is paid weekly
is adjusted every quarter
is re-examined from year to year
is fixed for the whole contract period
第1题:
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.
第2题:

A. The statement produces an error at line 1.
B. The statement produces an error at line 3.
C. The statement produces an error at line 6.
D. The statement returns the employee name, salary, department ID, and maximum salary earned in the department of the employee for all departments that pay less salary then the maximum salary paid in the company.
E. The statement returns the employee name, salary, department ID, and maximum salary earned in the department of the employee for all employees who earn less than the maximum salary in their department.
第3题:
回收用户U1,U2和U3在关系employee的salary属性上的UPDATE限的语句是
A.REVOKE UPDATE(salary) ON employee(U1,U2,U3)
B.REVOKE UPDATE(salary)ON employee FROM U1,U2,U3
C.REVOKE UPDATE ON employee FROM U1,U2,U3
D.REVOKE UPDATE(salary) FROM U1,U2,U3
第4题:
Click the Exhibit button to examine the data of the EMPLOYEES table.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.SELECT employee_id "Emp_id", emp_name "Employee", salary, employee_id "Mgr_id", emp_name "Manager" FROM employees WHERE salary > 4000;
B.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m WHERE e.mgr_id = m.mgr_id AND e.salary > 4000;
C.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m ON (e.mgr_id = m.employee_id) AND e.salary > 4000;
D.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id", m.emp_name "Manager" FROM employees e SELF JOIN employees m WHERE e.mgr_id = m.employee_id AND e.salary > 4000;
E.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id" m.emp_name "Manager" FROM employees e JOIN employees m USING (e.employee_id = m.employee_id) AND e.salary > 4000;
第5题:
() is the operation to check the quantity, quality and package of the goods according to the contract and the specific standard
第6题:
Evaluate the SQL statement: SELECT LPAD(salary,10,*) FROM EMP WHERE EMP_ID = 1001; If the employee with the EMP_ID 1001 has a salary of 17000, what is displayed?()
第7题:
Evaluate the SQL statement: SELECT LPAD (salary,10,*) FROM EMP WHERE EMP _ ID = 1001; If the employee with the EMP_ID 1001 has a salary of 17000, what is displayed?()
第8题:
CREATE INDEX my_idx_1 ON employee(salary*1.12)
CREATE UNIQUE INDEX my_idx_1 ON employee(salary)
CREATE BITMAP INDEX my_idx_1 ON employee(salary)
CREATE INDEX my_idx_1 ON employee(salary)REVERSE
第9题:
SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary BETWEEN t.min _ salary AND t.max_salary
SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary > t.min_salary, tax_percent
SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE MIN(e.salary) = t.min_salary AND MAX(e.salary) = t.max_salary
You cannot find the information because there is no common column between the two tables.
第10题:
The statement produces an error at line 1.
The statement produces an error at line 3.
The statement produces an error at line 6.
The statement returns the employee name, salary, department ID, and maximum salary earned in the department of the employee for all departments that pay less salary then the maximum salary paid in the company.
The statement returns the employee name, salary, department ID, and maximum salary earned in the department of the employee for all employees who earn less than the maximum salary in their department.
第11题:
Flashback Drop
Flashback Table
Flashback Database
Flashback Version Query
第12题:
SELECT employee_id Emp_id, emp_name Employee, salary, employee_id Mgr_id, emp_name Manager FROM employees WHERE salary > 4000;
SELECT e.employee_id Emp_id, e.emp_name Employee,
salary,
employee_id Mgr_id, m.emp_name Manager FROM employees e, employees m WHERE e.mgr_id = m.mgr_id AND e.salary > 4000;
第13题:
Examine the data in the EMPLOYEES and EMP_HIST tables:EMPLOYEESNAME 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 6500EMP HISTEMPLOYEE_ID NAME JOB_ID SALARY101 Smith SA_CLERK 2000103 Chris IT_CLERK 2200104 John HR_CLERK 2000106 Smith AD_ASST 3000108 Jennifer HR_MGR 4500The EMP_HIST table is updated at the end of every year. The employee ID, name, job ID, and salary of each existing employee are modified with the latest data. New employee details are added to the table.Which statement accomplishes this task?()
A. UPDATE emp_hist SET employee_id, name, job_id, salary = (SELECT employee_id, name, job_id, salary FROM employees) WHERE employee_id IN (SELECT employee_id FROM employees);
B. MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT VALUES (e.employee id, e.name, job id, e.salary);
C. MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE emp hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);
D. MERGE INTO emp_hist eh USING employees e WHEN MATCHED THEN UPDATE emp_hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);
第14题:
授予用户U1,U2和U3在关系employee的salary属性上的UPDATE权限的语句是
A.GRANT ON employee TOU1,U2,U3
B.GRANT UPDATE ON employee TO U1,U2,U3
C.GRANT UPDATE(salary) ON employee
D.GRANT UPDATE(salary)ON employee TO U1,U2,U3
第15题:
Click the Exhibit button to examine the data of the EMPLOYEES table. Evaluate this SQL statement:SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager"FROM employees e JOIN employees m ON (e.mgr_id = m.employee_id)AND e.salary > 4000;What is its output?()


A.A
B.B
C.C
D.D
E.E
第16题:
A.CREATE INDEX my_idx_1 ON employee(salary*1.12)
B.CREATE UNIQUE INDEX my_idx_1 ON employee(salary)
C.CREATE BITMAP INDEX my_idx_1 ON employee(salary)
D.CREATE INDEX my_idx_1 ON employee(salary)REVERSE
第17题:
Application A is designed to execute the following SQL statements within a single Unit of Work (UOW). UPDATE employee SET salary = salary * 1.1 WHERE empno='000010' UPDATE department SET deptname = 'NEW dept' WHERE deptno='A00'Application B is designed to execute the following SQL statements within a single Unit of Work (UOW). UPDATE department SET deptname = 'OLD DEPT' WHERE deptno='A00' UPDATE employee SET salary = salary * 0.5 WHERE empno='000010' Application A and application B execute their first SQL statement at the same time. When application A and application B try to execute their second SQL statement, a deadlock occurs. What will happen?()
第18题:
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) 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 HR_MGR 5000 106 Bryan 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 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?()
第19题:
You work as an application developer at Certkiller .com. You have been given the responsibility of creating a class named CalcSalary that will determine the salaries of Certkiller .com’s staff. The CalcSalary class includes methods to increment and decrement staff salaries. The following code is included in the CalcSalary class: public class CalcSalary { // for promotions public static bool IncrementSalary (Employee Emp, double Amount) { if (Emp.Status == QuarterlyReview.AboveGoals) Emp.Salary += Amount; return true; } else return false; } //for demotions public static bool DecrementSalary (Employee Emp, double Amount) { if (Emp.Status == QuarterlyReview.AboveGoals) Emp.Salary -= Amount; return true; } else return false; } } You would like to invoke the IncrementSalary and DecrementSalary methods dynamically at runtime from the sales manager application, and decide to create a delegate named SalaryDelegate to invoke them. You need to ensure that you use the appropriate code to declare the SalaryDelegate delegate.What is the correct line of code?()
第20题:
ALTER TABLE salary ADD CONSTRAINT fk_employee_id_01 FOREIGN KEY(employee_id)REFERENCES employees(employee_id)
ALTER TABLE salary ADD CONSTRAINT fk_employee_id_ FOREIGN KEY BETWEEN salary(employee_id)AND employees(employee_id)
ALTER TABLE salary FOREIGN KEY CONSTRAINT fk_employee_id_REFERENCES employees(employee_id)
ALTER TABLE salary ADD CONSTRAINT fk_employee_id_FOREIGN KEY salary(employee_id)=employees(employee_id)
第21题:
is paid weekly
is adjusted every quarter
is re-examined from year to year
is fixed for the whole contract period
第22题:
Inspection
Examination
Control
Test
第23题:
public delegate bool Salary (Employee Emp, double Amount);
public bool Salary (Employee Emp, double Amount);
public event bool Salary (Employee Emp, double Amount);
public delegate void Salary (Employee Emp, double Amount);
第24题:
UPDATE emp_hist SET employee_id, name, job_id, salary = (SELECT employee_id, name, job_id, salary FROM employees) WHERE employee_id IN (SELECT employee_id FROM employees);
MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT VALUES (e.employee id, e.name, job id, e.salary);
MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE emp hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);
MERGE INTO emp_hist eh USING employees e WHEN MATCHED THEN UPDATE emp_hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);