资料：HS ensures there is a clear career path for any employee, from any background. Three key levels in Harrods are the sales employee, department managers and senior managers. At each level, employees can benefit from the development programs in order to build a career.James is a sales Associate and one of HS first Sales Degree students. When an injure prevented him from following his previously chosen career in contemporary dance, he applied to HS. He has never looked back. HS training has given him transferable skills. He has been able to work within more than one department, providing the same high levels of experience.What are the three key levels in HS?A.the sales employees, department managers and senior managers.B.the sales employees,sales assistants and department managers.C.the senior managers,department managers and sales employees.D.not mentioned.
资料：HS ensures there is a clear career path for any employee, from any background. Three key levels in Harrods are the sales employee, department managers and senior managers. At each level, employees can benefit from the development programs in order to build a career.
James is a sales Associate and one of HS first Sales Degree students. When an injure prevented him from following his previously chosen career in contemporary dance, he applied to HS. He has never looked back. HS training has given him transferable skills. He has been able to work within more than one department, providing the same high levels of experience.
What are the three key levels in HS?
James is a sales Associate and one of HS first Sales Degree students. When an injure prevented him from following his previously chosen career in contemporary dance, he applied to HS. He has never looked back. HS training has given him transferable skills. He has been able to work within more than one department, providing the same high levels of experience.
What are the three key levels in HS?
A.the sales employees, department managers and senior managers.
B.the sales employees,sales assistants and department managers.
C.the senior managers,department managers and sales employees.
D.not mentioned.
B.the sales employees,sales assistants and department managers.
C.the senior managers,department managers and sales employees.
D.not mentioned.
参考解析
解析：本题考查的是细节理解。
【关键词】what three key levels; HS
【主题句】第1自然段Three key levels in Harrods are the sales employee, department managers and senior managers. 哈罗德百货公司的三个关键级别是销售人员、部门经理和高级经理。
【解析】本题的问题是“HS的三个关键级别是什么？”。根据主题句可知，HS的三个关键级别是销售人员、部门经理和高级经理，故选A。
【关键词】what three key levels; HS
【主题句】第1自然段Three key levels in Harrods are the sales employee, department managers and senior managers. 哈罗德百货公司的三个关键级别是销售人员、部门经理和高级经理。
【解析】本题的问题是“HS的三个关键级别是什么？”。根据主题句可知，HS的三个关键级别是销售人员、部门经理和高级经理，故选A。
相关考题：
(b) Identify and discuss the appropriateness of the cost drivers of any TWO expense values in EACH of levels (i)to (iii) above and ONE value that relates to level (iv).In addition, suggest a likely cause of the cost driver for any ONE value in EACH of levels (i) to (iii), andcomment on possible benefits from the identification of the cause of each cost driver. (10 marks)
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 keyDEPARTMENT_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.
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 ANG (SALARY));
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.
Examine the data of the EMPLOYEES table.EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID)Which statement lists the ID, name, and salary of the employee, and the ID and name of the employee‘s manager, for all the employees who have a manager and earn more than 4000?（）A.B.C.D.E.
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 table:LAST_NAME DEPARTMENT_ID SALARYGetz 10 3000Davis 20 1500Bill 20 2200Davis 30 5000...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 ANG (SALARY));
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));
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) ;
资料：HS ensures there is a clear career path for any employee, from any background. Three key levels in Harrods are the sales employee, department managers and senior managers. At each level, employees can benefit from the development programs in order to build a career.James is a sales Associate and one of HS first Sales Degree students. When an injure prevented him from following his previously chosen career in contemporary dance, he applied to HS. He has never looked back. HS training has given him transferable skills. He has been able to work within more than one department, providing the same high levels of experience.Why did James give up his previous career?A.because he wanted to make a change of his life.B.because ge got hurt and could't dance any more.C.because he wanted to take the courses.D.none above.
What type of constraint is used to ensure that each row inserted into the EMPLOYEE table with a value in the WORKDEPT column has a row with a corresponding value in the DEPTNO column of the DEPARTMENT table?（）A、A check constraint on the EMPLOYEE tableB、A unique constraint on the EMPLOYEE table WORKDEPT columnC、A foreign key reference from the DEPARTMENT tables DEPTNO column to the WORKDEPT column of the EMPLOYEE tableD、A foreign key reference from the EMPLOYEE tables WORKDEPT column to the DEPTNO column of the DEPARTMENT table
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables: EMPLOYEES EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE NEW_EMPLOYEES EMPLOYEE_ID NUMBER Primary Key NAME VARCHAR2 (60) Which DELETE statement is valid?（）A、DELETE FROM employees WHERE employee_id = (SELECT employee_id FROM employees);B、DELETE * FROM employees WHERE employee_id = (SELECT employee_id FROM new_ employees);C、DELETE FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = 'carrey');D、DELETE * FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = 'carrey');
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.
Examine the data in the EMPLOYEES table: LAST_NAME DEPARTMENT_ID SALARY Getz 10 3000 Davis 20 1500 Bill 20 2200 Davis 30 5000 ... 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 ANG (SALARY));
多选题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 ANG (SALARY));
单选题Examine the structure if the EMPLOYEES table: Column name Data Type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2(30) JOB_ID VARCHAR2(20) NOT NULL SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table You need to create a view called EMP_VU that allows the user to insert rows through the view. Which SQL statement, when used to create the EMP_VU view, allows the user to insert rows?（）ACREATE VIEW emp_Vu AS SELECT employee_id, emp_name, department_id FROM employees WHERE mgr_id IN (102, 120);BCREATE VIEW emp_Vu AS SELECT employee_id, emp_name, job_id department_id FROM employees WHERE mgr_id IN (102, 120);CCREATE VIEW emp_Vu AS SELECT department_id, SUM(sal) TOTALSAL FROM employees WHERE mgr_id IN (102, 120) GROUP BY department_ id;DCREATE VIEW emp_Vu AS SELECT employee_id, emp_name, job_id, DISTINCT department_id FROM employees;
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.
单选题您在公司的数据库中成功创建了名为SALARY的表。您现在要通过向引用EMPLOYEES表的匹配列的SALARY表添加FOREIGNKEY约束条件来建立EMPLOYEES表与SALARY表之间的父/子关系。尚未向SALARY表添加任何数据。应执行以下哪条语句（）AALTER TABLE salary ADD CONSTRAINT fk_employee_id_01 FOREIGN KEY（employee_id）REFERENCES employees（employee_id）BALTER TABLE salary ADD CONSTRAINT fk_employee_id_ FOREIGN KEY BETWEEN salary（employee_id）AND employees（employee_id）CALTER TABLE salary FOREIGN KEY CONSTRAINT fk_employee_id_REFERENCES employees（employee_id）DALTER TABLE salary ADD CONSTRAINT fk_employee_id_FOREIGN KEY salary（employee_id）=employees（employee_id）
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?（）AALTER VIEW EMP_dept_vu (ADD manger_id NUMBER);BMODIFY VIEW EMP_dept_vu (ADD manger_id NUMBER);CALTER 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;DMODIFY 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;ECREATE 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;FYou must remove the existing view first, and then run the CREATE VIEW command with a new column list to modify a view.
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);
单选题Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2(30) JOB_ID VARCHAR2(20) NOT NULL SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table You need to create a view called EMP_VU that allows the users to insert rows through the view. Which SQL statement, when used to create the EMP_VU view, allows the users to insert rows?（）ACREATE VIEW emp_vu AS SELECT employee_id, emp_name, department_id FROM employees WHERE mgr_id IN (102, 120);BCREATE VIEW emp_vu AS SELECT employee_id, emp_name, job_id, department_id FROM employees WHERE mgr_id IN (102, 120);CCREATE VIEW emp_vu AS SELECT department_id, SUM(sal) TOTALSAL FROM employees WHERE mgr_id IN (102, 120) GROUP BY department_id;DCREATE VIEW emp_vu AS SELECT employee_id, emp_name, job_id, DISTINCT department_id FROM employees;
单选题Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables: EMPLOYEES EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2 (25) LAST_NAME VARCHAR2 (25) HIRE_DATE DATE NEW EMPLOYEES EMPLOYEE_ID NUMBER Primary Key NAME VARCHAR2 (60) Which DELETE statement is valid? （）ADELETE FROM employees WHERE employee_id = (SELECT employee_id FROM employees);BDELETE * FROM employees WHERE employee_id = (SELECT employee_id FROM new_ employees);CDELETE FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = ('Carrey')'DDELETE * FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE last_ name = ('Carrey')'
单选题If Birmingham City Council plans to move an employee to a new job, it will definitely make sure that ______.Athere is continuity between the two jobsBno complaints from the employee occurCthe amount of work is reduced for the new jobDthe employee is prepared for any mental problems
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);
单选题What type of constraint is used to ensure that each row inserted into the EMPLOYEE table with a value in the WORKDEPT column has a row with a corresponding value in the DEPTNO column of the DEPARTMENT table?（）AA check constraint on the EMPLOYEE tableBA unique constraint on the EMPLOYEE table WORKDEPT columnCA foreign key reference from the DEPARTMENT tables DEPTNO column to the WORKDEPT column of the EMPLOYEE tableDA foreign key reference from the EMPLOYEE tables WORKDEPT column to the DEPTNO column of the DEPARTMENT table
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.