单选题Evaluate this SQL statement: SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME FROM EMP e, DEPARTMENT d WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID; In the statement, which capabilities of a SELECT statement are performed?（）A Selection, p

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You need to create a view EMP_VU. The view should allow the users to manipulate the records of only the employees that are working for departments 10 or 20.Which SQL statement would you use to create the view EMP_VU?（）

• A、CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20);
• B、CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) WITH READ ONLY;
• C、CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) WITH CHECK OPTION;
• D、CREATE FORCE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20);
• E、CREATE FORCE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) NO UPDATE;

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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? （）

• A、SELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);
• B、SELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;
• C、SELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;
• D、SELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);

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You need to display the last names of those employees who have the letter "A" as the second character in their names. Which SQL statement displays the required results? （）

• A、SELECT last_name FROM EMP WHERE last_ name LIKE '_A%';
• B、SELECT last_name FROM EMP WHERE last name ='*A%'
• C、SELECT last_name FROM EMP WHERE last name ='_A%';
• D、SELECT last_name FROM EMP WHERE last name LIKE '*A%'

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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?（）

• A、SELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);
• B、SELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;
• C、SELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;
• D、SELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);

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Evaluate this SQL statement: SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME FROM EMPLOYEES e, DEPARTMENTS d WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID; In the statement, which capabilities of a SELECT statement are performed?（）

• A、selection, projection, join
• B、difference, projection, join
• C、selection, intersection, join
• D、intersection, projection, join
• E、difference, projection, product

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Evaluate this SQL statement: SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME FROM EMP e, DEPARTMENT d WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID; In the statement, which capabilities of a SELECT statement are performed?（）

• A、Selection, projection, join
• B、Difference, projection, join
• C、Selection, intersection, join
• D、Intersection, projection, join
• E、Difference, projection, product

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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?（）

• A、SELECT last_name, department_name FROM employees NATURAL JOIN departments;
• B、SELECT last_name, department_name FROM employees JOIN departments ;
• C、SELECT last_name, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id);
• D、SELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);
• E、SELECT last_name, department_name FROM employees FULL JOIN departments ON (e.department_id = d.department_id);
• F、SELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

第16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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Evaluate this SQL statement: SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME FROM EMP e, DEPARTMENT d WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID; In the statement, which capabilities of a SELECT statement are performed?（）

• A、Selection, projection, join
• B、Difference, projection, join
• C、Selection, intersection, join
• D、Intersection, projection, join
• E、Difference, projection, product

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The user Sue issues this SQL statement: GRANT SELECT ON sue. EMP TO alice WITH GRANT OPTION; The user Alice issues this SQL statement: GRANT SELECT ON sue. EMP TO reena WITH GRANT OPTION; The user Reena issues this SQL statement: GRANT SELECT ON sue. EMP TO timber; The user Sue issues this SQL statement: REVOKE select on sue. EMP FROM alice; For which users does the revoke command revoke SELECT privileges on the SUE.EMP table?（）

• A、Alice only
• B、Alice and Reena
• C、Alice, Reena, and Timber
• D、Sue, Alice, Reena, and Timber

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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?（）

• A、SELECT last_name, department_name, location_id FROM employees , department ;
• B、SELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
• C、SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;
• D、SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;

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