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EMPLOYEES and DEPARTMENTS data: EMPLOYEES DEMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_I 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 On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID managers and refers to the EMPLOYEE_ID. On the DEPARTMENTS table DEPARTMENT_ID is the primary key. Evaluate this UPDATE statement. UPDATE employees SET mgr_id = (SELECT mgr_id FROM employees WHERE dept_id= (SELECT department_id FROM departments WHERE department_name = 'Administration')), Salary = (SELECT salary FROM employees WHERE emp_name = 'Smith') WHERE job_id = 'IT_ADMIN'; What happens when the statement is executed?（）
- A、The statement executes successfully, leaves the manager ID as the existing value, and changes the salary to 4000 for the employees with ID 103 and 105.
- B、The statement executes successfully, changes the manager ID to NULL, and changes the salary to 4000 for the employees with ID 103 and 105.
- C、The statement executes successfully, changes the manager ID to NULL, and changes the salary to 3000 for the employees with ID 103 and 105.
- D、The statement fails because there is more than one row matching the employee name Smith.
- E、The statement fails because there is more than one row matching the IT_ADMIN job ID in the EMPLOYEES table.
- F、The statement fails because there is no 'Administration' department in the DEPARTMENTS table.
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更多 “ EMPLOYEES and DEPARTMENTS data: EMPLOYEES DEMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_I 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 On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID managers and refers to the EMPLOYEE_ID. On the DEPARTMENTS table DEPARTMENT_ID is the primary key. Evaluate this UPDATE statement. UPDATE employees SET mgr_id = (SELECT mgr_id FROM employees WHERE dept_id= (SELECT department_id FROM departments WHERE department_name = 'Administration')), Salary = (SELECT salary FROM employees WHERE emp_name = 'Smith') WHERE job_id = 'IT_ADMIN'; What happens when the statement is executed?（）A、The statement executes successfully, leaves the manager ID as the existing value, and changes the salary to 4000 for the employees with ID 103 and 105.B、The statement executes successfully, changes the manager ID to NULL, and changes the salary to 4000 for the employees with ID 103 and 105.C、The statement executes successfully, changes the manager ID to NULL, and changes the salary to 3000 for the employees with ID 103 and 105.D、The statement fails because there is more than one row matching the employee name Smith.E、The statement fails because there is more than one row matching the IT_ADMIN job ID in the EMPLOYEES table.F、The statement fails because there is no 'Administration' department in the DEPARTMENTS table.” 相关考题
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在执行语句SELECT department_id FROM employees,departments WHERE employees.department_id= departments.department_id;时报错,原因是()。
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You added a PHONE_NUMBER column of NUMBER data type to an existing EMPLOYEES table. The EMPLOYEES table already contains records of 100 employees. Now, you want to enter the phone numbers of each of the 100 employees into the table. Some of the employees may not have a phone number available. Which data manipulation operation do you perform? （）A. MERGEB. INSERTC. UPDATED. ADDE. ENTERF. You cannot enter the phone numbers for the existing employee records.
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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.
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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?()
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Examine the structure of the EMPLOYEES and DEPARTMENTS tables:Evaluate this SQL statement:Which SQL statement is equivalent to the above SQL statement?()
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Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.EMPLOYEESNOT NULL,EMPLOYEE_ID NUMBERPrimary KeyVARCHAR2EMP_NAME(30)VARCHAR2JOB_ID(20)SALARY NUMBERReferencesMGR_ID NUMBEREMPLOYEE_IDcolumnDEPARTMENT_ID NUMBER Foreign key toDEPARTMENT_IDcolumn of theDEPARTMENTStableDEPARTMENTSNOT NULL, PrimaryDEPARTMENT_ID NUMBERKeyVARCHAR2DEPARTMENT_NAME(30)References NGR_IDMGR_ID NUMBERcolumn ofthe EMPLOYEES tableForeign key toLOCATION_ID NUMBERLOCATION_IDcolumn of theLOCATIONS tableLOCATIONSNOT NULL, PrimaryLOCATION_ID NUMBERKeyVARCHAR2CITY|30)Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?（）A. SELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary 10000;B. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary 10000;C. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary 10000;D. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary 10000;E. SELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary 10000;
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: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?（）
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 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
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You added a PHONE_NUMBER column of NUMBER data type to an existing EMPLOYEES table. The EMPLOYEES table already contains records of 100 employees. Now, you want to enter the phone numbers of each of the 100 employees into the table. Some of the employees may not have a phone number available. Which data manipulation operation do you perform?（）A、MERGEB、INSERTC、UPDATED、ADDE、ENTERF、You cannot enter the phone numbers for the existing employee records.
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
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A group of hosts are physically connected to the same switch. The hosts are used by employees of different departments and therefore do not need to directly exchange data on a regular basis.Which technology can a network administrator deploy to reduce unnecessary broadcast traffic between these hosts?（）A、Micro segmentationB、Transparent switchingC、peer-to-peer networkingD、port securityE、store-and-forward switchingF、virtual local area networks
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
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?（）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
单选题A group of hosts are physically connected to the same switch. The hosts are used by employees of different departments and therefore do not need to directly exchange data on a regular basis.Which technology can a network administrator deploy to reduce unnecessary broadcast traffic between these hosts?（）A
Micro segmentationB
Transparent switchingC
peer-to-peer networkingD
port securityE
store-and-forward switchingF
virtual local area networks
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You created a view called EMP_DEPT_VU that contains three columns from the EMPLOYEES and DEPARTMENTS tables: EMPLOYEE_ID, EMPLOYEE_NAME AND DEPARTMENT_NAME. The DEPARTMENT_ID column of the EMPLOYEES table is the foreign key to the primary key DEPARTMENT_ID column of the DEPARTMENTS table. You want to modify the view by adding a fourth column, MANAGER_ID of NUMBER data type from the EMPLOYEES tables. How can you accomplish this task?（）A
ALTER VIEW EMP_dept_vu (ADD manger_id NUMBER);B
MODIFY VIEW EMP_dept_vu (ADD manger_id NUMBER);C
ALTER VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employee e, departments d WHERE e.department _ id = d.department_id;D
MODIFY VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employees e, departments d WHERE e.department _ id = d.department_id;E
CREATE OR REPLACE VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employees e, departments d WHERE e.department _ id = d.department_id;F
You must remove the existing view first, and then run the CREATE VIEW command with a new column list to modify a view.
考题
多选题Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. EMPLOYEES NOT NULL, EMPLOYEE_ID NUMBER Primary 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, Primary DEPARTMENT_ID NUMBER Key VARCHAR2 DEPARTMENT_NAME (30) References NGR_ID MGR_ID NUMBER column of the EMPLOYEES table Foreign key to LOCATION_ID NUMBER LOCATION_ID column of the LOCATIONS table LOCATIONS NOT NULL, Primary LOCATION_ID NUMBER Key VARCHAR2 CITY |30) Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?（）ASELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary 10000;BSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary 10000;CSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary 10000;DSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary 10000;ESELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary 10000;
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You added a PHONE_NUMBER column of NUMBER data type to an existing EMPLOYEES table. The EMPLOYEES table already contains records of 100 employees. Now, you want to enter the phone numbers of each of the 100 employees into the table. Some of the employees may not have a phone number available. Which data manipulation operation do you perform?（）A
MERGEB
INSERTC
UPDATED
ADDE
ENTERF
You cannot enter the phone numbers for the existing employee records.
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
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?（）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, 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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