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