Kopieren und bearbeiten

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 mehr als 5 Jahre
1589
20
0

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;
Zusammenfassung anzeigen Zusammenfassung ausblenden

Möchten Sie mit GoConqr kostenlos Ihre eigenen Quiz erstellen? eigenen Mehr erfahren.

ähnlicher Inhalt

Breakdown of Philosophy
rlshindmarsh
Who did what now?...Ancient Greek edition
Chris Clark
Reason and Experience Plans
rlshindmarsh
The Cosmological Argument
Summer Pearce
AS Philosophy Exam Questions
Summer Pearce
Philosophy of Art
mccurryby
"The knower's perspective is essential in the pursuit of knowledge." To what extent do you agree?
nataliaapedraza
The Ontological Argument
daniella0128
Religious Experience
alexandramchugh9
Chapter 6: Freedom vs. Determinism Practice Quiz
Kristen Gardner
Environmental Ethics
Jason Edwards-Suarez
Bibliothek durchsuchen