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ГОС по БД #1

Beschreibung

God Philosophy Quiz am ГОС по БД #1, erstellt von хомяк убийца am 23/03/2019.
хомяк убийца
Quiz von хомяк убийца, aktualisiert more than 1 year ago
хомяк убийца
Erstellt von хомяк убийца vor etwa 5 Jahre
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Zusammenfassung der Ressource

Frage 1

Frage
Define a degree of a relation
Antworten
  • How many rows a table has
  • how long each tuple is, or how many columns the table has
  • how many different tuples there are
  • how many different datatypes table has

Frage 2

Frage
Define a cardinality of a relation
Antworten
  • how long each tuple is
  • how many columns the table has
  • how many different tuples there are, how many rows a table has
  • how many different datatypes table has

Frage 3

Frage
Which of the following refers to union-compatibility requirements ?
Antworten
  • Same number of columns
  • Same number of rows
  • Same number of tuples
  • Different domains

Frage 4

Frage
Which of the following refers to union-compatibility requirements ?
Antworten
  • Corresponding columns have the same domains
  • Same number of rows
  • Same number of tuples
  • Corresponding columns have the same fields

Frage 5

Frage
Define a domain
Antworten
  • restrict the possible values a tuple can assign to each attribute
  • relations to each other
  • uniquely identifies each tuple that appears in a relation
  • minimality of the attribute

Frage 6

Frage
Define a foreign key
Antworten
  • restrict the possible values a tuple can assign to each attribute
  • relations to each other
  • uniquely identifies each tuple that appears in a relation
  • minimality of the attribute

Frage 7

Frage
Define a primary key
Antworten
  • restrict the possible values a tuple can assign to each attribute
  • relations to each other
  • uniquely identifies each tuple that appears in a relation
  • datatypes of attributes

Frage 8

Frage
Define restrict
Antworten
  • stop the user from doing it
  • let the changes flow on
  • make referencing values the default for their column
  • make referencing values null

Frage 9

Frage
Define cascade
Antworten
  • stop the user from doing it
  • let the changes flow on
  • make referencing values the default for their column
  • make referencing values null

Frage 10

Frage
Define set default
Antworten
  • stop the user from doing it
  • let the changes flow on
  • make referencing values the default for their column
  • make referencing values null

Frage 11

Frage
Define set null
Antworten
  • stop the user from doing it
  • let the changes flow on
  • make referencing values the default for their column
  • make referencing values null

Frage 12

Frage
Define Data Definition Language
Antworten
  • Specify database format
  • Specify access controls (privileges)
  • Specify and retrieve database contents
  • Specify table attribute uniqueness

Frage 13

Frage
Define Data Control Language
Antworten
  • Specify access controls (privileges)
  • Specify database format
  • Specify and retrieve database contents
  • Specify table attribute uniqueness

Frage 14

Frage
Define Data Manipulation Language
Antworten
  • Specify table attribute uniqueness
  • Specify database format
  • Specify access controls (privileges)
  • Specify and retrieve database contents

Frage 15

Frage
Which of the following does not refer to DBMS tools
Antworten
  • Oracle
  • PostgreSQL
  • MySQL
  • Python

Frage 16

Frage
Which of the following is used to provide privilege to only a particular attribute?
Antworten
  • Grant select on employee to finance
  • Grant update(salary, rate) on employee to finance
  • Grant update(salary) on employee to finance
  • Grant delete to finance

Frage 17

Frage
Which of the following statement is used to remove the privilege from the user finance?
Antworten
  • Remove update on employee from finance
  • Delete select on employee from finance
  • Revoke update on employee from finance
  • Grant update on employee from finance

Frage 18

Frage
Which of the following is true regarding views?
Antworten
  • The user who creates a view cannot be given update authorization on a view without having update authorization on the relations used to define the view.
  • If a user creates a view on which no authorization can be granted, the system will allow the view creation request.
  • A user who creates a view receives all privileges on that view.

Frage 19

Frage
If we wish to grant a privilege and to allow the recipient to pass the privilege on to other users, we append the __________ clause to the appropriate grant command.
Antworten
  • With grant
  • Grant user
  • With grant option
  • Grant pass privelege

Frage 20

Frage
Which of the following is used to avoid cascading of authorizations from the user?
Antworten
  • Granted by current role
  • Revoke grant option for select on department from finance;
  • Revoke select on employee from finance, cashier restrict;
  • Revoke select on department from finance, cashier cascade;

Frage 21

Frage
Privileges are granted over some specified parts of a database, such as a
Antworten
  • Schema
  • Environment
  • Relation Or view
  • Query statement

Frage 22

Frage
Prevention of access to the database by unauthorized users is referred to as:
Antworten
  • Integrity
  • Productivity
  • Security
  • Reliability

Frage 23

Frage
Database Authentication refers to:
Antworten
  • methods of restricting user access to system
  • controlling access to portions of database
  • all of the answers mentioned
  • controlling the operation on the data

Frage 24

Frage
A set of possible data values is called
Antworten
  • attribute
  • degree
  • domain
  • tuple

Frage 25

Frage
A functional dependency between two or more non-key attributes is called
Antworten
  • Partial transitive dependency
  • Functional dependency
  • Transitive dependency
  • Partial functional dependency

Frage 26

Frage
__ refers to the correctness and completeness of the data in a database.
Antworten
  • Database security
  • Data constraint
  • Data integrity
  • Data independence

Frage 27

Frage
Which of the following creates a virtual relation for storing the query?
Antworten
  • Function
  • Procedure
  • View
  • None of the mentioned

Frage 28

Frage
Which of the following is the syntax for views where v is view name?
Antworten
  • Create view v as “query name”;
  • Create “query expression” as view;
  • Create view v as “query expression”;
  • Create view “query expression”;

Frage 29

Frage
Updating the value of the view
Antworten
  • Will not change the view definition
  • Will not affect the relation from which it is defined
  • Will affect the relation from which it is defined
  • Cannot determine

Frage 30

Frage
Create view faculty as: Select ID, name, dept name from instructor; Find the error in this query.
Antworten
  • Instructor
  • Select
  • None of the mentioned
  • View …as

Frage 31

Frage
Which of the following is a basic form of grant statement?
Antworten
  • Grant ‘privilege list’ on ‘user/role list’ to ‘relation name or view name’;
  • Grant ‘privilege list’ to ‘user/role list’;
  • Grant ‘privilege list’ on ‘relation name or view name’ to ‘user/role list’;
  • Grant ‘privilege list’ on ‘relation name or view name’ on ‘user/role list’;

Frage 32

Frage
Retrieve all data from the table OFFICE { id,room, name}
Antworten
  • Select *from office;
  • Select from office;
  • Select name from office;
  • Select *form office;

Frage 33

Frage
Retrieve office name from the table OFFICE {id, room, name}
Antworten
  • Select *from office;
  • Select from office;
  • Select name from office;
  • Select *form office;

Frage 34

Frage
Retrieve office id from the table OFFICE {id, room, name}
Antworten
  • Select *from office;
  • Select id from office;
  • Select name from office;
  • Select id form office;

Frage 35

Frage
Retrieve office id and room from the table OFFICE {id, room, name}
Antworten
  • Select *from office;
  • Select id, room from office;
  • Select room name from office;
  • Select id form office;

Frage 36

Frage
Retrieve quantity of offices in the office table - OFFICE {id, rom, name}
Antworten
  • Select avg (id) from office;
  • Select count (id) from office;
  • Select sum (id) from office;
  • Select max (id) from office;

Frage 37

Frage
Retrieve total number scholarship in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select avg (scholarship) from office;
  • Select count (scholarship) from office;
  • Select sum (scholarship) from office;
  • Select max (scholarship) from office;

Frage 38

Frage
Retrieve quantity of students in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select avg (id) from office;
  • Select count (id) from office;
  • Select sum (id) from office;
  • Select max (id) from office;

Frage 39

Frage
Sort students name by descending order - STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select name from students order by desc;
  • Select name from students group by desc;
  • Select *from students order by desc;
  • Select name form students order by desc;

Frage 40

Frage
Find an average scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select avg (scholarship) from students;
  • Select count (scholarship) from students;
  • Select sum (scholarship) from students;
  • Select max (scholarship) from students;

Frage 41

Frage
Find maximum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select avg (scholarship) from students;
  • Select count (scholarship) from students;
  • Select sum (scholarship) from students;
  • Select max (scholarship) from students;

Frage 42

Frage
Find minimum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Antworten
  • Select avg (scholarship) from students;
  • Select count (scholarship) from students;
  • Select min (scholarship) from students;
  • Select max (scholarship) from students;

Frage 43

Frage
Retrieve offices’ name and tutors’, who work in CSSE department. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, OFFICEID (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘CSSE’
  • Select o.name, t.name from office o join tutor t on o.id=t.id where o.name = ‘CSSE’
  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘csse’
  • Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name ‘CSSE’

Frage 44

Frage
Retrieve all offices’ name and tutors’, who work in departments. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • Select o.name, t.name from office o left join tutor t on o.id=t.officeid;
  • Select o.name, t.name from office o left join tutor t on o.id=t.id ;
  • Select o.name, t.name from office o right join tutor t on o.id=t.officeid;
  • Select o.name, t.name from office o right join tutor t on o.id=t.id ;

Frage 45

Frage
Retrieve students’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select s.name from students s where s.scholarship > (select avg (scholarship)from students);
  • select s.name from students s where s.scholarship > (select scholarship from students);
  • select s.name from students s where s.scholarship > avg (scholarship);
  • select s.name from students s where s.scholarship < avg (scholarship);

Frage 46

Frage
Retrieve students’ name, who registered in 10th of Jan 2016. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }.
Antworten
  • select name from students where registereddate = '2016-01-10';
  • select name from students where registereddate = '2016-10-10';
  • select name from students where registereddate = 2016-10-10;
  • select name from students where registereddate = '2016-01-01';

Frage 47

Frage
Retrieve tutors’ name, who have more than experience others along with their students name . Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select max (experience) from tutor);
  • select t.name, s.name, max(experience) from tutor t join students s on t.id=s.tutorid;
  • select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select max (experience) from tutor);

Frage 48

Frage
Retrieve tutors’ name, who have less experience than others along with their students name. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select min (experience) from tutor);
  • select t.name, s.name, min(experience) from tutor t join students s on t.id=s.tutorid;
  • select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select min (experience) from tutor);

Frage 49

Frage
Increase tutors experience and students’ scholarship twice. Retrieve experience and scholarship along with their names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.tutorid;
  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.id;
  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.tutorid;
  • select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.id;

Frage 50

Frage
Retrieve office, tutor and students name, but tutors name should have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';
  • select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name like '%y%';
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name like '%y%';

Frage 51

Frage
Retrieve office, tutor and students name, but tutors name should not have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutored (FK references tutor (id))}.
Antworten
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';
  • select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name not like '%y%';
  • select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name not like '%y%';

Frage 52

Frage
Retrieve students name and scholarship along with their department name, who have scholarship between 4000 and 5000. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;
  • select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship 4000 and 5000;
  • select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;
  • select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship >=4000

Frage 53

Frage
Retrieve information about office and their tutors as well. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.d
Antworten
  • select *from office o join tutor t on o.id=t.officeid;
  • select *from office o join tutor t in o.id=t.officeid;
  • select *form office o join tutor t on o.id=t.officeid;
  • select *from office o join tutor t on o.id=t.id;

Frage 54

Frage
Retrieve office name and tutor name, which have maximum experienced tutors. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select max (experience) from tutor);
  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select min (experience) from tutor);
  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select high (experience) from tutor);
  • select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select low (experience) from tutor);

Frage 55

Frage
Retrieve all tutors and students names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select t.name, s.name from tutor t full join students s on t.id=s.tutorid;
  • select t.name, s.name from tutor t full join students s t.id=s.tutorid;
  • select t.name, s.name from tutor t full join students s t.id=s.id;
  • select t.name, s.name from tutor t full join students s on t.id=s.id;

Frage 56

Frage
Retrieve tutors’ name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutoridgroup by t.name;
  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutorid;
  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.idgroup by t.name;
  • select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.id;

Frage 57

Frage
Retrieve office name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Antworten
  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o on o.id=t.officeid group by o.name;
  • select count (s.name) as numberofstudents, o.name from students s on join office o on o.id=s.id group by o.name;
  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o o.id=t.officeid group by o.name;
  • select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.idjoin office o on o.id=t.id group by o.name;

Frage 58

Frage
Retrieve all students’ name and scholarship, tutors name except students who have scholarship 3000$. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship <> 3000;
  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship = 3000;
  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship < 3000;
  • select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship is 3000;

Frage 59

Frage
How many types of anomalies exist?
Antworten
  • 1
  • 2
  • 3
  • 4

Frage 60

Frage
Which of the following anomaly does not exist?
Antworten
  • Creation anomaly
  • Deletion anomaly
  • Insertion anomaly
  • Modification anomaly

Frage 61

Frage
When an insertion anomaly occurs?
Antworten
  • we are prevented from inserting some data into a relation until other data can be supplied
  • deletion leads to an unintended loss of data
  • it is possible that not all data needs to be changed will always be changed
  • it does not occur

Frage 62

Frage
When a modification anomaly occurs?
Antworten
  • it is possible that not all data needs to be changed will always be changed
  • we are prevented from inserting some data into a relation until other data can be supplied
  • deletion leads to an unintended loss of data
  • it does not occur

Frage 63

Frage
When a deletion anomaly occurs?
Antworten
  • deletion leads to an unintended loss of data
  • it is possible that not all data needs to be changed will always be changed
  • we are prevented from inserting some data into a relation until other data can be supplied
  • it does not occur

Frage 64

Frage
Define inner join
Antworten
  • selects all rows from both tables as long as there is a match between the columns in both tables
  • returns all rows from the left table (1), with the matching rows in the right table (2)
  • returns all rows from the right table (2), with the matching rows in the right table (1)
  • returns all rows from the left table (1) and from the right table (2)

Frage 65

Frage
Define left join
Antworten
  • returns all rows from the left table (1), with the matching rows in the right table (2)
  • selects all rows from both tables as long as there is a match between the columns in both tables
  • returns all rows from the right table (2), with the matching rows in the right table (1)
  • returns all rows from the left table (1) and from the right table (2)

Frage 66

Frage
Define right join
Antworten
  • returns all rows from the right table (2), with the matching rows in the left table (1)
  • returns all rows from the left table (1), with the matching rows in the right table (2)
  • selects all rows from both tables as long as there is a match between the columns in both tables
  • returns all rows from the left table (1) and from the right table (2)

Frage 67

Frage
Define full join
Antworten
  • returns all rows from the left table (1) and from the right table (2)
  • returns all rows from the right table (2), with the matching rows in the right table (1)
  • returns all rows from the left table (1), with the matching rows in the right table (2)
  • selects all rows from both tables as long as there is a match between the columns in both tables

Frage 68

Frage
Define an union
Antworten
  • сombines the result set of two or more select statements
  • returns all rows from the left table (1) and from the right table (2)
  • returns all rows from the right table (2), with the matching rows in the right table (1)
  • returns all rows from the left table (1), with the matching rows in the right table (2)

Frage 69

Frage
INNER JOIN and JOIN are the same
Antworten
  • True
  • False

Frage 70

Frage
Define self join
Antworten
  • returns all rows from the table that references to yourself
  • returns all rows from the right table (2), with the matching rows in the right table (1)
  • selects all rows from both tables as long as there is a match between the columns in both tables
  • returns all rows from the left table (1) and from the right table (2)

Frage 71

Frage
How many types of functional dependencies exist
Antworten
  • 1
  • 2
  • 3
  • 0

Frage 72

Frage
Which of the following does not to the functional dependency
Antworten
  • equational
  • full
  • transitive
  • partial

Frage 73

Frage
Which of the following represent a full dependency?
Antworten
  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y
  • all above mentioned

Frage 74

Frage
Which of the following represent a partial dependency?
Antworten
  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y
  • all above mentioned

Frage 75

Frage
Which of the following represent a transitive dependency?
Antworten
  • x->y dependency in relation R and x, y , z are columns in R. X->Y and Y->Z in R. Final: X->Y
  • if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
  • it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
  • all above mentioned

Frage 76

Frage
When was first normalization developed?
Antworten
  • 1970
  • 1972
  • 1973
  • 1974

Frage 77

Frage
When was second normalization developed?
Antworten
  • 1971
  • 1972
  • 1973
  • 1974

Frage 78

Frage
When was Boyce–Codd normalization developed?
Antworten
  • 1974
  • 1972
  • 1973
  • 1971

Frage 79

Frage
Who is an inventor of relational model?
Antworten
  • Edgar F.Codd
  • Raymond Boyce
  • Marine Jone
  • John Salamondor

Frage 80

Frage
Which of the following refers to the requirement of 1NF
Antworten
  • Each cell should be single valued
  • Entries in a column are same type
  • Rows uniquely identified
  • All above mentioned

Frage 81

Frage
Which of the following refers to the requirement of 2NF
Antworten
  • All attributes (non-key columns) dependent on the key
  • Each cell should be single valued
  • All fields (columns) can be determined only by the key in the table and no other column
  • All above mentioned

Frage 82

Frage
Which of the following refers to the requirement of 3NF
Antworten
  • All fields (columns) can be determined only by the key in the table and no other column
  • All attributes (non-key columns) dependent on the key
  • Each cell should be single valued
  • All above mentioned

Frage 83

Frage
Define avg function
Antworten
  • Returns average value
  • Returns total value
  • Returns the first value
  • Converts to lowercase

Frage 84

Frage
Which function is used to retrieve quantity of rows
Antworten
  • count
  • sum
  • max
  • avg

Frage 85

Frage
Retrieve avg scholarship of students. Tables: office {id (PK), locations, name}, tutor {id (PK), name, officeid (FK references office (id)), experience}, students {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select avg (scholarship) from students;
  • select aveg (scholarship) from students;
  • select avr (scholarship) from students;
  • select avgr (scholarship) from students;

Frage 86

Frage
Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)
  • select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)
  • select s.name, s.scholarship, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)
  • select s.name, avg(s.scholarship), t.name from students s, tutor t

Frage 87

Frage
Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship AND increase those students twice. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)
  • select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)
  • select s.name, s.scholarship*2, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)
  • select s.name, avg(s.scholarship)*2, t.name from students s, tutor t

Frage 88

Frage
Retrieve students and mentors name, but mentors registration date should be before students registration date. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate<m.registereddate;
  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate>m.registereddate;
  • select s.name, m.name from students s join students m on s.tutorid=m.tutorid;
  • select s.name, m.name from students s join students m on s.registereddate=m.registereddate;

Frage 89

Frage
Retrieve the highest experience from tutor. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Antworten
  • select max (experience) from tutor;
  • select min (experience) from tutor;
  • select max (t.experience) from tutor t group by t.name;
  • select min (t.experience) from tutor t group by t.name;
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