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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?（）
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);
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更多 “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?（）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);” 相关考题
考题
在执行语句SELECT department_id FROM employees,departments WHERE employees.department_id= departments.department_id;时报错,原因是()。
A.没有给表employees和表departments加别名B.没有给列department_id加别名C.不能用employees.department_id=departments.department_id作为条件D.SELECT后面的department_id没有指定是哪个表
考题
ExhibitExamine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their manager‘s last names and their department names. Which query would you use?（）A. SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN department d ON (e.department_id = d.department_id);B. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.managaer_id = m.employee_id) LEFT OUTER JOIN department d ON (e.department_id = d.department_id);C. SELECT e.last_name, m.last_name, department_name FROM employees e RIGT OUTER JOIN employees m on ( e.manager_id = m.employee_id) FULL OUTER JOIN department d ON (e.department_id = d.department_id);D. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGT OUTER JOIN department d ON (e.department_id = d.department_id);E. SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id)F. SELECT last_name, manager_id, department_name FROM employees e JOIN department d ON (e.department_id = d.department_id);
考题
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.
考题
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.
考题
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.
考题
Examine the data in the EMPLOYEES and DEPARTMENTS tables.EMPLOYEESLAST_NAME DEPARTMENT_ID SALARYGetz 10 3000Davis 20 1500Bill 20 2200Davis 30 5000Kochhar 5000DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Sales20 Marketing30 Accounts40 AdministrationYou 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 , departments(+);B. SELECT last_name, department_name FROM employees JOIN departments(+);C. SELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dD. 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(+) , 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);
考题
Click the Exhibit button and examine the data in the EMPLOYEES table.Which three subqueries work? ()
A.SELECT * FROM employees where salary (SELECT MIN(salary) FROM employees GROUP BY department_id);B.SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);C.SELECT distinct department_id FROM employees WHERE salary ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);D.SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);E.SELECT last_name FROM employees WHERE salary ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);F.SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));
考题
Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their managers‘ last names and their department names. Which query would you use?（）A.SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN departments d ON (e.department_id = d.department_id);B.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);C.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);D.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);E.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);F.SELECT last_name, manager_id, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id) ;
考题
Click the Exhibit button to examine the structures 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 employeesB.to list the name, job_id, and manager name for all the employeesC.to find the name, salary, and the department name of employees who are not working with SmithD.to find the number of employees working for the Administrative department and earning less than 4000E.to display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned
考题
Click the Exhibit button and examine the data from the EMP table.The COMMISSION column shows the monthly commission earned by the employee.Which two tasks would require subqueries or joins in order to be performed in a single step? （）A.listing the employees who earn the same amount of commission as employee 3B.finding the total commission earned by the employees in department 10C.finding the number of employees who earn a commission that is higher than the average commission of the companyD.listing the departments whose average commission is more than 600E.listing the employees who do not earn commission and who are working for department 20 in descending order of the employee IDF.listing the employees whose annual commission is more than 6000
考题
在执行语句SELECT department_id FROM employees，departments WHERE employees.department_id= departments.department_id；时报错，原因是（）。A、没有给表employees和表departments加别名B、没有给列department_id加别名C、不能用employees.department_id=departments.department_id作为条件D、SELECT后面的department_id没有指定是哪个表
考题
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?（）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);
考题
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? （）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);
考题
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 );
考题
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?（）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);
考题
Click the Exhibit button and examine the data in the EMPLOYEES table. Which three subqueries work?（）A、SELECT * FROM employees where salary (SELECT MIN(salary) FROM employees GROUP BY department_id);B、SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);C、SELECT distinct department_id FROM employees WHERE salary ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);D、SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);E、SELECT last_name FROM employees WHERE salary ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);F、SELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));
考题
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?（）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.
考题
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?（）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;
考题
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? （）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);
考题
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?（）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;
考题
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?（）A
SELECT last_name, department_name FROM employees , departments(+);B
SELECT last_name, department_name FROM employees JOIN departments(+);C
SELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dD
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(+) , 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);
考题
单选题在执行语句SELECT department_id FROM employees，departments WHERE employees.department_id= departments.department_id；时报错，原因是（）。A
没有给表employees和表departments加别名B
没有给列department_id加别名C
不能用employees.department_id=departments.department_id作为条件D
SELECT后面的department_id没有指定是哪个表
考题
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?（）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);
考题
多选题Click the Exhibit button and examine the data in the EMPLOYEES table. Which three subqueries work?（）ASELECT * FROM employees where salary (SELECT MIN(salary) FROM employees GROUP BY department_id);BSELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);CSELECT distinct department_id FROM employees WHERE salary ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);DSELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);ESELECT last_name FROM employees WHERE salary ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);FSELECT department_id FROM employees WHERE salary ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));
考题
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 );
考题
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?（）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);
考题
单选题Exhibit: Examine the data in the EMPLOYEES table. Examine the subquery: SELECT last_name FROM employees WHERE salary IN (SELECT MAX(salary) FROM employees GROUP BY department_id); Which statement is true?（）A
The SELECT statement is syntactically accurate.B
The SELECT statement does not work because there is no HAVING clause.C
The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.D
The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.
考题
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?（）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.
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