Database midterm

Question 1

ER diagram looks like the following:

Accepted Answer

this one

Answer

not this one

Question 2

Bubble diagram looks like the following:

Accepted Answer

this one

Answer

not this one

Question 3

Entity: An entity is any object in the system that we want to model and store information about.

Answer

true

Question 4

Entity class is essentially a table of data within the database.

Answer

true

Question 5

Metadata: data that describes the properties and context of user data

Answer

true

Question 6

Attributes: Properties or characteristics of an entity or relationship type

Answer

true

Question 7

Types of attributes:

[blank_start]Required[blank_end] Attribute: A required attribute is an attribute that must have a data value.

[blank_start]optional[blank_end]: An ... attribute may not have a value in it and can be left blank.

[blank_start]Simple[blank_end]: The attributes which cannot be divided into smaller subparts are called ... (age of employee)

[blank_start]composite[blank_end]: Attribute can be subdivided into two or more other Attribute. (Name can be divided into First name, Middle name and Last name)

[blank_start]single[blank_end] value: An attribute, that has one value for a particular entity. (age of a employee)

[blank_start]multi[blank_end] value: An attributes that may have many values for the same entity. (colors of a car)

[blank_start]stored[blank_end]: An attribute, which cannot be derived from other attributes (BirthDate of employee)

[blank_start]derived[blank_end]: Attributes ... from other stored attributes. (Date of Birth and Today’s date)

Accepted Answer

Required

Accepted Answer

optional

Accepted Answer

Simple

Accepted Answer

composite

Accepted Answer

single

Accepted Answer

multi

Accepted Answer

stored

Accepted Answer

derived

Question 8

Types of identifiers (keys) are:

[blank_start]Key[blank_end]: An attribute or combination of attributes that uniquely identifies individual instances of an entity type.

[blank_start]Primary[blank_end] key:  A ... key is a one or more fields that uniquely identifies a row in a table

[blank_start]Foreign[blank_end] key:  A ... key is a relationship between columns in two database tables

[blank_start]Composite[blank_end] key: A primary key composed of one or more columns (table STAFF contains the fields FNAME and LNAME for first and last names)

Accepted Answer

Key

Accepted Answer

Primary

Accepted Answer

Foreign

Accepted Answer

Composite

Question 9

Database relationship: A relationship is established between two database tables when one table has a foreign key that references the primary key of another table

Answer

true

Question 10

Fill in the relationship type

Accepted Answer

one to one

Accepted Answer

One to many

Accepted Answer

many to many

Question 11

Label the relationships:

Accepted Answer

One

Accepted Answer

Many

Accepted Answer

Mandatory One

Accepted Answer

Zero to one

Accepted Answer

One to many

Accepted Answer

Zero to many

Question 12

Degree of relationships
[blank_start]Unary[blank_end] - One entity related to another of the same entity type
[blank_start]Binary[blank_end] - Entities of two different types related to each other
[blank_start]Ternary[blank_end] - Entities of three different types related to each other

Accepted Answer

Unary

Accepted Answer

Binary

Accepted Answer

Ternary

Question 13

Example of an ER diagram

Accepted Answer

this one

Answer

not this one

Question 14

Normalize the following table (type next in blank for correct):
Do this on separate page or application and compare with the following questions.

Accepted Answer

Question 15

0NF (list all the fields, data in [ ] brackets are multi value type)

Order(Order#, OrderDate, Customer#, CustomerName, CustAddr, Custcity, CustProv, CustPostal, [Product#, ProdDesc, QtyOrdered, UnitPrice])

Answer

true

Question 16

1NF: Repeating groups are removed into a new table, new table will have a concatenated name and concatenated key
(separate the [ ] data into another table, link it to main table with a primary key, first value of the [ ] data becomes a primary key in new table)

Order(Order# (PK), OrderDate, Customer#, CustomerName, CustAddr, Custcity, CustProv, CustPostal)

OrderProduct(Order# (PK), Product# (PK), ProdDesc, QtyOrdered, UnitPrice)

Answer

true

Question 17

2NF: Partial dependencies are removed, only table from [ ] one is inspected
(Further break down the created table from 1NF into separate tables with data the can be uniquely identified by a key, IF YOU CAN, this is not always possible)

Order(Order# (PK), OrderDate, Customer#, CustomerName, CustAddr, Custcity, CustProv, CustPostal)

OrderProduct(Order# (PK), Product# (PK), QtyOrdered, UnitPrice)

Product(Product# (PK), ProdDesc)

Answer

true

Question 18

3NF: Transitive dependencies are removed,  
(Typically you separate customer data in this step)

Order(Order# (PK), OrderDate, Customer# (FK))

Customer(Customer# (PK), CustomerName, CustAddr, Custcity, CustProv, CustPostal)

OrderProduct(Order# (PK), Product# (PK), QtyOrdered, UnitPrice)

Product(Product# (PK), ProdDesc)

Answer

true

Question 19

Normalize the following table (type next in blank for correct):
Do this on separate page or application and compare with the following questions.

Accepted Answer

Question 20

0NF

DRevRep(Date, [Pat#, Loc, PatName, [FinSource, [CC, [ItemCode, Desc, Charge]]], Total])

Answer

true

Question 21

1NF

DrevRep(Date(PK))

DRevRepPat(Date(PK), Pat#(PK), Loc, PatName, Total)

DRevRepPatFin(Date(PK), Pat#(PK), FinSource(PK))

DRevRepPatFinCC(Date(PK), Pat#(PK), FinSource(PK), CC(PK))

DRevRepPatFinCCItem(Date(PK), Pat#(PK), FinSource(PK), CC(PK), ItemCode(PK), Desc, Charge)

Answer

true

Question 22

2NF

DrevRep(Date(PK))

DRevRepPat(Date(PK), Pat#(PK), Loc, Total)

DRevRepPatFin(Date(PK), Pat#(PK), FinSource(PK))

DRevRepPatFinCC(Date(PK), Pat#(PK), FinSource(PK), CC(PK))

DRevRepPatFinCCItem(Date(PK), Pat#(PK), FinSource(PK), CC(PK), ItemCode(PK))

Patient(Pat#(PK), PatName)

Item(ItemCode(PK), Desc, Charge)

Answer

true

Question 23

3NF (no change)

DrevRep(Date(PK))

DRevRepPat(Date(PK), Pat#(PK), Loc, Total)

DRevRepPatFin(Date(PK), Pat#(PK), FinSource(PK))

DRevRepPatFinCC(Date(PK), Pat#(PK), FinSource(PK), CC(PK))

DRevRepPatFinCCItem(Date(PK), Pat#(PK), FinSource(PK), CC(PK), ItemCode(PK))

Patient(Pat#(PK), PatName)

Item(ItemCode(PK), Desc, Charge)

Answer

true

Question 24

write SQL command to get data from s9.tbl_customer:

[blank_start]SELECT * FROM s9.tbl_customer;[blank_end]

Accepted Answer

SELECT * FROM s9.tbl_customer;

Question 25

write SQL command to get custnum and custname column data from s9.tbl_customer:

SELECT [blank_start]custnum, custname FROM s9.tbl_customer;[blank_end]

Accepted Answer

custnum, custname FROM s9.tbl_customer;

Question 26

write SQL command to get custnum from s9.tbl_customer and display column name as cust_no

SELECT [blank_start]custnum AS cust_no[blank_end] FROM s9.tbl_customer;

Accepted Answer

custnum AS cust_no

Question 27

write SQL command to get a total value price of saleprice and quantity from s9.tbl_orderline and display as totalprice

SELECT [blank_start]saleprice * quantity AS totalprice[blank_end] FROM s9.tbl_orderline;

(tip: for more complex SELECTS where you have to calculate multiple fields, you can separate by brackets)

Accepted Answer

saleprice * quantity AS totalprice

Question 28

write SQL command to get the difference in salary from max_salary and min_salary from hr.jobs and display as difference

SELECT [blank_start]max_salary - min_salary AS difference[blank_end] FROM hr.jobs;

Accepted Answer

max_salary - min_salary AS difference

Question 29

When formatting strings, you can use:
CHR(10) for new line
CHR(9) for tab space

Answer

true

Question 30

write SQL command to get first_name and last_name seperated by a tab space in hr.employees

SELECT [blank_start]first_name || CHR(9) || last_name[blank_end] FROM hr.employees;

Accepted Answer

first_name || CHR(9) || last_name

Question 31

TO_CHAR converts a number, date, or expression to TEXT in a specified format

Answer

true

Question 32

SYSDATE returns todays date and time

Answer

true

Question 33

ROUND will round a number to closest whole and can specific number of decimal places

SELECT ROUND(123.456, 2) FROM table;

Answer

true

Question 34

write SQL command to get the day, month and year from orderdate and display as separate columns from s9.tbl_orders, display the column names as d, m, y 
(hint: use EXTRACT)

SELECT [blank_start]EXTRACT (day FROM orderdate) AS d[blank_end],
[blank_start]EXTRACT (month FROM orderdate) AS m[blank_end],
[blank_start]EXTRACT (year FROM orderdate) AS y[blank_end],
FROM s9.tbl_orders;

Accepted Answer

EXTRACT (day FROM orderdate) AS d

Accepted Answer

EXTRACT (month FROM orderdate) AS m

Accepted Answer

EXTRACT (year FROM orderdate) AS y

Question 35

write SQL command to get ordernum, orderdate, and days passed from current date and orderdate displayed as numofdays from s9.tbl_orders;

SELECT ordernum, orderdate, [blank_start]ROUND(SYSDATE - orderdate) AS numofdays[blank_end] FROM s9.tbl_orders;

Accepted Answer

ROUND(SYSDATE - orderdate) AS numofdays

Question 36

write SQL command to get first_name, last_name and years the employee has been working displayed as "Years" from hr.employees
(hint: use hire_date)

SELECT first_name, last_name, [blank_start]ROUND((SYSDATE - hire_date)/365)[blank_end] AS "Years" FROM hr.employees;

Accepted Answer

ROUND((SYSDATE - hire_date)/365)

Question 37

write SQL command to select all employees with department_id of 100 and salary greater than or equal to 7000 from hr.employees

SELECT * FROM hr.employees WHERE [blank_start]department_id = 100 AND salary >= 7000;[blank_end]

Accepted Answer

department_id = 100 AND salary >= 7000;

Question 38

write SQL command to get all customers who live in Toronto from s9.tbl_customer

SELECT * FROM s9.tbl_customer [blank_start]WHERE UPPER(city) = 'TORONTO';[blank_end]

(since Oracle data is case-sensitive, you must use UPPER)

Accepted Answer

WHERE UPPER(city) = 'TORONTO';

Question 39

Accepted Answer

=

Accepted Answer

>

Accepted Answer

>=

Accepted Answer

<

Accepted Answer

<=

Accepted Answer

<>

Question 40

Other types of comparisons are:
BETWEEN .. AND ..
IN(set)
LIKE
IS NULL

Answer

true

Question 41

write SQL command to get all employees with salary between 2000 and 4000 from hr.employees

SELECT * FROM hr.employees [blank_start]WHERE salary BETWEEN 2000 AND 4000;[blank_end]

Accepted Answer

WHERE salary BETWEEN 2000 AND 4000;

Question 42

write SQL command to get all customers with creditlimit greater than 0 and live in any city starting with Tor from s9.tbl_customer

SELECT * FROM s9.tbl_customer WHERE [blank_start]creditlimit > 0 AND city LIKE 'Tor%';[blank_end]

Accepted Answer

creditlimit > 0 AND city LIKE 'Tor%';

Question 43

write SQL command to get all unique salary fields from hr.employee

SELECT [blank_start]DISTINCT salary[blank_end] FROM hr.employees;

Accepted Answer

DISTINCT salary

Question 44

write SQL command to get all employees with department_id of 100 and salary is greater or equal to 7000, order by first_name from hr.employees

SELECT * FROM hr.employees
WHERE [blank_start]department_id = 100 AND salary >= 7000[blank_end]
[blank_start]ORDER BY first_name;[blank_end]

Accepted Answer

department_id = 100 AND salary >= 7000

Accepted Answer

ORDER BY first_name;

Question 45

The aggregate functions of SQL are:
AVG() - calculates the average)
SUM() - calculates the total
MIN() - gets the lowest
MAX() - gets the highest
COUNT() - gets the number of times a value appears

Answer

true

Question 46

write SQL command to get the employee with the highest salary from hr.employees

SELECT [blank_start]MAX(salary)[blank_end] FROM hr.employees;

Accepted Answer

MAX(salary)

Question 47

write SQL command to get the amount of customers displayed as numofcust from s9.tbl_customer

SELECT [blank_start]COUNT(*) AS numofcust[blank_end] FROM s9.tbl_customer;

Accepted Answer

COUNT(*) AS numofcust

Question 48

write SQL command to get the average salary of all employees with department_id of 100, round the result to 2 decimal from hr.employees

SELECT [blank_start]ROUND(AVG(salary), 2)[blank_end]
FROM hr.employees
[blank_start]WHERE department_id = 100;[blank_end]

Accepted Answer

ROUND(AVG(salary), 2)

Accepted Answer

WHERE department_id = 100;

Question 49

In SQL you can subqueries SELECT statements:
WHERE field = (SELECT ... return single value)
WHERE field NOT IN (SELECT ... use DISTINCT if needed)
WHERE field IN (SELECT ... use DISTINCT if needed)

Answer

true

Question 50

write SQL command to get which employee_id left the company on 2007-12-31 from hr.employees and hr.job_history

SELECT * FROM hr.employees
[blank_start]WHERE employee_id IN[blank_end]
(SELECT [blank_start]employee_id FROM hr.job_history[blank_end] WHERE [blank_start]end_date = '2007-12-31'[blank_end]);

Accepted Answer

WHERE employee_id IN

Accepted Answer

employee_id FROM hr.job_history

Accepted Answer

end_date = '2007-12-31'

Question 51

write SQL command to get which unique productnum has never been ordered from s9.tbl_orderline

SELECT * FROM s9.tbl_product
WHERE [blank_start]productnum NOT IN[blank_end]
(SELECT [blank_start]DISTINCT productnum[blank_end] FROM s9.tbl_orderline);

Accepted Answer

productnum NOT IN

Accepted Answer

DISTINCT productnum

Question 52

write SQL command to get which product has the most expensive msrp price from s9.tbl_product

SELECT * FROM s9.tbl_product 
[blank_start]WHERE msrp =[blank_end] (SELECT [blank_start]MAX(msrp)[blank_end] FROM s9.tbl_product);

Accepted Answer

WHERE msrp =

Accepted Answer

MAX(msrp)

Database midterm

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