Examine the data in the EMPLOYEES table:Which three subqueries work?()
A.
B.
C.
D.
E.
F.
A.
B.
C.
D.
E.
F.
A.INSERT INTO employees VALUES (NULL, ‘John‘, ‘Smith‘);
B.INSERT INTO employees(first_name, last_name) VALUES(‘John‘, ‘Smith‘);
C.INSERT INTO employees VALUES (‘1000‘, ‘John‘, NULL);
D.INSERT INTO employees (first_name, last_name, employee_id) VALUES (1000, ‘John‘, ‘Smith‘);
E.INSERT INTO employees (employee_id) VALUES (1000);
F.INSERT INTO employees (employee_id, first_name, last_name) VALUES (1000, ‘John‘, ‘ ‘);
Examine the description of the EMPLOYEES table:EMP_ID NUMBER(4) NOT NULLLAST_NAME VARCHAR2(30) NOT NULLFIRST_NAME VARCHAR2(30)DEPT_ID NUMBER(2)Which statement produces the number of different departments that have employees with last name Smith? ()
A. SELECT COUNT (*) FROM employees WHERE last _name=‘smith‘;
B. SELECT COUNT (dept_id) FROM employees WHERE last _name=‘smith‘;
C. SELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name=‘smith‘;
D. SELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name=‘smith‘;
E. SELECT UNIQE (dept_id) FROM employees WHERE last _name=‘smith‘;
A. INSERT INTO employees VALUES (NULL, ‘John‘, ‘smith‘);
B. INSERT INTO employees (first_name, last_name) VALUES (‘John‘, ‘smith‘);
C. INSERT INTO employees VALUES (‘1000, ‘John‘, ‘smith‘);
D. INSERT INTO employees (first_name, last_name, employee_id) VALUES (1000, ‘John‘, ‘smith‘);
E. INSERT INTO employees (employee_id) VALUES (1000);
F. INSERT INTO employees (employee_id, first_name, last_name, ) VALUES (1000, ‘John‘,‘‘);
A. INSERT INTO employees VALUES (NULL, ‘JOHN‘,‘Smith‘);
B. INSERT INTO employees(first_name, last_name) VALUES (‘JOHN‘,‘Smith‘);
C. INSERT INTO employees VALUES (‘1000‘,‘JOHN‘,‘NULL‘);
D. INSERT INTO employees(first_name,last_name, employee_id) VALUES (‘1000, ‘john‘,‘Smith‘);
E. INSERT INTO employees (employee_id) VALUES (1000);
F. INSERT INTO employees (employee_id, first_name, last_name) VALUES (1000, ‘john‘,");
A.SELECT COUNT(*) FROM employees WHERE last_name=‘Smith‘;
B.SELECT COUNT(dept_id) FROM employees WHERE last_name=‘Smith‘;
C.SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name=‘Smith‘;
D.SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name=‘Smith‘;
E.SELECT UNIQUE(dept_id) FROM employees WHERE last_name=‘Smith‘;
A.backupthecontrolfiletotrace
B.bringdatabasetotheMOUNTstate
C.openaconnectiontotheRMANrecoverycatalog,whichcontainstheRMANmetadataforthetargetdatabase
D.setthedatabaseID(DBID),butonlyiftheDB_NAMEparameterassociatedwiththetargetdatabaseisuniqueintherecoverycatalog
A. You obtain the results retrieved from the public synonym HR created by the database administrator.
B. You obtain the results retrieved from the HR table that belongs to your schema.
C. You get an error message because you cannot retrieve from a table that has the same name as a public synonym.
D. You obtain the results retrieved from both the public synonym HR and the HR table that belongs to your schema, as a Cartesian product.
E. You obtain the results retrieved from both the public synonym HR and the HR table that belongs to your schema, as a FULL JOIN.
A. You obtain the results retrieved from the public synonym HR created by the database administrator.
B. You obtain the results retrieved from the HR table that belongs to your schema.
C. You get an error message because you cannot retrieve from a table that has the same name as a public synonym.
D. You obtain the results retrieved from both the public synonym HR and the HR table that belongs to your schema, as a Cartesian product.
E. You obtain the results retrieved from both the public synonym HR and the HR table that belongs to your schema, as a FULL JOIN.
A.Setset=newTreeSet();
B.Setset=newHashSet();
C.Setset=newSortedSet();
D.Listset=newSortedList();
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