17415233
Description
Quiz by хомяк убийца, updated more than 1 year ago
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Created by хомяк убийца
over 5 years ago
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Question 1
Question
Retrieve the lowest 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))
Answer
-
select min (experience) from tutor;
-
select max (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;
Question 2
Question
Retrieve book name, number of books, price and publisher, for those books which have more than 20 copies. Tables : BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}
Answer
-
select count (b.bookid),b.title, b.price, b.publisher_name from book b join book_copies bc on b.bookid=bc.bookid where bc.no_of_copies>20 group by b.title, b.price, b.publisher_name;
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select count (b.bookid),b.title, b.price, b.publisher_name from book b join book_copies bc on b.bookid=bc.bookid where bc.no_of_copies>20;
-
select count (b.bookid),b.title, b.price, b.publisher_name from book b join book bc on b.bookid=bc.bookid where bc.no_of_copies>20 group by b.title, b.price, b.publisher_name;
-
select b.bookid,b.title, b.price, b.publisher_name from book b join book_copies bc on b.bookid=bc.bookid where bc.no_of_copies>20 group by b.title, b.price, b.publisher_name;
Question 3
Question
Retrieve card number, book name, borrower name, who have borrowed in 12 July, bookid (references Book (bookid)), libraryid (references Library (libraryid)), cardno 2013. Tables: BOOK {bookid, title, publisher_name, price}. BOOKLOANS {bookloans, cardno(references Borrower (cardno)),bookid (references Book (bookid)) dateout, duedate}. BORROWER {cardno, c_name, b_address, phoneno}
Answer
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select bl.cardno, b.title, br.c_name from book b join bookloans bl on bl.bookid=b.bookid join borrower br on br.cardno=bl.cardno where bl.dateout='12-07-2013'
-
select bl.cardno, b.title, br.c_name from book b join bookloans bl on bl.bookid=b.bookid join borrower br on br.cardno=bl.cardno where bl.duedate='12-07-2013'
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select bl.cardno, b.title, br.c_name from book b join bookloans bl on bl.bookid=b.bookid where bl.dateout='12-07-2013'
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select bl.cardno, b.title, br.c_name from book b join borrower br on br.cardno=bl.cardno where bl.dateout='12-07-2013'
Question 4
Question
Retrieve max price for Fariza's publication. BOOK {bookid, title, publisher_name, price}.BOOK_AUTHORS {author_name, bookid (references Book (bookid))}
Answer
-
select max (b.price)from book b join book_authors ba on b.bookid=ba.bookid where ba.author_name = 'Fariza';
-
select max (b.price)from book b join book_authors ba on b.bookid=ba.bookid where ba.author_name = Fariza;
-
select max (b.price)from book b join book_authors ba b.bookid=ba.bookid where ba.author_name = 'Fariza';
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select max (b.price)from book b join book_authors ba on b.bookid=ba.bookid; #subToPewDiePie
Question 5
Question
Retrieve book name, total number of book copies. BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}
Answer
-
select b.title, sum (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid group by b.title;
-
select b.title, sum (no_of_copies) from book b join book_copies bc b.bookid=bc.bookid group by b.title;
-
select b.title, count(no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid group by b.title;
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select b.title, count(no_of_copies) from book b join book_copies bc b.bookid=bc.bookid group by b.title;
Question 6
Question
Retrieve book title and total number of book copies, which has id 1. BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}
Answer
-
select b.title, sum (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid where b.bookid=1 group by b.title;
-
select b.title, sum (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid where b.bookid=1;
-
select b.title, count (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid where b.bookid=1 group by b.title;
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select b.title, count (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid where b.bookid=1;
Question 7
Question
Retrieve borrower name and number of books that each borrower has. BOOKLOANS {bookloans, bookid (references Book (bookid)), libraryid (references Library (libraryid)), cardno (references Borrower (cardno)), dateout, duedate}. BORROWER {cardno, c_name, b_address, phoneno}
Answer
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select br.c_name, count(bl.bookid) from borrower br join bookloans bl on br.cardno = bl.cardno group by br.c_name;
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select br.c_name, count(bl.bookid) from borrower br join bookloans bl on br.cardno = bl.cardno;
-
select br.c_name, sum(bl.bookid) from borrower br join bookloans bl on br.cardno = bl.cardno group by br.c_name;
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select br.c_name, sum(bl.bookid) from borrower br join bookloans bl on br.cardno = bl.cardno;
Question 8
Question
Retrieve books name and prices. Increase price twice more for books, which have more than 20 copies;BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}
Answer
-
select b.title, b.price, b.price *2 as increasedprice from book b join book_copies bc on bc.bookid=b.bookid where bc.no_of_copies > 20;
-
select b.title, b.price book b join book_copies bc on bc.bookid=b.bookid where bc.no_of_copies > 20;
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impossible to solve
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select b.title, b.price, b.price *2 as increasedprice from book b join book_copies bc on bc.bookid=b.bookid;
Question 9
Question
Retrieve library name, book name, number of copies, which have more than 25 copies. BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}. LIBRARY (libraryid)), cardno (references Borrower (cardno)), dateout, duedate}.
Answer
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select l.libraryname, b.title, bc.no_of_copies from book b join book_copies bc on b.bookid=bc.bookid join library l on l.libraryid = bc.libraryid where bc.no_of_copies>25;
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select l.libraryname, b.title, bc.no_of_copies from book b join book_copies bc on b.bookid=bc.bookid where bc.no_of_copies>25;
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select l.libraryname, b.title, bc.no_of_copies from book join library l on l.libraryid = bc.libraryid where bc.no_of_copies>25;
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select l.libraryname, b.title, bc.no_of_copies from book b join book_copies bc on b.bookid=bc.bookid join library l l.libraryid = bc.libraryid where bc.no_of_copies>25;
Question 10
Question
Retrieve the number of departments in the department table. Tables: DEPARTMENT {id, name}. EMPLOYEE {id, name, salary, dep_id (references Department (id))}
Answer
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select count(id) from department
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select sum (id) from department;
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select coun(id) from department
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select id from department
Question 11
Question
Retrieve total salaries payable to employee. Tables: DEPARTMENT {id, name}. EMPLOYEE {id, name, salary, dep_id (references Department (id))}
Answer
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select sum (salary) from employee;
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select count (salary) from employee;
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select sum (id) from employee;
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select max (salary) from employee;
Question 12
Question
Retrieve office name and tutor name, who have experience equal or more than five years. 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))
Answer
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select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience >=5;
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select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience >5;
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select o.name, t.name from office o join tutor t o.id=t.officeid where t.experience >=5;
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select o.name, t.name from office join tutor on o.id=t.officeid where t.experience >=5;
Question 13
Question
Retrieve tutor name and their office locations, who refer to CSSE and Management. 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))
Answer
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select t.name, o.locations from office o join tutor t on o.id=t.officeid where o.name in ('CSSE', 'Management');
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select t.name, o.locations from office o join tutor t on o.id=t.officeid where o.name in ('CSSE' or 'Management');
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select t.name, o.locations from office o join tutor t o.id=t.officeid where o.name in ('CSSE', 'Management');
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select t.name, o.locations from office o join tutor t on o.id=t.officeid where o.name in ('CSSE') or ('Management');
Question 14
Question
Retrieve student's name, which starts from A. 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))
Answer
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select name from students where name like 'A_%';
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select name from students where name like 'A';
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select name from students where name like 'a%';
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select name from students where name like '%A%'
Question 15
Question
Retrieve all students name with the department name, but excluding department CSSE. 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))
Answer
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name <> 'CSSE';
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid;
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name = 'CSSE';
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name, s.name;
Question 16
Question
Which of the following query is incorrect for retrieving all students name with the department name, but excluding department CSSE. 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))
Answer
-
select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name <= 'CSSE';
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name <> 'CSSE';
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name not like 'CSSE';
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select s.name, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where o.name not in ('CSSE');
Question 17
Question
Retrieve students name, scholarship and their office name, who have scholarship ranging from 3200 to 4500.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))
Answer
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select s.name, s.scholarship, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where s.scholarship between 3200 and 4500;
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select s.name, s.scholarship, o.name from students s join tutor t on t.id=s.tutorid where s.scholarship between 3200 and 4500;
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select s.name, s.scholarship, o.name from students join office o on o.id=t.officeid where s.scholarship between 3200 and 4500;
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select s.name, s.scholarship, o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid where s.scholarship 3200 and 4500;
Question 18
Question
What kind symbol is used to find one character in wildcards?
Answer
-
-
-
?
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$
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in
Question 19
Question
Find an average scholarship of students. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }
Answer
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select avg (scholarship) from students ;
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select averg (scholarship) from students ;
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select avrg (scholarship) from students ;
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select av (scholarship) from students ;
Question 20
Question
Retrieve average scholarship of students for each 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))
Answer
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select avg (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select avg (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid ;
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select avg (scholarship), o.name from students s join tutor t on t.id=s.tutorid group by o.name;
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select avg (scholarship), o.name from students s d join office o on o.id=t.officeid group by o.name;
Question 21
Question
Retrieve student number in each 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))
Answer
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select count (s.id), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select sum (s.id), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select count (s.id), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid;
Question 22
Question
Retrieve student number in each 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))
Answer
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select count (s.id), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select sum (s.id), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select count (s.id), o.name from students s join tutor t
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on t.id=s.tutorid join office o on o.id=t.officeid;
Question 23
Question
Retrieve total amount of scholarship of each 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))
Answer
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select sum (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select count (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid group by o.name;
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select sum (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid;
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select count (scholarship), o.name from students s join tutor t on t.id=s.tutorid join office o on o.id=t.officeid;
Question 24
Question
SQL stands for
Answer
-
Sequence Question Language
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Structured Query Language
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Structured Querty Language
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Selection Query Language
Question 25
Question
You can add a row using SQL in a database with which of the following
Answer
-
ADD
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CREATE
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INSERT
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UPDATE
Question 26
Question
The command to remove rows from a table “Customer”
Answer
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Remove from Customer
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Delete from Customer
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Drop from Customer
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Update row from Customer
Question 27
Question
The SQL WHERE clause
Answer
-
Condition that limits the column data that are returned
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Condition that limits the row data are returned
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Clause that returns all rows
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Clause that returns nothing
Question 28
Question
The command to eliminate a table from a database
Answer
-
Drop
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Delete
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Remove
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Update
Question 29
Question
Which of the following is correct order of keywords for SQL Select statement?
Answer
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SELECT, FROM, WHERE
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WHERE, FROM, SELECT
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FROM, WHERE, SELECT
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SELECT, WHERE, FROM
Question 30
Question
SQL data definition commands make up a(n)
Answer
-
DDC
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DML
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DDL
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DDD
Question 31
Question
In a relation, the columns are also called attributes
Answer
- True
- False
Question 32
Question
Find the SQL statement that is equal to: SELECT NAME FROM CUSTOMER WHERE STATE = 'VA';
Answer
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SELECT NAME IN CUSTOMER WHERE STATE IN 'VA';
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SELECT NAME IN CUSTOMER WHERE STATE = 'VA';
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SELECT NAME FROM CUSTOMER WHERE STATE IN 'VA';
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SELECT NAME IN CUSTOMER WHERE STATE = 'V';
Question 33
Question
In E\R diagrams, we will represent Entities as
Answer
-
Boxes with rounded corners
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Links between two entities
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Ovals
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Diamond box
Question 34
Question
In E\R diagrams, we will represent Relationships as
Answer
-
Boxes with rounded corners
-
Links between two entities
-
Ovals
-
Diamond box
Question 35
Question
In E\R diagrams, we will represent Attributes as
Answer
-
Boxes with rounded corners
-
Links between two entities
-
Ovals
-
Diamond box
Question 36
Question
Many to many relationships are difficult to represent in database, so we need to
Answer
-
Split many to many relationship into two one to many relationships
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Split one to many relationship into two one to many relationships
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Split many to many relationship into one to many relationships
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Split many to many relationship into three one to many relationships
Question 37
Question
The result of a SQL SELECT statement is a(n)
Answer
-
Report
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Table
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Form
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File
Question 38
Question
Which of the following are the five built-in functions provided by SQL?
Answer
-
COUNT, SUM, AVG, MAX, MULT
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SUM, AVG, MIN, MAX, NAME
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SUM, AVG, MULT, DIV, MIN
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COUNT, SUM, AVG, MAX, MIN
Question 39
Question
In a relation, the order of the rows matters
Answer
- True
- False
Question 40
Question
Given the functional dependency R → (S,T) , then it is also true that R → S
Answer
- True
- False
Question 41
Question
Given the functional dependency R → (S,T) , then it is also true that R → T
Answer
- True
- False
Question 42
Question
SQL can be used to
Answer
-
create database structures only
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query database data only
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modify database data only
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All of the these can be done by SQL
Question 43
Question
The SQL statement that queries or reads data from a table is _
Answer
-
READ
-
QUERY
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SELECT
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NONE
Question 44
Question
A subquery in an SQL SELECT statement
Answer
-
can only be used with two tables
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has a distinct form that cannot be duplicated by a join
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can always be duplicated by a join
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cannot have its results sorted using ORDER BY
Question 45
Question
The SQL keyword BETWEEN is used
Answer
-
for ranges
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as a wildcard
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to limit the columns displayed
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None of the above
Question 46
Question
Sometimes you want to change the structure of an existing table, what are your options?
Answer
-
Change table
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Drop Table
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Alter Table
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Create Table
Question 47
Question
how to Add a new column into existing table?
Answer
-
ALTER TABLE Student MODIFY COLUMN sDegree CHAR(64) NOT NULL
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ALTER TABLE Student ADD COLUMN sDegree VARCHAR(64) NOT NULL
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ALTER TABLE S ADD COLUMN s VACHAR(64) NOT NULL
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ALTER TABLE Student DROP COLUMN sDegree VARCHAR(64) NOT NULL
Question 48
Question
how to rename a column in existing table?
Answer
-
ALTER TABLE Student MODIFY COLUMN sDegree CHAR(64) NOT NULL
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ALTER TABLE Student ADD COLUMN sDegree VARCHAR(64) NOT NULL
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ALTER TABLE Student RENAME COLUMN s to SS VACHAR(64) NOT NULL
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ALTER TABLE Student DROP COLUMN sDegree VARCHAR(64) NOT NULL
Question 49
Question
how to change the row(s) in a table
Answer
-
insert
-
update
-
change
-
delete
Question 50
Question
how to remove the row(s) from a table
Answer
-
insert
-
update
-
change
-
delete
Question 51
Question
A SELECT statement can be nested inside another query to form a
Answer
-
Subselect
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Subresults
-
Subquery
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Query in query
Question 52
Question
SQL uses privileges to control access to tables and other database objects, so which is NOT?
Answer
-
Select privilege
-
Update privilege
-
Insert privilege
-
Drop privilege
Question 53
Question
To convert any relation into _______, split any nonatomic values
Answer
-
First normal form
-
Second normal form
-
Third normal form
-
Fourth normal form
Question 54
Question
A functional dependency (FD) is a
Answer
-
link between three sets of attributes in a relation
-
link between all sets of attributes in a relation
-
link between four sets of attributes in a relation
-
link between two sets of attributes in a relation
Question 55
Question
which if the following does not refer to redundancy problems?
Answer
-
INSERT anomalies
-
CREATE anomalies
-
UPDATE anomalies
-
DELETE anomalies
Question 56
Question
To convert any relation into _______, remove transitive dependency
Answer
-
First normal form
-
Second normal form
-
Third normal form
-
Fourth normal form
Question 57
Answer
-
Relation
-
Tuples
-
Attributes
-
Relationships
Question 58
Answer
-
Relation
-
Tuples
-
Attributes
-
Relationships
Question 59
Answer
-
Relation
-
Tuples
-
Attributes
-
Relationships
Question 60
Answer
-
Union
-
Difference
-
Itersection
-
Product
Question 61
Answer
-
Union
-
Difference
-
Intersection
-
Product
Question 62
Answer
-
Difference
-
Union
-
Intersection
-
Product
Question 63
Answer
-
Union
-
Difference
-
Intersection
-
Product
Question 64
Answer
-
yes
-
no
Question 65
Question
Which of the following is not correct?
INSERT INTO Employee(ID, Name, Salary) VALUES
Image:
9 (binary/octet-stream)
Answer
-
INSERT INTO Employee(ID, Name, Salary) VALUES(2,‘Mary’,26);
-
INSERT INTO Employee (Name,ID) VALUES (‘Mary’,2);
-
INSERT INTO Employee VALUES (2, ‘Mary’,26000);
-
INSERT INTO Employe VALUES(26, ‘Mary’,26);
Question 66
Answer
-
UPDATE Employee SET Salary = Salary*0.5
-
UPDATE Employee SET Salary = Salary * 0.1
-
UPDATE Employee SET Salary = Salary * 1.1
-
UPDATE Employee Update Salary = Salary*0.1
Question 67
Answer
-
DELETE FROM Employee WHERE Salary >=22000
-
DELETE FROM Employe WHERE Salary => 22000;
-
REMOVE FROM Employee WHERE Salary >22000;
-
DELETE FROM Employee WHERE Salary = 22000;
Question 68
Question
πsName,sAddress(Student) is equal to?
Answer
-
SELECT Sname FROM Students
-
SELECT Sname, SAddress FORM Students
-
SELECT Sname and SAddress FROM Students
-
SELECT Sname, SAddress FROM Student
Question 69
Question
SQL query to find a list of the ID numbers and Marks for students who have passed IAI
Image:
11 (binary/octet-stream)
Answer
-
Select ID, Mark from Grade Where code = 'IAI' and Mark > 50;
-
Select ID, Mark from Grade Where code = 'AIA' and Mark > 50;
-
Select ID, Mark, Code from Grade Where code = 'IAI';
-
Select ID, Mark from Grade Where code = 'IAI' and Mark >=50;
Question 70
Answer
-
Select First,Last from Student,Grade Where Code='PR1'OR'PR2'
-
Select First,Last from Student Natural Join Grade Where Code like'PR%'
-
Select First Last from Student Natural Join Grade Where Code like'PR%'
-
Select First from Student Natural Join Grade Where Code = 'PR%'
Question 71
Question
How to use privileges in SQL?
Answer
-
ON<objects> TO<users> GRANT<privileges>
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GRANT<privileges> ON<objects> TO<users>
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GRANT<privileges> TO<users> ON<objects>
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GRANT<tables> ON<objects> TO<users>
Question 72
Question
If Admin’ grants ALL privileges to ‘Manager’, and SELECT to ‘Finance’ with grant option, So..
Image:
13 (binary/octet-stream)
Answer
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‘Manager’ grants ALL to ‘Personnel’
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‘Manager’ grants SELECT to ‘Personnel’
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‘Finance’ grants SELECT to ‘Manager’
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‘Finance’ grants ALL to ‘Manager’
Question 73
Question
If Admin’ grants ALL privileges to ‘Manager’, and SELECT to ‘Finance’ with grant option, So..
Image:
13 (binary/octet-stream)
Answer
-
‘Finance’ grants SELECT to ‘Personnel’
-
‘Manager’ grants SELECT to ‘Personnel’
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‘Finance’ grants SELECT to ‘Manager’
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‘Finance’ grants ALL to ‘Manager’
Question 74
Answer
-
‘Personnel’ still has ALL privileges from ‘Finance’
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‘Finance’ still has SELECT privileges from ‘Personnel’
-
‘Personnel’ still has SELECT privileges from ‘Finance’
-
‘Admin’ still has ALL privileges from ‘Finance’
Question 75
Answer
-
‘Personnel’ still has ALL privileges from ‘Finance’
-
‘Finance’ still has SELECT privileges from ‘Personnel’
-
‘Personnel’ still has ALL privileges from ‘Manager’
-
‘Admin’ still has ALL privileges from ‘Finance’
Question 76
Answer
-
‘Personnel’ still has ALL privileges from ‘Finance’
-
‘Finance’ still has SELECT privileges from ‘Personnel’
-
‘Personnel’ still has ALL privileges from ‘Manager’
-
‘Admin’ still has ALL privileges from ‘Finance’
Question 77
Answer
-
‘Personnel’ still has ALL privileges from ‘Finance’
-
‘Finance’ still has SELECT privileges from ‘Personnel’
-
‘Finance still has SELECT privileges from ‘Admin’
-
‘Admin’ still has ALL privileges from ‘Finance’
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