130 lines
3.2 KiB
Markdown
130 lines
3.2 KiB
Markdown
## Lecture 1 (13:00) - Conceptual Modelling
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### Role of Conceptual Modelling
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- Real world problem -> Conceptual Model Data -> Logical Data Model -> Physical Model (DBMS)
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- Identify important concepts and data needs, creating a conceptual model.
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- Many different ways to get to the same solution.
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- Iterative process - if one method does not work, it can be revised.
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### Conceptual Data Model
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- Abstract view of situation.
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- Identifies important elements and relationships between them. (ex. Library: Books, Members, ~~Carpets~~)
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- Human interact-able, not computer terms.
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- Implementation Independent.
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- Useful for client brief.
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### Reasons for Using ER Modelling
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- Simpler and easier to understand than database tables
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- Helps discussions with clients and colleagues
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- Allows for segmentation of work.
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- Modelling real world solution
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- Designing DB tables
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- Most large organisations will require it
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### Obtaining an ER Model
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- Given a spec, you need to identify:
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- entities - 'things' with physical or conceptual existence
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- relationships
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- attributes
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- constraints or assumptions
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#### Example
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Departments control many projects and each department has many employees. Each employee works on only one project at a time. A project’s start date must be before the project’s target completion date. Each employee has an NI number, name and address
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Entities:
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- departments
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- projects
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- employees
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Relationships:
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- **control** between departments and projects
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- **has between** departments and employees
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- **works on** between employees and projects
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Attributes:
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- startdate
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- completiondate
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- ninumber
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- name
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- address
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Constraints:
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- Start date must be before the target completion date
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### ER Diagram
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- Many different notations
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- One is "crows foot":
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## Lecture 3 (16:00)
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1. Generate a list of the ages of ATMs whose security level is normal. The list should be in descending order.
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```sql
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SELECT age
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FROM atm
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WHERE security-level = "normal"
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ORDER BY age DESC;
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```
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1. List the name of each bank which has a branch whose address starts with the word Eccles. The list should not include any duplicates.
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```sql
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SELECT DISTINCT name
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FROM bank, branch
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WHERE Bank.bankID = Branch.bankID
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AND address LIKE "Eccles*";
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```
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1. Calculate the average age of the ATMs with low or normal levels of security. Name the results column “Average age”.
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```sql
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SELECT AVG(age) AS "Average Age"
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FROM atm
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WHERE security-level = "low" OR security-level = "normal";
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```
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1. Remove the tuple from the relation ATM whose identifier is atm04.
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```sql
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DELETE FROM atm
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WHERE atmID = "atm04";
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```
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1. Add a row for an ATM, whose atmID is atm05, whose security level is high and which is located at branch T1.
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```sql
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INSERT INTO atm
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VALUES ("atm05", "high", "0", "T1");
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```
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1. Add an attribute called telephone number to the relation Branch.
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```sql
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ALTER TABLE branch
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ADD telephone CHAR(15)
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```
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1. Set the address of every branch to the value Salford.
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```sql
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UPDATE branch
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SET address = "Salford";
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```
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1. Bank LEFT OUTER JOIN Branch
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```sql
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SELECT *
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FROM bank
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INNER JOIN
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```
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