CPT103-Introduction to Databases小豪的笔记

这里通过 XAMPP来用 PhpMyAdmin 创建数据库

1. Table and Data

1.1 Creating a Database

• First, we need to create a schema

CREATE SCHEMA name;

CREATE DATABASE name;

• If you want to create tables in this schema, you need to tell MySQL to “enter” into this schema, type:

Use name;

1.2 Creating Tables

• Syntax

CREATE TABLE [IF NOT EXISTS] name (
    col-name datatype [col-options],
    :
    col-name datatype [col-options],
    [constraint-1],
    :
    [constraint-n]
);
-- Contents between [] are optional

###1.3 Data Types

• For floating-point and fixed-point data types, M is the total number of digits that can be stored.

Data Type	属性
SMALLINT[(M)] [UNSIGNED] [ZEROFILL]	2 Bytes, signed range: -32768 to 32767
MEDIUMINT[(M)] [UNSIGNED] [ZEROFILL]	3 Bytes
INT[(M)] [UNSIGNED] [ZEROFILL]	4 Bytes
BIGINT[(M)] [UNSIGNED] [ZEROFILL]	8 Bytes
DECIMAL / DEC / NUMERIC / FIXED[(M[,D])] [UNSIGNED] [ZEROFILL]	M大小 1~65
FLOAT§ [UNSIGNED] [ZEROFILL]	automatically choose single precision or double precision based on the value of p
CHAR[(M)]	1 Byte, M is the length, 0 ~ 255
VARCHAR(M)	A variable-length string, M is 0 to 65,535

• Things between [] are optional

• Date and Time

数据类型	描述
DATE	YYYY-MM-DD
DATETIME[(fsp)]	‘YYYY-MM-DD hh:mm:ss[.fraction]’
TIMESTAMP	UTC time

###1.4 Column Options

col-name datatype [col-options]

• NOT NULL
  • values of this column cannot be null.
• UNIQUE
  • each value must be unique (candidate key on a single attribute)
• DEFAULT value
• AUTO_INCREMENT = baseValue
  • Must be applied to a key column (primary key, unique key)
  • a value (usually max(col) + 1) is automatically inserted when data is added.
  • You can also manually provide values to override this behaviour.
  • ALTER TABLE Persons AUTO_INCREMENT = 100;

1.5 Tuple Manipulation

1.5.1 INSERT

INSERT INTO tablename (col1, col2, …)
    VALUES (val1, val2, …),
    	:
    	(val1,val2,val3);

• If you are adding a value to every column, you don’t have to list them

INSERT INTO tablename VALUES (val1, val2, …);

1.5.2 UPDATE

UPDATE table-name
    SET col1 = val1 [,col2 = val2…]
    [WHERE condition]

1.5.3 DELETE

DELETE FROM
    table-name
    [WHERE condition]

• • If no condition is given then ALL rows are deleted.

2. Table Constraints

CREATE TABLE name (
    col-name datatype [col-options],
    :
    col-name datatype [col-options],
    [constraint-1],
    :
    [constraint-n]
);

2.1 Syntax of Constraints

CONSTRAINT name TYPE details;

• Constraint name is created so that later this constraint can be removed by referring to its name.
  • If you don’t provide a name, one will be generated.
• MySQL provides following constraint types
  • PRIMARY KEY
  • UNIQUE
  • FOREIGN KEY
  • INDEX

2.2 Domain Constraints

• A domain constraint can be defined along with the column or separately:

CREATE TABLE People (
    id INTEGER PRIMARY KEY,
    name VARCHAR(100) NOT NULL,
    sex CHAR NOT NULL CHECK (sex IN ('M','F')),
    CONSTRAINT id_positive CHECK (id > 0)
);

2.2.1 UNIQUE

CONSTRAINT name UNIQUE (col1, col2, …)

2.2.2 Primary Key

CONSTRAINT name PRIMARY KEY (col1, col2 …)

2.2.3 Foreign Key

CONSTRAINT name
    FOREIGN KEY
        (col1, col2, ...)
    REFERENCES
        table-name
    (col1, col2, ...)
    [ON UPDATE ref_opt
    ON DELETE ref_opt]

-- ref_opt: RESTRICT | CASCADE | SET NULL | SET DEFAULT

• The Binary Keyword
• The BINARY keyword instructs MySQL to compare the characters in the string using their underlying ASCII values rather than just their letters.

CREATE TABLE `branch` (
    `branchNo` char(4) BINARY NOT NULL,
    PRIMARY KEY (`branchNo`),
...);

2.3 Reference Options

• RESTRICT – stop the user from doing it

  • The default option

• CASCADE – let the changes flow on

• SET NULL – make referencing values null

• SET DEFAULT – make referencing values the default for their column

• These options can be applied to one or both kinds of the table updates:

  • ON DELETE
  • ON UPDATE

CONSTRAINT `FK_staff_branchNo`
    FOREIGN KEY (`branchNo`)
    REFERENCES `branch` (`branchNo`)
        ON DELETE SET NULL
        ON UPDATE CASCADE

3. Altering Table

3.1 Add column

ALTER TABLE table_name
	ADD column_name datatype [options like UNIQUE …];

3.2 Drop colum

ALTER TABLE table_name DROP COLUMN column_name;

3.3 Modify column name and definition

ALTER TABLE table_name
    CHANGE COLUMN
    col_name new_col_name datatype [col_options];

3.4 Modify column definition only

ALTER TABLE table_name
    MODIFY COLUMN
    column_name datatype [col_options]; 

###3.5 Adding Constraints

ALTER TABLE table-name
    ADD CONSTRAINT name definition;

3.6 Removing Constraints

ALTER TABLE table-name
DROP INDEX name | DROP FOREIGN KEY name | DROP PRIMARY KEY
-- INDEX - Can be used to drop unique keys

3.7 Deleting Tables

DROP TABLE [IF EXISTS] table-name1, table-name2…;

4. SQL Select Ⅰ

SELECT [DISTINCT | ALL]
    column-list FROM table-names
    [WHERE condition]
    [ORDER BY column-list]
    [GROUP BY column-list]
    [HAVING condition]

4.1 SELECT

SELECT col1[,col2…] FROM table-name;

4.2 DISTINCT and ALL

• Using DISTINCT after the SELECT keyword removes duplicates

4.3 Expressions in SELECT

select a, b, a+b as sum
	from dup_test;

4.4 Where

SELECT * FROM table-name
	WHERE predicate;

• Asterisk (*) means getting all columns of that table.

4.5 Word Search

4.5.1 LIKE

• We can use the LIKE keyword to perform string comparisons in queries

SELECT * FROM books
	WHERE bookName LIKE '%crypt%'
;

• Like is not the same as ‘=’ because it allows wildcard characters

• It is NOT normally case sensitive

• The % character can represent any number of characters, including none

• The _ character represents exactly one character

4.5.2 Dealing with Date and Time

• like numbers

SELECT * FROM table-name
	WHERE date-of-event < '2012-01-01'; 

• like a string

SELECT * FROM table-name
	WHERE date-of-event LIKE '2014-11-%';

4.6 Select and Cartesian Product

SELECT * FROM Table1, Table2;

• If the tables have columns with the same name

TableName.ColumnName

4.7 Aliases

• Column alias

SELECT column [AS] new-col-name

• Table alias

SELECT * FROM table [AS] new-table-name

Note: You cannot use a column alias in a WHERE clause

• Aliases and ‘Self-Joins’

SELECT A.Name FROM
    Employee A,
    Employee B
WHERE
	A.Dept = B.Dept
AND
	B.Name = 'Andy';

4.8 Subqueries

SELECT col1 FROM tablename
	WHERE col2 = (
		SELECT col FROM tablename
			WHERE condition)

• The first FROM part is evaluated first

4.8.1 IN

SELECT columns FROM tables
	WHERE col IN set;
	
SELECT columns FROM tables
	WHERE col NOT IN set;

4.8.2 EXISITS

• Using EXISTS we can see whether there is at least one element in a given set.

SELECT columns
FROM tables
WHERE EXISTS set; 

• NOT EXISTS is true if the set is empty

SELECT columns
FROM tables
WHERE NOT EXISTS set;

• The set is always given by a subquery

4.8.3 ANY and ALL

• ANY and ALL compare a single value to a set of values
• They are used with comparison operators like = , >, <, <>, >=, <=

val = ANY (set)

• is true if there is at least one member of the set equal to value

val = ALL (set)

• is true if all members of the set are equal to the value

5. SQL Select II

5.1 Joins

SELECT * FROM A CROSSJOINB;
-- same as
SELECT * FROM A,B;

5.2 INNER JOIN

SELECT * FROM A INNERJOINBON condition

• Can also use a USING clause that will output rows with equal values in the specified columns

SELECT * FROM A INNERJOINBUSING (col1, col2)

• col1 and col2 must appear in both A andB

5.3 NATURAL JOIN

SELECT * FROM A NATURALJOINB;

• A NATURAL JOIN is effectively a special caseof anINNERJOIN where the USING clause has specifiedall identicallynamed columns.

5.4 OUTER JOIN

SELECT cols FROM
    table1 type OUTER JOINtable2ON condition;

• Where type is one of LEFT, RIGHT or FULL

Full Outer Join in MySQL

(SELECT * FROM Student LEFT OUTER JOINEnrolment ON Student.ID = Enrolment.ID)
UNION
(SELECT * FROM Student RIGHT OUTER JOINEnrolment ON Student.ID = Enrolment.ID);

5.5 Order By

SELECT columns FROM tables 
    WHERE condition 
    ORDER BY cols [ASC|DESC]

• The ORDER BY clause sorts the results of a query
  • You can sort in ascending (default) or descending order
  • Multiple columns can be given

5.6 Aggregate Functions

• COUNT: The number of rows
• SUM: The sum of the entries in the column
• AVG: The average entry in a column
• MIN, MAX: The minimum/maximum entries in a column

SELECT COUNT | SUM | AVG | MIN | MAX (*) AS C FROM tablename;

• You cannot use aggregate functions in the WHERE clause

The use of aggregate functions leads to all rows afterthefirst row being truncated.

• But you can use them in the subqueries in the WHERE clause

5.7 GROUP BY

SELECT column_set1 FROM tablesWHERE 	predicate
	GROUP BY column_set2;

• Every entry in ‘column_set2’ should be in ‘column_set1’,beaconstant, or be an aggregate function

5.8 Having

• HAVING is like a WHERE clause, except that it onlyappliesto the results of a GROUP BY query
• It can be used to select groups which satisfyagivencondition

SELECT Name,
    AVG(Mark) AS Average
    FROM Grades
    GROUP BY Name
    HAVING
    	AVG(Mark) >= 40;

• HAVING refers to the groups of rows, and socannotusecolumns or aggregate functions that does not exist afterthestep of column selection (see below).

5.9 UNION, INTERSECT and EXCEPT

• Only UNION is supported in MySQL. The other two can be simulated with subqueries.
• They all combine the results from two select statements

Lab1

drop table if exists `activity`, `module_enrollment`, `student`, `modules`, `teachers`;

Task 1: Creating tables

create table `teachers`
(
    `id` int, -- we will add primary key later. NOT NULL and UNIQUE is not needed.
    `name` varchar(200) not null,
    `tel_no` varchar(40),  -- not using INT here as we have numbers like +44
    `office` varchar(15) not null
);

create table `modules`
(
    `code` varchar(10), -- we will add primary key later. NOT NULL and UNIQUE is not needed.
    `title` varchar(100) not null,
    `teacher_id` int
);

create table `student`
(
    `id` int(6), -- we will add primary key later. NOT NULL and UNIQUE is not needed.
    `name` varchar(200) not null ,
    `email` varchar(100) not null,
    `enrolled_modules` varchar(255)
);

Task 2: Primary keys

alter table `teachers` add primary key (`id`);
alter table `modules` add primary key (`code`);
alter table `student` add primary key (`id`);
-- The one below is a super key. Not a good choice:
--          alter table `student` add primary key (`student_id`, `student_name`);

Task 3: Foreign key

alter table `modules` add 
constraint fk_module_teacher foreign key (`teacher_id`) 
references `teachers` (`id`)

Task 4: Inserting data

insert into `teachers` values (6503399, 'John Drake', '12022017202020', 'SD-766');
insert into `modules` values ('HCI-101', 'Human Computer Interaction', 6503399);

insert into `teachers` values (7614411, 'Felicia Gomez', '1024', 'BES-207');
insert into `modules` values ('HSB', 'Haskell for Beginners', 7614411);

insert into `teachers` values (5899114, 'John Cartwright', '12345 ext 1212', 'BES-201');
insert into `modules` values ('MC1', 'Mathematics', 5899114);

-- Another way to insert data, if the module is not assigned with any teacher yet,
-- you can set it to be null and update it later
insert into `modules` values ('MC2', 'Advanced Mathematics', null); -- module is not assigned with teacher yet
insert into `teachers` values (7099543, 'Dave Moe', 'BES-205', '65432 ext 2121');
update `modules` set `teacher_id` = 7099543 where `code` = 'MC2'; -- update the teacher reference

Task 5: Foreign key for student

alter table `student` add constraint fk_student_module
    foreign key (`enrolled_modules`) references `modules` (`code`);

insert into `student` values (764411, 'Daryl', '[email protected]', 'MC1'); -- first module for daryl

Task 6: Second module for Dary

-- insert into `student` values (764411, 'Daryl', '[email protected]', 'MC2');
-- You will get error:  [23000][1062] (conn=40) Duplicate entry '764411' for key 'PRIMARY'

Task 7: Create the activity relation

create table activity
(
    `name` varchar(100) primary key,
    `student_id` int(6),
    `description` varchar(255),
    constraint fk_activity_student foreign key (`student_id`) references `student` (`id`)
);

-- remove the primary key for student
-- alter table `student` drop primary key;

-- now try to add foreign key to `activity`.`student_id`
-- alter table `activity` add constraint fk_activity_student
--    foreign key (`student_id`) references `student` (`id`);
--
-- Does not work: Can't create table `jianjun`.`activity` (errno: 150 "Foreign key constraint is incorrectly formed")
-- Reason explained in the lab sheet

Task 8: Redesigns

drop table if exists module_enrollment, activity, student, modules;

create table `modules`
(
    `code` varchar(10) primary key,
    `title` varchar(100) not null,
    `teacher_id` int,
     constraint fk_module_teacher foreign key (`teacher_id`) references `teachers` (`id`)
);

create table `student`
(
    `id` int(6) primary key,
    `name` varchar(200) not null,
    `email` varchar(100) not null
);

create table module_enrollment
(
    `enrollment_id` int primary key, -- this column is optional.
    `module_code` varchar(10),
    `student_id` int(6),
    constraint fk_enrollment_module foreign key (`module_code`) references `modules` (`code`),
    constraint fk_enrollment_student foreign key (`student_id`) references `student` (`id`)
);

lab2

###Data

drop table if exists `activity`, `module_enrollment`, `student`, `modules`, `teachers`;

create table `teachers`
(
    `id` int primary key, -- we will add primary key later. NOT NULL and UNIQUE is not needed.
    `name` varchar(200) not null,
    `tel_no` varchar(40),  -- not using INT here as we have numbers like +44
    `office` varchar(15) not null
);

create table `modules`
(
    `code` varchar(10) primary key,
    `title` varchar(100) not null,
    `teacher_id` int,
     constraint fk_module_teacher foreign key (`teacher_id`) references `teachers` (`id`)
);

create table `student`
(
    `id` int(6) primary key,
    `name` varchar(200) not null,
    `email` varchar(100) not null
);

create table module_enrollment
(
    `enrollment_id` int primary key, -- this column is optional.
    `module_code` varchar(10),
    `student_id` int(6),
    constraint fk_enrollment_module foreign key (`module_code`) references `modules` (`code`),
    constraint fk_enrollment_student foreign key (`student_id`) references `student` (`id`)
);

create table activity
(
    `name` varchar(100) primary key,
    `student_id` int(6),
    `description` varchar(255),
    constraint fk_activity_student foreign key (`student_id`) references `student` (`id`)
);

insert into `teachers` values (6503399, 'John Drake', '12022017202020', 'SD-766');
insert into `modules` values ('HCI-101', 'Human Computer Interaction', 6503399);

insert into `teachers` values (7614411, 'Felicia Gomez', '1024', 'BES-207');
insert into `modules` values ('HSB', 'Haskell for Beginners', 7614411);

insert into `teachers` values (5899114, 'John Cartwright', '12345 ext 1212', 'BES-201');
insert into `modules` values ('MC1', 'Mathematics', 5899114);

insert into `teachers` values (7099543, 'Dave Moe', 'BES-205', '65432 ext 2121');
insert into `modules` values ('MC2', 'Advanced Mathematics', 7099543);

insert into `student` values 
(156123,'Nuno Bloggs','[email protected]'),
(156897,'John Trump','[email protected]'),
(123987,'Lidia Elliott','[email protected]'),
(777123,'Alicia Smith','[email protected]'),
(127845,'Sophie Johns','[email protected]');


insert into `module_enrollment` values
(1,'HSB',156123),
(2,'HCI-101',156123),
(3,'HSB',156897),
(4,'MC1',156897),
(5,'MC2',156897),
(6,'MC2',777123),
(7,'HSB',127845),
(8,'HCI-101',127845),
(9,'MC1',127845),
(10,'MC2',127845);

Task 1:

1. Print the Names of all Teachers.

select name from teachers;

2. Retrieve all students and all of the information held about them (excluding enrolment).

select * from student;

3. Retrieve the student ID’s who study MC2.

select distinct id from student, module_enrollment 
	where student.id = module_enrollment.student_id and module_code = 'MC2';

4. Retrieve the list of student ID’s, names and email address for students who study HCI-101.

--         Your results should not include duplicates or incorrect information.
select distinct id, name, email 
	from student s, module_enrollment me 
	where s.id = me.student_id and me.module_code = 'HCI-101';

5. Retrieve the Names of students who do not take the module ‘MC1’

select distinct s.name from student s left outer join module_enrollment me on s.id = me.student_id
	where s.name not in (
		select s.name from student s, module_enrollment me
			where s.id = me.student_id and me.module_code = 'MC1'
	)

6. Retrieve the Names and Emails of students who study both ‘MC1’ and ‘MC2’.

select s.name, s.email from student s, module_enrollment me 
	where s.id = me.student_id and me.module_code = 'MC1'
	and s.name in (
		select s.name from student s, module_enrollment me
			where s.id = me.student_id and me.module_code = 'MC2'
	);

select s.name, s.email from student s, module_enrollment me1, module_enrollment me2
	where s.id = me1.student_id and s.id = me2.student_id
    and me1.module_code = 'MC1'
	and me2.module_code = 'MC2'

7. Retrieve the Names and Telephone Numbers of lecturers whose office is in the ‘BES’ building.

select name, tel_no from teachers where office like 'BES%';

Task 2:

8. Retrieve the IDs of students who take the module ‘MC1’ OR ‘MC2’.

select id from module_enrollment where module_code in ('MC1', 'MC2');

9. Retrieve the Names and Emails of students who study either in ‘MC1’ OR ‘MC2’.

select distinct s.name, s.email from student s, module_enrollment me 
	where s.id = me.student_id and me.module_code = 'MC2'
	or s.name in (
		select s.name from student s, module_enrollment me
			where s.id = me.student_id and me.module_code = 'MC1'
	);

10. Retrieve the Names and Emails of students whose email is not ended with ‘@gmail.com’.

select name, email from student where email not like '%gmail.com';

11. Identify if there is any student who enrolled in HSB, HCI-101 and MC2 and Retrieve the Names and ID of them.

select distinct s.name, s.email from student s, module_enrollment me1, module_enrollment me2, module_enrollment me3
	where s.id = me1.student_id and s.id = me2.student_id and s.id=me3.student_id
    and me1.module_code = 'HSB'
	and me2.module_code = 'MC2'
    and me3.module_code = 'HCI-101'

12. Retrieve the ID and Names of students whose name includes ‘ia’

select id, name from student where name like '%ia%';

13. Retrieve the Names and Emails of students who study not in ‘MC1’ OR ‘MC2’ OR ‘HCI-101’.

select distinct s.name from student s left outer join module_enrollment me on s.id = me.student_id
	where s.name not in (
		select s.name from student s, module_enrollment me
			where s.id = me.student_id and me.module_code = 'MC1' OR s.id = me.student_id and me.module_code = 'MC2' OR s.id = me.student_id and me.module_code = 'HCI-101'
	)

lab3

Data

drop table if exists cd, artist;

create table `artist` (
	`artid` int primary key,
	`artname` varchar(100)
);

insert into `artist` values
(6,'Animal Collective'),
(3,'Deadmau5'),
(7,'Kings of Leon'),
(4,'Mark Ronson'),
(5,'Mark Ronson & The Business Intl'),
(8,'Maroon 5'),
(2,'Mr Scruff'),
(1,'Muse');

create table `cd` (
	`cdid` int primary key,
	`artid` int,
	`cdtitle` varchar(100),
	`cdprice` double,
	`cdgenre` varchar(50),
	`cdnumtracks` int,
	constraint fk_cd foreign key (`artid`) references `artist` (`artid`)
);

insert into `cd` values
(1,1,'Black Holes and Revelations',9.99,'Rock',NULL),
(2,1,'The Resistance',11.99,'Rock',NULL),
(3,2,'Ninja Tuna',9.99,'Electronica',NULL),
(4,3,'For Lack of a Better Name',9.99,'Electro House',NULL),
(5,4,'Version',12.99,'Pop',NULL),
(6,5,'Record Collection',11.99,'Alternative Rock',NULL),
(7,6,'Merriweather Post Pavilion',12.99,'Electronica',NULL),
(8,7,'7 Only By The Night',9.99,'Rock',NULL),
(9,7,'Come Around Sundown',12.99,'Rock',NULL),
(10,8,'Hands All Over',11.99,'Pop',NULL);

Task 1:

a. List the titles and prices of CDs in order of price from highest to lowest.

select cdtitle, cdprice from cd order by cdprice desc;

b. List the Artist Name, Titles and the Price of CDs in alphabetical order by artist name.

-- The Price of the CD should be returned in a column called ‘Full Price’ with tax (20%) included - the cd price in the database is not inclusive of tax.
select artname, cdtitle, (cdprice * 1.2) as `full price` from cd, artist where cd.artid = artist.artid;

c. List the titles, genres and prices CDs in alphabetical order by genre

-- then by price from the highest price to the lowest one.
select cdtitle, cdgenre, cdprice from cd order by cdgenre ASC, cdprice DESC;

Task 2:

a. Find the lowest price of any CD. The result should be presented in a column named ‘Cheapest CD

select cdtitle as `Cheapest CD` from cd where cdprice <= all(select cdprice from cd);

select min(cdprice) as `Cheapest CD` from cd; -- in case you have a different understanding of this question.

b. Identify the difference between the most expensive and cheapest CD.

--  The result should be presented in a column named ‘CD Price Range’.
select (max(cdprice) - min(cdprice)) as 'CD Price Range' from cd;

c. Find the number of CDs costing 9.99. The result should be presented in a column named ‘Count of £9.99 CD’s’

select (count(cdtitle)) as 'Count of $9.99 CD\'s' from cd where cdprice = 9.99;

d. Find the title of the most expensive Electronica CD(s).

select cdtitle from cd where cdprice >= all (select cdprice from cd where cdgenre = 'Electronica') and

####e. Find the number of different Prices in the CD table.

select count(distinct cdprice) from cd;

f. List all the information about the cheapest (lowest priced) CDs.

select * from cd, artist where cd.artid = artist.artid and cdprice <= all(select cdprice from cd);

Task 3:

a. Find a list of artist names, the number of CDs they have produced, and the average price for their CDs.

--      Only return results for artists with more than one CD. The results should be in the following format:
--      ||   Artist |  Average CD Price  | Number of CD’s  ||
select artist.artname as `Artist`, AVG(cdprice) as `Average CD price`, COUNT(cdid) as `Number of CDs`
	from cd, artist where cd.artid = artist.artid group by artist.artname having `Number of CDs` > 1;

b. Find a list of artist names, the number of CDs by that artist and the average price for their CDs

--      but not including ‘Electronica’ albums (you might like to use a WHERE in this one too).
--      The results should be in the following format:
--      ||  Artist | Average CD Price | Number of CD’s ||
select artist.artname as `Artist`, AVG(cdprice) as `Average CD price`, COUNT(cdid) as `Number of CDs`
	from cd, artist where cd.artid = artist.artid and cdgenre <> 'Electronica' group by artist.artname;