17415233
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Quiz von хомяк убийца, aktualisiert more than 1 year ago
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Erstellt von хомяк убийца
vor mehr als 5 Jahre
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Frage 1
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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))
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select min (experience) from tutor;
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select max (experience) from tutor;
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select max (t.experience) from tutor t group by t.name;
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select min (t.experience) from tutor t group by t.name;
Frage 2
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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)]}
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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 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;
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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;
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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;
Frage 3
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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}
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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'
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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.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'
Frage 4
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Retrieve max price for Fariza's publication. BOOK {bookid, title, publisher_name, price}.BOOK_AUTHORS {author_name, bookid (references Book (bookid))}
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select max (b.price)from book b join book_authors ba on 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 where ba.author_name = Fariza;
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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
Frage 5
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Retrieve book name, total number of book copies. BOOK {bookid, title, publisher_name, price}. BOOKCOPIES {bookid, libraryid,noOfCopies [references Book (bookid)]}
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select b.title, sum (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, sum (no_of_copies) from book b join book_copies bc 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 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;
Frage 6
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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)]}
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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;
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select b.title, sum (no_of_copies) from book b join book_copies bc on b.bookid=bc.bookid where b.bookid=1;
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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 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;
Frage 7
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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}
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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;
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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;
Frage 8
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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)]}
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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 where bc.no_of_copies > 20;
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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;
Frage 9
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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}.
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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;
Frage 10
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Retrieve the number of departments in the department table. Tables: DEPARTMENT {id, name}. EMPLOYEE {id, name, salary, dep_id (references Department (id))}
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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
Frage 11
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Retrieve total salaries payable to employee. Tables: DEPARTMENT {id, name}. EMPLOYEE {id, name, salary, dep_id (references Department (id))}
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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;
Frage 12
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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))
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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;
Frage 13
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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))
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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');
Frage 14
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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))
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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%'
Frage 15
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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))
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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;
Frage 16
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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))
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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 <> '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');
Frage 17
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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))
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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;
Frage 18
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What kind symbol is used to find one character in wildcards?
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-
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?
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$
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in
Frage 19
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Find an average scholarship of students. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }
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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 ;
Frage 20
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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))
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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;
Frage 21
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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))
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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;
Frage 22
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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))
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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;
Frage 23
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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))
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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;
Frage 24
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SQL stands for
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Sequence Question Language
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Structured Query Language
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Structured Querty Language
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Selection Query Language
Frage 25
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You can add a row using SQL in a database with which of the following
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ADD
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CREATE
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INSERT
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UPDATE
Frage 26
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The command to remove rows from a table “Customer”
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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
Frage 27
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The SQL WHERE clause
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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
Frage 28
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The command to eliminate a table from a database
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Drop
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Delete
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Remove
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Update
Frage 29
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Which of the following is correct order of keywords for SQL Select statement?
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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
Frage 30
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SQL data definition commands make up a(n)
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DDC
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DML
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DDL
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DDD
Frage 31
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In a relation, the columns are also called attributes
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- True
- False
Frage 32
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Find the SQL statement that is equal to: SELECT NAME FROM CUSTOMER WHERE STATE = 'VA';
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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';
Frage 33
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In E\R diagrams, we will represent Entities as
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Boxes with rounded corners
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Links between two entities
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Ovals
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Diamond box
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In E\R diagrams, we will represent Relationships as
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Boxes with rounded corners
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Links between two entities
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Ovals
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Diamond box
Frage 35
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In E\R diagrams, we will represent Attributes as
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Boxes with rounded corners
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Links between two entities
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Ovals
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Diamond box
Frage 36
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Many to many relationships are difficult to represent in database, so we need to
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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
Frage 37
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The result of a SQL SELECT statement is a(n)
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Report
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Table
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Form
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File
Frage 38
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Which of the following are the five built-in functions provided by SQL?
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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
Frage 39
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In a relation, the order of the rows matters
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- True
- False
Frage 40
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Given the functional dependency R → (S,T) , then it is also true that R → S
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- True
- False
Frage 41
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Given the functional dependency R → (S,T) , then it is also true that R → T
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- True
- False
Frage 42
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SQL can be used to
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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
Frage 43
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The SQL statement that queries or reads data from a table is _
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READ
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QUERY
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SELECT
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NONE
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A subquery in an SQL SELECT statement
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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
Frage 45
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The SQL keyword BETWEEN is used
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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
Frage 46
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Sometimes you want to change the structure of an existing table, what are your options?
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Change table
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Drop Table
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Alter Table
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Create Table
Frage 47
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how to Add a new column into existing table?
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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
Frage 48
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how to rename a column in existing table?
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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
Frage 49
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how to change the row(s) in a table
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insert
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update
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change
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delete
Frage 50
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how to remove the row(s) from a table
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insert
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update
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change
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delete
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A SELECT statement can be nested inside another query to form a
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Subselect
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Subresults
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Subquery
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Query in query
Frage 52
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SQL uses privileges to control access to tables and other database objects, so which is NOT?
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Select privilege
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Update privilege
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Insert privilege
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Drop privilege
Frage 53
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To convert any relation into _______, split any nonatomic values
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First normal form
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Second normal form
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Third normal form
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Fourth normal form
Frage 54
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A functional dependency (FD) is a
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link between three sets of attributes in a relation
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link between all sets of attributes in a relation
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link between four sets of attributes in a relation
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link between two sets of attributes in a relation
Frage 55
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which if the following does not refer to redundancy problems?
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INSERT anomalies
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CREATE anomalies
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UPDATE anomalies
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DELETE anomalies
Frage 56
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To convert any relation into _______, remove transitive dependency
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First normal form
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Second normal form
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Third normal form
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Fourth normal form
Frage 57
Antworten
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Relation
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Tuples
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Attributes
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Relationships
Frage 58
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Relation
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Tuples
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Attributes
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Relationships
Frage 59
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Relation
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Tuples
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Attributes
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Relationships
Frage 60
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Union
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Difference
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Itersection
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Product
Frage 61
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Union
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Difference
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Intersection
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Product
Frage 62
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Difference
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Union
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Intersection
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Product
Frage 63
Antworten
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Union
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Difference
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Intersection
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Product
Frage 64
Antworten
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yes
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no
Frage 65
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Which of the following is not correct?
INSERT INTO Employee(ID, Name, Salary) VALUES
Image:
9 (binary/octet-stream)
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INSERT INTO Employee(ID, Name, Salary) VALUES(2,‘Mary’,26);
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INSERT INTO Employee (Name,ID) VALUES (‘Mary’,2);
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INSERT INTO Employee VALUES (2, ‘Mary’,26000);
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INSERT INTO Employe VALUES(26, ‘Mary’,26);
Frage 66
Antworten
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UPDATE Employee SET Salary = Salary*0.5
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UPDATE Employee SET Salary = Salary * 0.1
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UPDATE Employee SET Salary = Salary * 1.1
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UPDATE Employee Update Salary = Salary*0.1
Frage 67
Antworten
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DELETE FROM Employee WHERE Salary >=22000
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DELETE FROM Employe WHERE Salary => 22000;
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REMOVE FROM Employee WHERE Salary >22000;
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DELETE FROM Employee WHERE Salary = 22000;
Frage 68
Frage
πsName,sAddress(Student) is equal to?
Antworten
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SELECT Sname FROM Students
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SELECT Sname, SAddress FORM Students
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SELECT Sname and SAddress FROM Students
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SELECT Sname, SAddress FROM Student
Frage 69
Frage
SQL query to find a list of the ID numbers and Marks for students who have passed IAI
Image:
11 (binary/octet-stream)
Antworten
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Select ID, Mark from Grade Where code = 'IAI' and Mark > 50;
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Select ID, Mark from Grade Where code = 'AIA' and Mark > 50;
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Select ID, Mark, Code from Grade Where code = 'IAI';
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Select ID, Mark from Grade Where code = 'IAI' and Mark >=50;
Frage 70
Antworten
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Select First,Last from Student,Grade Where Code='PR1'OR'PR2'
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Select First,Last from Student Natural Join Grade Where Code like'PR%'
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Select First Last from Student Natural Join Grade Where Code like'PR%'
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Select First from Student Natural Join Grade Where Code = 'PR%'
Frage 71
Frage
How to use privileges in SQL?
Antworten
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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>
Frage 72
Frage
If Admin’ grants ALL privileges to ‘Manager’, and SELECT to ‘Finance’ with grant option, So..
Image:
13 (binary/octet-stream)
Antworten
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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’
Frage 73
Frage
If Admin’ grants ALL privileges to ‘Manager’, and SELECT to ‘Finance’ with grant option, So..
Image:
13 (binary/octet-stream)
Antworten
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‘Finance’ grants SELECT 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’
Frage 74
Antworten
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‘Personnel’ still has ALL privileges from ‘Finance’
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‘Finance’ still has SELECT privileges from ‘Personnel’
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‘Personnel’ still has SELECT privileges from ‘Finance’
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‘Admin’ still has ALL privileges from ‘Finance’
Frage 75
Antworten
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‘Personnel’ still has ALL privileges from ‘Finance’
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‘Finance’ still has SELECT privileges from ‘Personnel’
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‘Personnel’ still has ALL privileges from ‘Manager’
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‘Admin’ still has ALL privileges from ‘Finance’
Frage 76
Antworten
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‘Personnel’ still has ALL privileges from ‘Finance’
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‘Finance’ still has SELECT privileges from ‘Personnel’
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‘Personnel’ still has ALL privileges from ‘Manager’
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‘Admin’ still has ALL privileges from ‘Finance’
Frage 77
Antworten
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‘Personnel’ still has ALL privileges from ‘Finance’
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‘Finance’ still has SELECT privileges from ‘Personnel’
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‘Finance still has SELECT privileges from ‘Admin’
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‘Admin’ still has ALL privileges from ‘Finance’
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