the string “100 %” like ‘ 100 \% ' escape ' \ ' SQL supports a variety of string operations such as concatenation (using “||”) converting from upper to lower case (and vice versa) finding string length, extracting substrings, etc.

19. ©Silberschatz, Korth and Sudarshan3.19Database System Concepts - 6 th Edition Ordering the Display of Tuples List in alphabetic order the names of all instructors select distinct name from instructor order by name We may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default. Example: order by name desc Can sort on multiple attributes Example: order by dept_name, name

20. ©Silberschatz, Korth and Sudarshan3.20Database System Concepts - 6 th Edition Where Clause Predicates SQL includes a between comparison operator Example: Find the names of all instructors with salary between $90,000 and $100,000 (that is,  $90,000 and  $100,000) select name from instructor where salary between 90000 and 100000 Tuple comparison select name, course_id from instructor, teaches where (instructor.ID, dept_name) = (teaches.ID, ’Biology’);

21. ©Silberschatz, Korth and Sudarshan3.21Database System Concepts - 6 th Edition Set Operations Find courses that ran in Fall 2009 or in Spring 2010 Find courses that ran in Fall 2009 but not in Spring 2010 (select course_id from section where sem = ‘Fall’ and year = 2009) union (select course_id from section where sem = ‘Spring’ and year = 2010) Find courses that ran in Fall 2009 and in Spring 2010 (select course_id from section where sem = ‘Fall’ and year = 2009) intersect (select course_id from section where sem = ‘Spring’ and year = 2010) (select course_id from section where sem = ‘Fall’ and year = 2009) except (select course_id from section where sem = ‘Spring’ and year = 2010)

22. ©Silberschatz, Korth and Sudarshan3.22Database System Concepts - 6 th Edition Set Operations Set operations union, intersect, and except Each of the above operations automatically eliminates duplicates To retain all duplicates use the corresponding multiset versions union all, intersect all and except all.

23. ©Silberschatz, Korth and Sudarshan3.23Database System Concepts - 6 th Edition Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null Example: 5 + null returns null The predicate is null can be used to check for null values. Example: Find all instructors whose salary is null. select name from instructor where salary is null

24. ©Silberschatz, Korth and Sudarshan3.24Database System Concepts - 6 th Edition Aggregate Functions These functions operate on the multiset of values of a column of a relation, and return a value avg: average value min: minimum value max: maximum value sum: sum of values count: number of values

25. ©Silberschatz, Korth and Sudarshan3.25Database System Concepts - 6 th Edition Aggregate Functions (Cont.) Find the average salary of instructors in the Computer Science department select avg (salary) from instructor where dept_name= ’Comp. Sci.’; Find the total number of instructors who teach a course in the Spring 2010 semester select count (distinct ID) from teaches where semester = ’Spring’ and year = 2010 Find the number of tuples in the course relation select count (*) from course;

26. ©Silberschatz, Korth and Sudarshan3.26Database System Concepts - 6 th Edition Aggregate Functions – Group By Find the average salary of instructors in each department select dept_name, avg (salary) from instructor group by dept_name; avg_salary

27. ©Silberschatz, Korth and Sudarshan3.27Database System Concepts - 6 th Edition Aggregation (Cont.) Attributes in select clause outside of aggregate functions must appear in group by list /* erroneous query */ select dept_name, ID, avg (salary) from instructor group by dept_name;

28. ©Silberschatz, Korth and Sudarshan3.28Database System Concepts - 6 th Edition Aggregate Functions – Having Clause Find the names and average salaries of all departments whose average salary is greater than 42000 Note: predicates in the having clause are applied after the formation of groups whereas predicates in the where clause are applied before forming groups select dept_name, avg (salary) from instructor group by dept_name having avg (salary) > 0;

29. ©Silberschatz, Korth and Sudarshan3.29Database System Concepts - 6 th Edition Null Values and Aggregates Total all salaries select sum (salary ) from instructor Above statement ignores null amounts Result is null if there is no non-null amount All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes What if collection has only null values? count returns 0 all other aggregates return null

30. ©Silberschatz, Korth and Sudarshan3.30Database System Concepts - 6 th Edition Nested Subqueries SQL 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.

31. ©Silberschatz, Korth and Sudarshan3.31Database System Concepts - 6 th Edition Example Query Find courses offered in Fall 2009 and in Spring 2010 Find courses offered in Fall 2009 but not in Spring 2010 select distinct course_id from section where semester = ’Fall’ and year= 2009 and course_id in (select course_id from section where semester = ’Spring’ and year= 2010); select distinct course_id from section where semester = ’Fall’ and year= 2009 and course_id not in (select course_id from section where semester = ’Spring’ and year= 2010);

32. ©Silberschatz, Korth and Sudarshan3.32Database System Concepts - 6 th Edition Example Query Find the total number of (distinct) students who have taken course sections taught by the instructor with ID 10101 Note: Above query can be written in a much simpler manner. The formulation above is simply to illustrate SQL features. select count (distinct ID) from takes where (course_id, sec_id, semester, year) in (select course_id, sec_id, semester, year from teaches where teaches.ID= 10101);

33. ©Silberschatz, Korth and Sudarshan3.33Database System Concepts - 6 th Edition Set Comparison Find names of instructors with salary greater than that of some (at least one) instructor in the Biology department. Same query using > some clause select name from instructor where salary > some (select salary from instructor where dept name = ’Biology’); select distinct T.name from instructor as T, instructor as S where T.salary > S.salary and S.dept name = ’Biology’;

34. ©Silberschatz, Korth and Sudarshan3.34Database System Concepts - 6 th Edition Definition of Some Clause F some r  t  r  such that (F t ) Where can be:      0 5 6 (5 < some) = true 0 5 0 ) = false 5 0 5 (5  some) = true (since 0  5) (read: 5 < some tuple in the relation) (5<some ) = true (5 = some (= some)  in However, (  some)  not in

35. ©Silberschatz, Korth and Sudarshan3.35Database System Concepts - 6 th Edition Example Query Find the names of all instructors whose salary is greater than the salary of all instructors in the Biology department. select name from instructor where salary > all (select salary from instructor where dept name = ’Biology’);

36. ©Silberschatz, Korth and Sudarshan3.36Database System Concepts - 6 th Edition Definition of all Clause F all r  t  r  (F t) 0 5 6 (5 < all) = false 6 10 4 ) = true 5 4 6 (5  all ) = true (since 5  4 and 5  6) (5 < all ) = false (5 = all (  all)  not in However, (= all)  in

37. ©Silberschatz, Korth and Sudarshan3.37Database System Concepts - 6 th Edition Test for Empty Relations The exists construct returns the value true if the argument subquery is nonempty. exists r  r  Ø not exists r  r = Ø

38. ©Silberschatz, Korth and Sudarshan3.38Database System Concepts - 6 th Edition Not Exists Find all studentswho have taken all courses offered in the Biology department. select distinct S.ID, S.name from student as S where not exists ( (select course_id from course where dept_name ’Biology’) except (select T.course_id from takes as T where S.ID = T.ID)); Note that X – Y = Ø  X  Y Note: Cannot write this query using = all and its variants

39. ©Silberschatz, Korth and Sudarshan3.39Database System Concepts - 6 th Edition Test for Absence of Duplicate Tuples The unique construct tests whether a subquery has any duplicate tuples in its result. Find all courses that were offered at most once in 2009 select T.course_id from course as T where unique (select R.course_id from section as R where T.course_id= R.course_id and R.year = 2009);

40. ©Silberschatz, Korth and Sudarshan3.40Database System Concepts - 6 th Edition Derived Relations SQL allows a subquery expression to be used in the from clause Find the average instructors’ salaries of those departments where the average salary is greater than $42,000.” select dept_name, avg_salary from (select dept_name, avg (salary) as avg_salary from instructor group by dept_name) where avg_salary > 42000; Note that we do not need to use the having clause Another way to write above query select dept_name, avg_salary from (select dept_name, avg (salary) from instructor group by dept_name) as dept_avg (dept_name, avg_salary) where avg_salary > 42000;

41. ©Silberschatz, Korth and Sudarshan3.41Database System Concepts - 6 th Edition Derived Relations (Cont.) And yet another way to write it: lateral clause select name, salary, avg_salary from instructor I1, lateral (select avg(salary) as avg_salary from instructor I2 where I2.dept_name= I1.dept_name);

42. ©Silberschatz, Korth and Sudarshan3.42Database System Concepts - 6 th Edition With Clause The 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 departments with the maximum budget with max_budget (value) as (select max(budget) from department) select budget from department, max_budget where department.budget = max_budget.value;

43. ©Silberschatz, Korth and Sudarshan3.43Database System Concepts - 6 th Edition Complex Queries using With Clause Find all departments where the total salary is greater than the average of the total salary at all departments with dept _total (dept_name, value) as (select dept_name, sum(salary) from instructor group by dept_name), dept_total_avg(value) as (select avg(value) from dept_total) select dept_name from dept_total, dept_total_avg where dept_total.value >= dept_total_avg.value;

44. ©Silberschatz, Korth and Sudarshan3.44Database System Concepts - 6 th Edition Scalar Subquery select dept_name, (select count(*) from instructor where department.dept_name = instructor.dept_name) as num_instructors from department;

45. ©Silberschatz, Korth and Sudarshan3.45Database System Concepts - 6 th Edition Modification of the Database – Deletion Delete all instructors delete from instructor Delete all instructors from the Finance department delete from instructor where dept_name= ’Finance’; Delete all tuples in the instructor relation for those instructors associated with a department located in the Watson building. delete from instructor where dept name in (select dept name from department where building = ’Watson’);

46. ©Silberschatz, Korth and Sudarshan3.46Database System Concepts - 6 th Edition Example Query Delete all instructors whose salary is less than the average salary of instructors delete from instructor where salary< (select avg (salary) from instructor); Problem: as we delete tuples from deposit, the average salary changes Solution used in SQL: 1. First, compute avg salary and find all tuples to delete 2. Next, delete all tuples found above (without recomputing avg or retesting the tuples)

47. ©Silberschatz, Korth and Sudarshan3.47Database System Concepts - 6 th Edition Modification of the Database – Insertion Add a new tuple to course insert into course values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4); or equivalently insert into course (course_id, title, dept_name, credits) values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4); Add a new tuple to student with tot_creds set to null insert into student values (’3003’, ’Green’, ’Finance’, null);

48. ©Silberschatz, Korth and Sudarshan3.48Database System Concepts - 6 th Edition Modification of the Database – Insertion Add all instructors to the student relation with tot_creds set to 0 insert into student select ID, name, dept_name, 0 from instructor The 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)

49. ©Silberschatz, Korth and Sudarshan3.49Database System Concepts - 6 th Edition Modification of the Database – Updates Increase salaries of instructors whose salary is over $100,000 by 3%, and all others receive a 5% raise Write two update statements: update instructor set salary = salary * 1.03 where salary > 100000; update instructor set salary = salary * 1.05 where salary <= 100000; The order is important Can be done better using the case statement (next slide)

50. ©Silberschatz, Korth and Sudarshan3.50Database System Concepts - 6 th Edition Case Statement for Conditional Updates Same query as before but with case statement update instructor set salary = case when salary <= 100000 then salary * 1.05 else salary * 1.03 end

51. ©Silberschatz, Korth and Sudarshan3.51Database System Concepts - 6 th Edition Updates with Scalar Subqueries Recompute and update tot_creds value for all students update student S set tot_cred = ( select sum(credits) from takes natural join course where S.ID= takes.ID and takes.grade <> ’F’ and takes.grade is not null); Sets tot_creds to null for students who have not taken any course Instead of sum(credits), use: case when sum(credits) is not null then sum(credits) else 0 end

52. ©Silberschatz, Korth and Sudarshan4.52Database System Concepts - 6th Edition Joined Relations n Join operations take two relations and return as a result another relation. n A join operation is a Cartesian product which requires that tuples in the two relations match (under some condition). It also specifies the attributes that are present in the result of the join. n The join operations are typically used as subquery expressions in the from clause.

53. ©Silberschatz, Korth and Sudarshan4.53Database System Concepts - 6th Edition Join operations – Example n Relation course n Relation prereq n Note: prereq information missing for CS-315 and course information missing for CS-437.

54. ©Silberschatz, Korth and Sudarshan4.54Database System Concepts - 6th Edition Outer Join n An extension of the join operation that avoids loss of information. n Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join. n Uses null values.

55. ©Silberschatz, Korth and Sudarshan4.55Database System Concepts - 6th Edition Left Outer Join n course natural left outer join prereq Note: read prere_id as prereq_id

56. ©Silberschatz, Korth and Sudarshan4.56Database System Concepts - 6th Edition Right Outer Join n course natural right outer join prereq

57. ©Silberschatz, Korth and Sudarshan4.57Database System Concepts - 6th Edition Full Outer Join n course natural full outer join prereq

58. ©Silberschatz, Korth and Sudarshan4.58Database System Concepts - 6th Edition Joined Relations – Examples n course inner join prereq on course.course_id = prereq.course_id n course left outer join prereq on course.course_id = prereq.course_id

59. ©Silberschatz, Korth and Sudarshan4.59Database System Concepts - 6th Edition Joined Relations – Examples n course natural right outer join prereq n course right outer join prereq using (course_id)

60. ©Silberschatz, Korth and Sudarshan4.60Database System Concepts - 6th Edition Views n In 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.) n Consider a person who needs to know an instructors name and department, but not the salary. This person should see a relation described, in SQL, by select ID, name, dept_name from instructor n A view provides a mechanism to hide certain data from the view of certain users. n Any relation that is not of the conceptual model but is made visible to a user as a “virtual relation” is called a view.

61. ©Silberschatz, Korth and Sudarshan4.61Database System Concepts - 6th Edition Example Views n A view of instructors without their salary create view faculty as select ID, name, dept_name from instructor n Find all instructors in the Biology department select name from faculty where dept_name = ‘Biology’ n Create a view of department salary totals create view departments_total_salary(dept_name, total_salary) as select dept_name, sum (salary) from instructor group by dept_name;

62. ©Silberschatz, Korth and Sudarshan4.62Database System Concepts - 6th Edition Views Defined Using Other Views n create view physics_fall_2009 as select course.course_id, sec_id, building, room_number from course, section where course.course_id = section.course_id and course.dept_name = ’Physics’ and section.semester = ’Fall’ and section.year = ’2009’; n create view physics_fall_2009_watson as select course_id, room_number from physics_fall_2009 where building= ’Watson’;

63. ©Silberschatz, Korth and Sudarshan4.63Database System Concepts - 6th Edition View Expansion n Expand use of a view in a query/another view create view physics_fall_2009_watson as (select course_id, room_number from (select course.course_id, building, room_number from course, section where course.course_id = section.course_id and course.dept_name = ’Physics’ and section.semester = ’Fall’ and section.year = ’2009’) where building= ’Watson’;

64. ©Silberschatz, Korth and Sudarshan4.64Database System Concepts - 6th Edition Views Defined Using Other Views n One view may be used in the expression defining another view, n A view relation v 1 is said to depend directly on a view relation v2 if v2 is used in the expression defining v1 n A 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 n A view relation v is said to be recursive if it depends on itself.

65. ©Silberschatz, Korth and Sudarshan4.65Database System Concepts - 6th Edition View Expansion n A way to define the meaning of views defined in terms of other views. n Let view v 1 be defined by an expression e1 that may itself contain uses of view relations. n 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 e1 n As long as the view definitions are not recursive, this loop will terminate.

66. ©Silberschatz, Korth and Sudarshan4.66Database System Concepts - 6th Edition Update of a View n Add a new tuple to faculty view which we defined earlier insert into faculty values (’30765’, ’Green’, ’Music’); This insertion must be represented by the insertion of the tuple (’30765’, ’Green’, ’Music’, null) into the instructor relation.

67. ©Silberschatz, Korth and Sudarshan4.67Database System Concepts - 6th Edition Some Updates cannot be Translated Uniquely n create view instructor_info as select ID, name, building from instructor, department where instructor.dept_name= department.dept_name; n insert into instructor info values (’69987’, ’White’, ’Taylor’);  which department, if multiple departments in Taylor?  what if no department is in Taylor? n Most SQL implementations allow updates only on simple views l The from clause has only one database relation. l The select clause contains only attribute names of the relation, and does not have any expressions, aggregates, or distinct specification. l Any attribute not listed in the select clause can be set to null l The query does not have a group by or having clause.

68. ©Silberschatz, Korth and Sudarshan4.68Database System Concepts - 6th Edition And Some Not at All n create view history_instructors as select * from instructor where dept_name= ’History’; n Insert (’25566’, ’Brown’, ’Biology’, 100000) into history_instructors

69. ©Silberschatz, Korth and Sudarshan4.69Database System Concepts - 6th Edition Transactions n Unit of work n Atomic transaction l either fully executed or rolled back as if it never occurred n Isolation from concurrent transactions n Transactions begin implicitly l Ended by commit work or rollback work n But default on most databases: each SQL statement commits automatically l Can turn off auto commit for a session (e.g. using API) l In SQL:1999, can use: begin atomic …. end

70. ©Silberschatz, Korth and Sudarshan4.70Database System Concepts - 6th Edition Referential Integrity n Ensures that a value that appears in one relation for a given set of attributes also appears for a certain set of attributes in another relation. l Example: If “Biology” is a department name appearing in one of the tuples in the instructor relation, then there exists a tuple in the department relation for “Biology”. n Let A be a set of attributes. Let R and S be two relations that contain attributes A and where A is the primary key of S. A is said to be a foreign key of R if for any values of A appearing in R these values also appear in S.

71. ©Silberschatz, Korth and Sudarshan4.71Database System Concepts - 6th Edition Cascading Actions in Referential Integrity n create table course ( course_id char(5) primary key, title varchar(20), dept_name varchar(20) references department ) n create table course ( … dept_name varchar(20), foreign key (dept_name) references department on delete cascade on update cascade,... ) n alternative actions to cascade: set null, set default

72. ©Silberschatz, Korth and Sudarshan4.72Database System Concepts - 6th Edition Integrity Constraint Violation During Transactions n E.g., create table person ( ID char(10), name char(40), mother char(10), father char(10), primary key ID, foreign key father references person, foreign key mother references person) n How to insert a tuple? n What if mother or father is declared not null? l constraint father_ref foreign key father references person, constraint mother_ref foreign key mother references person) l set constraints father_ref, mother_ref deferred

73. ©Silberschatz, Korth and Sudarshan4.73Database System Concepts - 6th Edition Complex Check Clauses n check (time_slot_id in (select time_slot_id from time_slot)) l why not use a foreign key here? n Every section has at least one instructor teaching the section. l how to write this? n Unfortunately: subquery in check clause not supported by pretty much any database l Alternative: triggers (later) n create assertion check ; l Also not supported by anyone

74. ©Silberschatz, Korth and Sudarshan4.74Database System Concepts - 6th Edition Built-in Data Types in SQL n date: Dates, containing a (4 digit) year, month and date l Example: date ‘2005-7-27’ n time: Time of day, in hours, minutes and seconds. l Example: time ‘09:00:30’ time ‘09:00:30.75’ n timestamp: date plus time of day l Example: timestamp ‘2005-7-27 09:00:30.75’ n interval: period of time l Example: interval ‘1’ day l Subtracting a date/time/timestamp value from another gives an interval value l Interval values can be added to date/time/timestamp values

75. ©Silberschatz, Korth and Sudarshan4.75Database System Concepts - 6th Edition Other Features n create table student (ID varchar (5), name varchar (20) not null, dept_name varchar (20), tot_cred numeric (3,0) default 0, primary key (ID)) n create index studentID index on student(ID) n Large objects l book review clob(10KB) l image blob(10MB) l movie blob(2GB)

76. ©Silberschatz, Korth and Sudarshan4.76Database System Concepts - 6th Edition User-Defined Types n create type construct in SQL creates user-defined type create type Dollars as numeric (12,2) final l create table department (dept_name varchar (20), building varchar (15), budget Dollars);

77. ©Silberschatz, Korth and Sudarshan4.77Database System Concepts - 6th Edition Domains n create domain construct in SQL-92 creates user-defined domain types create domain person_name char(20) not null n Types and domains are similar. Domains can have constraints, such as not null, specified on them. n create domain degree_level varchar(10) constraint degree_level_test check (value in (’Bachelors’, ’Masters’, ’Doctorate’));

78. ©Silberschatz, Korth and Sudarshan4.78Database System Concepts - 6th Edition Large-Object Types n Large objects (photos, videos, CAD files, etc.) are stored as a large object: l blob: binary large object -- object is a large collection of uninterpreted binary data (whose interpretation is left to an application outside of the database system) l clob: character large object -- object is a large collection of character data l When a query returns a large object, a pointer is returned rather than the large object itself.

79. ©Silberschatz, Korth and Sudarshan4.79Database System Concepts - 6th Edition Authorization Forms of authorization on parts of the database: n Read - allows reading, but not modification of data. n Insert - allows insertion of new data, but not modification of existing data. n Update - allows modification, but not deletion of data. n Delete - allows deletion of data. Forms of authorization to modify the database schema n Index - allows creation and deletion of indices. n Resources - allows creation of new relations. n Alteration - allows addition or deletion of attributes in a relation. n Drop - allows deletion of relations.

80. ©Silberschatz, Korth and Sudarshan4.80Database System Concepts - 6th Edition Authorization Specification in SQL n The grant statement is used to confer authorization grant on to n is: l a user-id l public, which allows all valid users the privilege granted l A role (more on this later) n Granting a privilege on a view does not imply granting any privileges on the underlying relations. n The grantor of the privilege must already hold the privilege on the specified item (or be the database administrator).

81. ©Silberschatz, Korth and Sudarshan4.81Database System Concepts - 6th Edition Privileges in SQL n select: allows read access to relation, or the ability to query using the view l Example: grant users U1, U2, and U3 select authorization on the branch relation: grant select on instructor to U1, U2, U3 n insert: the ability to insert tuples. n update: the ability to update using the SQL update statement. n delete: the ability to delete tuples. n all privileges: used as a short form for all the allowable privileges.

82. ©Silberschatz, Korth and Sudarshan4.82Database System Concepts - 6th Edition Revoking Authorization in SQL n The revoke statement is used to revoke authorization. revoke on from n Example: revoke select on branch from U1, U2, U3 n may be all to revoke all privileges the revokee may hold. n If includes public, all users lose the privilege except those granted it explicitly. n If the same privilege was granted twice to the same user by different grantees, the user may retain the privilege after the revocation. n All privileges that depend on the privilege being revoked are also revoked.

83. ©Silberschatz, Korth and Sudarshan4.83Database System Concepts - 6th Edition Roles n create role instructor; n Privileges can be granted to roles: l grant select on takes to instructor; n Roles can be granted to users, as well as to other roles l create role student l grant instructor to Amit; l create role dean; l grant instructor to dean; l grant dean to Satoshi;

84. ©Silberschatz, Korth and Sudarshan4.84Database System Concepts - 6th Edition Authorization on Views n create view geo_instructor as (select * from instructor where dept_name = ’Geology’); n grant select on geo_instructor to staff n Suppose that a staff member issues l select * from geo_instructor; n What if l staff does not have permissions on instructor? l creator of view did not have some permissions on instructor?

85. ©Silberschatz, Korth and Sudarshan4.85Database System Concepts - 6th Edition Other Authorization Features n references privilege to create foreign key l grant reference (dept_name) on department to Mariano; l why is this required? n transfer of privileges l grant select on department to Amit with grant option; l revoke select on department from Amit, Satoshi cascade; l revoke select on department from Amit, Satoshi restrict;

86. ©Silberschatz, Korth and Sudarshan4.86Database System Concepts - 6th Edition LOADING DATA FROM EXTERNAL FILES External file must be put into directory ‘/tmp’ * To unload relation into a file use the following command Select * from into OUTFILE ‘/tmp/ ’; * To load a relation from an external file use the following command: LOAD DATA INFILE ‘/tmp/ ’ into table

87. ©Silberschatz, Korth and Sudarshan4.87Database System Concepts - 6th Edition EXAMPLE * Suppose that you want to populate your relation instructor in your database with the data from relation Instructor of database S10_yuri. You are all granted select priviliges on S10_yuri You submit the following commands: use S10_yuri select * from instructor into OUTFILE ‘/tmp/Instructor.txt’; use S10_mchesnes delete from instructor ; // to avoid duplicate keys LOAD DATA INFILE ‘/tmp/Instructor.txt’ into table instructor; * You can create a command file to get the data for every relation in S10_yuri and unload them into external files stored in directory /tmp and then load it into your relations.

88. ©Silberschatz, Korth and Sudarshan4.88Database System Concepts - 6th Edition PHP INTERFACE with MYSQL Connect database using command $db=new mysqli (‘hercules’, ‘ ’, ‘ ’, ‘ ’) $query = “ select * from where “.$searchtype.” like ‘%”.$searchterm.”%’;” Retrieve row by using $row=$result->fetch_assoc(); Select columns that are in original query Finally, close database using $result->free() followed by $db->close();