单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use?（）ASELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ;BSELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id;CSELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d;DSELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );
单选题
Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use?（）
A
SELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ;
B
SELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id;
C
SELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d;
D
SELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );
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Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables.For which situation would you use a nonequijoin query?() A. To find the tax percentage for each of the employees.B. To list the name, job id, and manager name for all the employees.C. To find the name, salary, and department name of employees who are not working with Smith.D. To find the number of employees working for the Administrative department and earning less then 4000.E. To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.
Examine the structure of the EMPLOYEES table:You need to update the records of employees 103 and 115. The UPDATE statement you specify should update the rows with the values specified below:Which UPDATE statement meets the requirements?（）A.B.C.D.E.
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Examine the structure of the EMPLOYEES table:You issue these statements:At the end of this transaction, what is true?() A. You have no rows in the table.B. You have an employee with the name of James.C. You cannot roll back to the same savepoint more than once.D. Your last update fails to update any rows because employee ID 180 was already eleted.
Examine the data from the EMP table:The COMMISSION column shows the monthly commission earned by the employee.Which three tasks would require subqueries or joins in order to perform in a single step?（）A. Deleting the records of employees who do not earn commission.B. Increasing the commission of employee 3 by the average commission earned in department 20.C. Finding the number of employees who do NOT earn commission and are working for department 20.D. Inserting into the table a new employee 10 who works for department 20 and earns a commission that is equal to the commission earned by employee 3.E. Creating a table called COMMISSION that has the same structure and data as the columns EMP_ID and COMMISSIONS of the EMP table.F. Decreasing the commission by 150 for the employees who are working in department 30 and earning a commission of more then 800.
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Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?() A.B.C.D.
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Examine the data in the EMPLOYEES and DEPARTMENTS tables:Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables:On the EMPLOYEES table, EMPLOYEE_ID is the primary key.MGR_ID is the ID of managers and refers to the EMPLOYEE_ID.DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table.On the DEPARTMENTS table, DEPARTMENT_ID is the primary key.Examine this DELETE statement:What happens when you execute the DELETE statement?（）A. Only the row with department ID 40 is deleted in the DEPARTMENTS table.B. The statement fails because there are child records in the EMPLOYEES table with department ID 40.C. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.D. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 106 and 110 and the employees working under employee 110 are deleted from the EMPLOYEES table.E. The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.F. The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
Examine the data in the EMPLOYEES and DEPARTMENTS tables:EMPLOYEESEMP_NAME DEPT_ID MGR_ID JOB_ID SALARYEMPLOYEE_ID101 Smith 20 120 SA_REP 4000102 Martin 10 105 CLERK 2500103 Chris 20 120 IT_ADMIN 4200104 John 30 108 HR_CLERK 2500105 Diana 30 108 IT_ADMIN 5000106 Smith 40 110 AD_ASST 3000108 Jennifer 30 110 HR_DIR 6500110 Bob 40 EX_DIR 8000120 Ravi 20 110 SA*DIR 6500DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Admin20 Education30 IT40 Human ResourcesAlso examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables:CREATE TABLE departments(department_id NUMBER PRIMARY KEY,department _ name VARCHAR2(30));CREATE TABLE employees(EMPLOYEE_ID NUMBER PRIMARY KEY,EMP_NAME VARCHAR2(20),DEPT_ID NUMBER REFERENCESdepartments(department_id),MGR_ID NUMBER REFERENCESemployees(employee id),MGR_ID NUMBER REFERENCESemployees(employee id),JOB_ID VARCHAR2(15).SALARY NUMBER);ON the EMPLOYEES,On the EMPLOYEES table, EMPLOYEE_ID is the primary key.MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table. On the DEPARTMENTS table, DEPARTMENT_ID is the primary key.Examine this DELETE statement:DELETEFROM departmentsWHERE department id = 40;What happens when you execute the DELETE statement?（）
Examine the data in the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_ID 101 Smith 20 120 SA_REP 4000 102 Martin 10 105 CLERK 2500 103 Chris 20 120 IT_ADMIN 4200 104 John 30 108 HR_CLERK 2500 105 Diana 30 108 IT_ADMIN 5000 106 Smith 40 110 AD_ASST 3000 108 Jennifer 30 110 HR_DIR 6500 110 Bob 40 EX_DIR 8000 120 Ravi 20 110 SA*DIR 6500 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Admin 20 Education 30 IT 40 Human Resources Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables: CREATE TABLE departments (department_id NUMBER PRIMARY KEY, department _ name VARCHAR2(30)); CREATE TABLE employees (EMPLOYEE_ID NUMBER PRIMARY KEY, EMP_NAME VARCHAR2(20), DEPT_ID NUMBER REFERENCES departments(department_id), MGR_ID NUMBER REFERENCES employees(employee id), MGR_ID NUMBER REFERENCES employees(employee id), JOB_ID VARCHAR2(15). SALARY NUMBER); ON the EMPLOYEES, On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table. On the DEPARTMENTS table, DEPARTMENT_ID is the primary key. Examine this DELETE statement: DELETE FROM departments WHERE department id = 40; What happens when you execute the DELETE statement?（）A、Only the row with department ID 40 is deleted in the DEPARTMENTS table.B、The statement fails because there are child records in the EMPLOYEES table with department ID 40.C、The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.D、The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 106 and 110 and the employees working under employee 110 are deleted from the EMPLOYEES table.E、The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.F、The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
单选题Examine the statement:Create synonym emp for hr. employees;What happens when you issue the statement?（）AAn error is generated.BYou will have two identical tables in the HR schema with different names.CYou create a table called employees in the HR schema based on you EMP table.DYou create an alternative name for the employees table in the HR schema in your own schema.
单选题You own a table called EMPLOYEES with this table structure: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE What happens when you execute this DELETE statement? DELETE employees; （）AYou get an error because of a primary key violation.BThe data and structure of the EMPLOYEES table are deleted.CThe data in the EMPLOYEES table is deleted but not the structure.DYou get an error because the statement is not syntactically correct.
单选题Examine the data in the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_ID 101 Smith 20 120 SA_REP 4000 102 Martin 10 105 CLERK 2500 103 Chris 20 120 IT_ADMIN 4200 104 John 30 108 HR_CLERK 2500 105 Diana 30 108 IT_ADMIN 5000 106 Smith 40 110 AD_ASST 3000 108 Jennifer 30 110 HR_DIR 6500 110 Bob 40 EX_DIR 8000 120 Ravi 20 110 SA*DIR 6500 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Admin 20 Education 30 IT 40 Human Resources Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables: CREATE TABLE departments (department_id NUMBER PRIMARY KEY, department _ name VARCHAR2(30)); CREATE TABLE employees (EMPLOYEE_ID NUMBER PRIMARY KEY, EMP_NAME VARCHAR2(20), DEPT_ID NUMBER REFERENCES departments(department_id), MGR_ID NUMBER REFERENCES employees(employee id), MGR_ID NUMBER REFERENCES employees(employee id), JOB_ID VARCHAR2(15). SALARY NUMBER); ON the EMPLOYEES, On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table. On the DEPARTMENTS table, DEPARTMENT_ID is the primary key. Examine this DELETE statement: DELETE FROM departments WHERE department id = 40; What happens when you execute the DELETE statement?（）AOnly the row with department ID 40 is deleted in the DEPARTMENTS table.BThe statement fails because there are child records in the EMPLOYEES table with department ID 40.CThe row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.DThe row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 106 and 110 and the employees working under employee 110 are deleted from the EMPLOYEES table.EThe row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.FThe statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?（）ASELECT last_name, department_name, location_id FROM employees , department ;BSELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;CSELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;DSELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
单选题You own a table called EMPLOYEES with this table structure: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE What happens when you execute this DELETE statement? DELETE employees; （）AYou get an error because of a primary key violation.BThe data and structure of the EMPLOYEES table are deleted.CThe data in the EMPLOYEES table is deleted but not the structure.DYou get an error because the statement is not syntactically correct.
单选题Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith? （）ASELECT COUNT (*) FROM employees WHERE last _name='smith';BSELECT COUNT (dept_id) FROM employees WHERE last _name='smith';CSELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';DSELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';ESELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2 (30) JOB_ID VARCHAR2 (20) SALARY NUMBER MGR_ID NUMBER References EMPLOYEE_ID COLUMN DEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID column of the DEPARTMENTS table DEPARTMENTSColumn name Data type Remarks DEPARTMENT_ID NUMBER NOT NULL, Primary Key DEPARTMENT_NAME VARCHAR2(30) MGR_ID NUMBER References MGR_ID column of the EMPLOYEES table Evaluate this SQL statement: SELECT employee_id, e.department_id, department_name, salary FROM employees e, departments d WHERE e. department_id = d.department_id; Which SQL statement is equivalent to the above SQL statement? （）ASELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);BSELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;CSELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;DSELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);
多选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2(30) JOB_ID NUMBER/ SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table You created a sequence called EMP_ID_SEQ in orderto populate sequential values for the EMPLOYEE_ID column of the EMPLOYEES table. Which two statements regarding the EMP_ID_SEQ sequence are true? （）AYou cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.BThe EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.CThe EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.DAny other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.EThe EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.FThe EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
多选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements inserts a row into the table? （）AINSERT INTO employees VALUES (NULL, 'JOHN','Smith');BINSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');CINSERT INTO employees VALUES ('1000','JOHN','NULL');DINSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');EINSERT INTO employees (employee_id) VALUES (1000);FINSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',);
单选题Examine the structure of the EMP_DEPT_VU view: Column Name Type Remarks EMPLOYEE_ID NUMBER From the EMPLOYEES table EMP_NAME VARCHAR2(30) From the EMPLOYEES table JOB_ID VARCHAR2(20) From the EMPLOYEES table SALARY NUMBER From the EMPLOYEES table DEPARTMENT_ID NUMBER From the DEPARTMENTS table DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table Which SQL statement produces an error?（）ASELECT * FROM emp_dept_vu;BSELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department _ id;CSELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department _ id, job_id;DSELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10,20) GROUP BY job_id HAVING SUM (salary) 20000ENone of the statements produce an error; all are valid.
单选题Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees, whether or not they have matching departments in the departments table. Which query would you use?（）ASELECT last_name, department_name FROM employees NATURAL JOIN departments;BSELECT last_name, department_name FROM employees JOIN departments ;CSELECT last_name, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id);DSELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);ESELECT last_name, department_name FROM employees FULL JOIN departments ON (e.department_id = d.department_id);FSELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);
单选题Examine the data in the EMPLOYEES and DEPARTMENTS tables. EMPLOYEES LAST_NAME DEPARTMENT_ID SALARY Getz 10 3000 Davis 20 1500 Bill 20 2200 Davis 30 5000 Kochhar 5000 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Sales 20 Marketing 30 Accounts 40 Administration You want to retrieve all employees, whether or not they have matching departments in the departments table. Which query would you use?（）ASELECT last_name, department_name FROM employees , departments(+);BSELECT last_name, department_name FROM employees JOIN departments(+);CSELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dDSELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);ESELECT last_name, department_name FROM employees(+) , departments ON (e. department _ id = d. department _id);FSELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e. department _ id = d. department _id);
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use?（）ASELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ;BSELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id;CSELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d;DSELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );
多选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table You created a sequence called EMP_ID_SEQ in order to populate sequential values for the EMPLOYEE_ID column of the EMPLOYEES table. Which two statements regarding the EMP_ID_SEQ sequence are true? （）AYou cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.BThe EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.CThe EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.DAny other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.EThe EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.FThe EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
单选题Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. EMPLOYEES NOT NULL, Primary EMPLOYEE_ID NUMBER Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, DEPARTMENT_ID NUMBER Primary Key VARCHAR2 DEPARTMENT_NAME |30| References MGR_ID column MGR_ID NUMBER of the EMPLOYEES table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER For which situation would you use a nonequijoin query?（）ATo find the tax percentage for each of the employees.BTo list the name, job id, and manager name for all the employees.CTo find the name, salary, and department name of employees who are not working with Smith.DTo find the number of employees working for the Administrative department and earning less then 4000.ETo display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.