考题
You create a Web Form that displays a GridView. The GridViews data source is a DataSet named dsOrders. The DataSet contains two DataTables named Orders and OrderDetails. You create a relation between the two DataTables using the following code segment. (Line numbers are included for reference only.)01 dtOrders = dsOrders.Tables[“Orders”]; 02 dtOrderDetails = dsOrders.Tables[“OrderDetail”]; 03 colParent = dtOrders.Columns[“OrderID”]; 04 colChild = dtOrderDetails.Columns[“ParentOrderID”]; 05 dsOrders.Relations.Add(“Rell”, colParent, colChild, false); You need to find the cause of the exception being raised in line 05. What should you do? ()A. Ensure that the child column and the parent column have the same names.B. Ensure that the child table and the parent table have the same names.C. Ensure that the child column and the parent column have the same data types.D. Ensure that each row in the child table has a corresponding row in the parent table.E. Ensure that the tables have an explicit relationship defined by a foreign key constraint in the database.
考题
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【 】UPDATE(salary) ON employee TO U1,U2,U3
考题
Evaluate this SQL statement:SELECT e.employee_id, (.15* e.salary) + (.5 * e.commission_pct) + (s.sales amount * (.35 * e.bonus)) AS CALC_VALUEFROM employees e, sales sWHERE e.employee_id = s.emp_id;What will happen if you remove all the parentheses from the calculation? ()A. The value displayed in the CALC_VALUE column will be lower.B. The value displayed in the CALC_VALUE column will be higher.C. There will be no difference in the value displayed in the CALC_VALUE column.D. An error will be reported.
考题
The EMPLOYEES table contains these columns:EMPLOYEE_ID NUMBER(4)LAST_NAME VARCHAR2 (25)JOB_ID VARCHAR2(10)You want to search for strings that contain ‘SA_‘ in the JOB_ID column. Which SQL statement do you use?()A. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE ‘%SA\_%‘ESCAPE‘\‘;B. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE ‘%SA_‘;C. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE ‘%SA_‘ESCAPE‘\‘;D. SELECT employee_id, last_name, job_id FROM employees WHERE job_id ‘%SA_‘;
考题
阅读以下说明C++代码,将应填入(n)处的字句写在对应栏内。[说明]本程序实现了雇员信息管理功能,其中封装了雇员信息及其设置、修改、删除操作。已知当输入为“Smith 31 2960.0”时,程序的输出是:姓名:Smith 年龄:31 工资:2960姓名:Smith 年龄:31 工资:3500姓名:Mary 年龄:23 工资:2500[C++程序]include <iostream.h>include <string.h>class employee{char *name; //雇员姓名short age; //年龄float salary;//工资public:employee();void set_name(char *);void set_age(short a) {age=a;}void set_salary(float s) {salary=s;}(1);~ employee(){delete[] name;}};employee::employee() { name="";age=0;salary=0.0;void employee::set_name(char *n){ name=new char[strlen(n)+1];(2) (name,n);}void employee::print(){ cout<<"姓名":"<<name<<" 年龄:"<<agc<<" 工资:" <<salary<<endl;}void main(){ char *na;short ag=0;float sa=0;(3);na=new char[10];cin>>na>>ag>>sa;emp.set_name(na);emp.set_age(ag);emp.set_salary(sa);emp.print();(4) (3500.0);emp.print();(5);emp.set_name("Mary");emp.set_age(23);emp.set_salary(2500.0);emp.print();}
考题
回收用户U1,U2和U3在关系employee的salary属性上的UPDATE权限的语句是【 】UPDATE(salary) ON employee FROM Ul,U2,U3。
考题
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 ManagerFROM employees e JOIN employees m ON (e.mgr_id = m.employee_id)AND e.salary 4000;What is its output?()A.AB.BC.CD.DE.E
考题
Evaluate this SQL statement:e.employee_id, (.15* e.salary) + (.5 * e.commission_pct)+ (s.sales_amount * (.35 * e.bonus)) AS CALC_VALUEFROM employees e, sales sWHERE e.employee_id = s.emp_id;What will happen if you remove all the parentheses from the calculation?()A.The value displayed in the CALC_VALUE column will be lower.B.The value displayed in the CALC_VALUE column will be higher.C.There will be no difference in the value displayed in the CALC_VALUE column.D.An error will be reported.
考题
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;