최신Oracle Database 12c: SQL Fundamentals - 1z0-061무료샘플문제

A. only the CONNECT privilege

B. no privileges

C. all the privileges of a default user

D. only the SELECT privilege

A. All uncommitted transactions are committed.

B. Sequences used in the employees table become invalid.

C. The employees table can be recovered using the rollback command.

D. The space used by the employees table is reclaimed immediately.

E. The employees table is moved to the recycle bin.

F. All indexes and constraints defined on the table being dropped are also dropped.

A. It displays 5 percent of the products with the lowest amount sold.

B. It results in an error because the ORDER BY clause should be the last clause.

C. It displays the first 5 percent of the rows from the SALES table.

D. It displays 5 percent of the products with the highest amount sold.

A. SELECT student_name, SUM(subject1)FROM marksWHERE student_name LIKE 'R%';

B. SELECT student_name, subject1FROM marksWHERE subject1 > AVG(subject1);

C. SELECT SUM(DISTINCT NVL(subject1, 0)), MAX(subject1)FROM marksWHERE
subject1 > subject2;

D. SELECT SUM(subject1+subject2+subject3)FROM marksWHERE student_name IS NULL;

A. Only 'hiredate'

B. Only 'mgr_id'

C. None, because no input required

D. Both the substitution variables 'hiredate' and 'mgr_id\

A. UPDATE

B. RENAME

C. SELECT

D. DELETE

E. DESCRIBE

F. INSERT

A. It produces an error.

B. It displays all values.

C. It displays only the values A%_WQ123 and AB_WQ123 .

D. It displays only the values A%BWQ123 and AB_WQ123.

E. It displays only the values A%_WQ123 and A%BWQ123 .

A. 'Oder' and 'Aus'

B. an error because the REPLACE function specified is not valid

C. an error because the TRIM function specified is not valid

D. an error because the LOWER function specified is not valid

A. UPDATE employeeSET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180;

B. UPDATE employeesSET first_name = 'John'SET last_name = 'Smith' WHERE employee_id = 180;

C. UPDATE employeeSET first_name = 'John'AND last_name = 'Smith' WHERE employee_id = 180;

D. UPDATE employeesSET first_name = 'John',SET last_name = 'Smoth' WHERE employee_id = 180;

A. INSERT INTO employees(employee_id, first_name, last_name, Hire_date)VALUES (
1 000, 'John', 'Smith', To_date('01/01/01'));

B. INSERT INTO employees (employee_id, first_name, last_name, hire_date)VALUES (
1 000, 'John', 'Smith', '01/01/01');

C. INSERT INTO employees(employee_id, first_name, last_name, hire_date)VALUES (
1 000, 'John', 'Smith', '01 January 01');

D. INSERT INTO employees(employee_id, first_name, last_name, hire_date)VALUES (
1 000, 'John', 'Smith', 01-Jan-01);

A. The view has already done the work of joining the tables.

B. The view uses ISO standard join syntax, which is faster than the Oracle join syntax used in the second query.

C. There is no reason for the first query to be quicker.

D. The view is precompiled, so the first query requires less dynamic compilation than the second query.

A. INSERT INTO employees VALUES ( NULL, 'John', 'Smith');

B. INSERT INTO employees (first_name, last_name, employee_id) VALUES ( 1000, 'John', 'Smith');

C. INSERT INTO employees VALUES ( 1000, 'John', NULL);

D. INSERT INTO employees( first_name, last_name) VALUES( 'John', 'Smith');

E. INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'John', '');

F. INSERT INTO employees (employee_id) VALUES (1000);

A. UPDATE promotionsSET promo_cost = promo_cost+ 100WHERE promo_end_date > TO_DATE(SUBSTR('01-JAN-2000', 8));

B. UPDATE promotionsSET promo_cost = promo_cost+ 100WHERE
TO_CHAR(promo_end_date,'yyyy') > '2000';

C. SELECT promo_begin_dateFROM promotionsWHERE
TO_CHAR(promo_begin_date,'mon dd yy')='jul 01 98';

D. SELECT TO_CHAR(promo_begin_date,'dd/month')FROM promotionsWHERE
promo_begin_date IN (TO_DATE('JUN 01 98'), TO_DATE('JUL 01 98'));

A. NVL(NULLIF(list_price, min_price), 0)

B. COALESCE(NVL2(list_price, list_price, min_price), 0)

C. NVL2(COALESCE(list_price, min_price), min_price, 0)

D. NVL(COALESCE(list_price, min_price), 0)

A. SELECT prod_id FROM productsINTERSECTSELECT prod_id FROM sales;

B. SELECT DISTINCT p.prod_idFROM products p JOIN sales sON p.prod_id <> s.prod_id;

C. SELECT DISTINCT p.prod_idFROM products p JOIN sales sON p.prod_id=s.prod_id;

D. SELECT prod_id FROM productsMINUSSELECT prod_id FROM sales;