CSC443

Integrity

Set operations

List operations

Review

• Commands:

• Create

• Drop

• Alter

• Insert

• Select

Syllabus

• New syllabus in course info

• Reading this week - 4.1-3; 8.6-8; finishing chap 3

• Test: mostly on executing sql statements

Integrity - Primary key

• Primary key uniqueness was serious

• SQL> insert into odetailsvalues(1020, 10506, 1);  2insert into odetails*ERROR at line 1:ORA-00001: unique constraint (S06_011.SYS_C0016803) violated

Integrity - Foreign key

• Foreign key existence was serious

SQL> insert into customers values (1113,'James','123 Park Ave',66666,'333-333-33

33');

insert into customers values (1113,'James','123 Park Ave',66666,'333-333-3333')

*

ERROR at line 1:

ORA-02291: integrity constraint (PEPPERK.SYS_C0016470) violated - parent key not found

Integrity - Check or Not Null qoh int check(qoh >= 0), -> set to -3

SQL> insert into parts values (3352,'bolt', -3,55.2,1);

insert into parts values (3352,'bolt', -3,55.2,1)

*

ERROR at line 1:

ORA-02290: check constraint (PEPPERK.SYS_C0016465) violated

Large Object Types

• Clob = characters

• Blob = binary Book_review clob(10KB)

Image blob(20MB)

Assertions and Domains

• Domain - a new data type with a pre-defined check

• Assertion - statement that the database always validates is true

• DOES NOT EXIST IN SQL

Triggers

• Procedure to keep tables in balance:

• First, created a reorder table:

SQL> create table reorder

2 (pno number(5) primary key,

3 qty number(38));

Triggers - cont

• Trigger example: create trigger reorder_rule

after update on parts

referencing new as nrow

for each row

when (nrow.qoh < 200)

begin

insert into reorder values (:nrow.pno, 300);

end reorder_rule;

Triggers - last one

• Result: SQL> update parts set qoh = 50 where qoh = 60;

1 row updated.

SQL> select * from reorder;

PNO QTY---------- ---------- 10509 300

Commit and Rollback

• Transactions - Need all together in order to be in sync.• Example: SQL> commit;

Commit complete.

SQL> update parts set qoh = 30 where qoh = 100;

1 row updated.

SQL> select * from parts where qoh = 100;

no rows selected

Commit & Rollback Cont

SQL> rollback;

Rollback complete.

SQL> select * from parts where qoh = 100;

PNO PNAME QOH PRICE OLEVEL

---------- ------------------------------ ---------- ---------- ----------

10900 Dr. Zhivago 100 24.99 30

Back into Query Swing

• List the first name of each student with a grade of less than 75 in any course.

• List the average grade of each student assuming every component equal weight

• Change the query above to exclude compname “programs” and to only show students with avg over 80.

Answers

• Select distinct fname from students s, scores g where s.sid = g.sid and points<75;

• Select avg(points), sid from scores group by sid;

• Select avg(points) as pavg, sid from scores where compname != 'programs' group by sid having avg(points) > 80;

Set Operations

• Union

• Intersect

• Except– Optional: all

Union

• Example: Show the max points from components mixed as another row in scores

• select * from (select * from scores union all (select 'MAX',term, lineno, compname, maxpoints from components)) order by lineno, compname, sid;

Intersect

• Find all sids with scores:

select sid from students intersect select sid from scores;

• Find all sids with no scores

select sid from students where sid not in (select sid from scores intersect select  sid from students);

Except - Minus

• Select all students with no grades

• Select distinct sid from students minus (select sid from scores);

SubQueries and Lists

Using a sub-query to get one value

Select distinct sid from scores where points > (select avg(points) from scores);

List Comparison - in

• Compare to values in a list - in or not in

• List student first names of sid 1111 and 2222

Select distinct fname from students where SID in (1111, 2222);

List Comparison - in

• Can use a subquery to create the list:

• List student first names who are enrolled in term f96

Select distinct fname from enrolls where term = ‘f96’);

You try: List first names of students in course 1030, 1031 and 1035

List Comparison - any, all, some

> ,< , <>,=, >=, <=

• Select scores greater than all in the list:

Select distinct sid from scores where points >= all (90, 340, 70)

Substitute a query listing all points

List Comparison - any, all, some

• Select scores greater than all in the list:

Select distinct sid from scores where points >= all (select points from scores);

1111

• You try:

List scores less than all in the list

List Comparison - any, all, some

Select distinct sid from scores where points <= all (select points from scores);

2222

Try one more:

List all scores which are greater than the average score

Sub-Query using outer

• Sub query acts for every row in the main (outer) query

• Use the fields from the outer query in the inner query

• List of courses in which all individual components grades are higher than B:

Select term, lineno, cno from courses where b <all (select points from scores where scores.lineno = courses.lineno and scores.term = courses.term);

Exists

• If subquery has any row in result, then exists is TRUE

• If no row results, then FALSESelect all courses in which 1111 is enrolled: select distinct cno from courses where exists

(select 'a' from enrolls where courses.lineno = enrolls.lineno and sid = ‘1111’);

You try: list all the fields from the student table where any score is greater than 90. (use exists)

Exists

• select * from students where exists (select ‘a’ from scores where students.sid = scores.sid and points > 90);

Important Stuff

• Database checks integrity

• Subqueries help make lists for where conditions

• In, All, Some, Any

• Exists