Slide 1 Chapter 7 – Part 2 Chapter 7 – Part 2 Constraints JOIN statement View.

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Chapter 7 – Part 2Chapter 7 – Part 2

ConstraintsConstraintsJOIN statementJOIN statement

ViewView

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Contents

A. Art Museum ProblemB. SolutionC. Other Requirements

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A. Art Museum Problem

1. Problem Description2. Examples

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1. Problem Description

DAWN museum where a lot of valuable works have been exhibited needs to manage and classify its works. Therefore, a database is one of the best solutions. The owner also requests all of works to be arranged by the artist who created these works. To identify the artist, the following information will be used: artist Id, artist name, nationality, birthday, deceased day. In addition, each artist may have many works or not. The work of each artist will be identified by work id, title, copy and description.

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Besides the general information, there are some special requirements such as: Artist’s nationality must be within these following countries:

Canadian, English, French, German, Mexican, Russian, Spanish and US.

Deceased day (if any) must be greater than artist’s birthday. Artist name must be unique Both Work title and work copy must be unique

As a database developer, you are requested to design the above database.

After that, write SQL statement to display artist name and the number of works (count of works) of each artist.

Finally, write SQL statement to display artist name and the number of works (count of works) of each artist and these work must be first copy.

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2. Examples Artist name: LEIGHTON, SIR FREDERICK. 1830-1896. Nationality: English Birthday: 1830 Deceased day: 1896 Some Chief Works:

First work:• Title: Hercules Wrestling with Death• Copy: First copy• Description: Painting

Second work;• Title: Daphnephoria• Copy: Original• Description: Painting

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B. Solution

1. Create Logical Diagram2. Create Physical Diagram3. Write SQL Statement to Create Tables 4. Create Constraints5. Insert data into tables6. Write SQL Statement to Display information

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1. Create Logical Diagram

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2. Create Physical Diagram

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3. Write SQL Statement to Create Tables

Create Table ARTISTS ( ART_ID int not null Primary

Key Identity,

ARTName varchar(50) not null Unique ,

ARTNationality varchar(50) not null, ARTBirthDay datetime not null, ARTDeceasedDay datetime);

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Create Table WORKS ( WOR_ID bigint not null Primary

Key Identity, ART_ID int not null, WORTitle varchar(100) not null, WORCopy varchar(20) not null, WORDescription varchar(200), Constraint AK_Title_Copy Unique (WORTitle,

WORCopy), Constraint FK_Works Foreign Key (ART_ID)

ReferencesARTISTS (ART_ID) on update cascade

);

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4. Create Constraints

4.1. Set Constraint For Nationality in Artist 4.2. Set Constraint Between Birth Date and

Deceased Date

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4.1. Set Constraint For Nationality in Artist

Artist’s nationality must be within these following countries: Canadian, English, French, German, Mexican, Russian, Spanish, US.

Alter Table ARTISTSAdd Constraint Chk_Nationality Check (ARTNationality in

('Canadian','English','French','German','Mexican','Russian‘, 'Spanish','US'));

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4.2. Set Constraint Between Birth Date and Deceased Date

Deceased day (if any) must be greater than artist’s birthday

Alter Table ARTISTSAdd Constraint ChkDeceasedDay Check

(ARTDeceasedDay > ARTBirthDay);

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5. Insert data into tables Exercise: Using DML, write SQL statements to

insert the following data into tables (Artists and Works):

Artists table:

Works table:

No. Name Nationality Birthday Deceased day

1 Artist 1 US 10/2/1925 30/6/1975

2 Artist 2 US 5/4/1955

3 Artist 3 French 6/7/1930 12/8/1990

4 Artist 4 Russian 8/9/1950

No. Title Copy Description Artist ID

1 Title 1 Original Painting 1

2 Title 2 First copy Photograph 1

3 Title 3 Original Painting 3

4 Title 4 First copy Painting 3

5 Title 5 Second copy Photograph 3

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6. Write SQL Statement to Display information

6.1. Count Number of Artist’s Work 6.2. Count Number of Artist’s Work which is

First Copy

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6.1. Count Number of Artist’s Work

Write SQL statement to display all artist name and the number of works (count of works) of each artist.

Using Joins:Select a.ArtName, count(w.Wor_ID) as numWorkFrom Artists a, Works wWhere a.Art_ID = w.Art_IDGroup by a.ArtName;

Result:

(in Artist table, there are 4 artists)

Name numWork

Artist 1 2

Artist 3 3

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Using Left Join:Select a.ArtName, count(w.Wor_ID) as

numWorkFrom Artists a LEFT JOIN Works w ON a.Art_ID

= w.Art_IDGroup by a.ArtName;

ResultName numWork

Artist 1 2

Artist 2 0

Artist 3 3

Artist 4 0

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6.2. Count Number of Artist’s Work which is First Copy

Write SQL statement to display all artist name and the number of works (count of works) of each artist and these work must be first copy.

Select a.ArtName, count(w.Wor_ID) as numWork

From Artists a LEFT JOIN Works w ON a.Art_ID = w.Art_ID

Where w.WorCopy =‘First copy’ or w.WorCopy is NULL

Group by a.ArtName; Result:

Name numWork

Artist 1 1

Artist 2 0

Artist 3 1

Artist 4 0

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C. Other Requirements

1. Creating View Problem2. Creating View Solution

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1. Creating View Problem Code a SQL statement to create a view that shows

NAME and NATIONALITY for all artist. Code a SQL statement to create a view that show

artist NAME and computes the number of works for each artist.

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2. Creating View Solution

2.1. What is view for?2.2. Create view

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2.1. What is view for? A SQL view is a virtual table that us constructed

from other tables or views. A view has no data of its own, but obtain data from tables or other views. View are constructed from SQL SELECT statements. Using views to hide columns and rows. Using views to display results of computed columns. Using views to hide complicated SQL syntax.

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2.2. Create view

2.2.1. Create View to Show Name and Nationality2.2.2. Create View to Show Artist Name and the

number of Artist Work

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2.2.1. Create View to Show Name and Nationality

Create View ArtistInfo AsSelect ArtName, ArtNationalityFrom Artists;

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2.2.2. Create View to Show Artist Name and the number of Artist Work

Create View ArtistsWorks AsSelect a.ArtName, Count(w.Wor_ID) as

numWorkFrom Artists a Left Join Works w on a.Art_ID

= w.Art_IDGroup by a.ArtName;

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