下面的XML片断中结构完整的是（）。A、＜customer name=”＜xml＞.con”＞＜address＞123 MainStreet＞＜/address＞＜/customer＞B、＜customer＞＜name＞Joe’s XML Works＜/name＞＜address＞New York＜/costomer＞C、＜ customer type=extemal＞＜name＞Partners Unlimited＜/name＞＜/customer＞D、＜customer name=”John Doe”

题目

下面的XML片断中结构完整的是（）。

A、＜customer name=”＜xml＞.con”＞＜address＞123 MainStreet＞＜/address＞＜/customer＞
B、＜customer＞＜name＞Joe’s XML Works＜/name＞＜address＞New York＜/costomer＞
C、＜ customer type=extemal＞＜name＞Partners Unlimited＜/name＞＜/customer＞
D、＜customer name=”John Doe”＞＜address＞123 Main Street＜/address＞＜zip code=”01837”/＞＜/customer＞

相似考题

1.假设需要给某个客户表Customer添加注释信息:客户表,可以使用如下()方式A.COMMENTONTABLECUSTOMERIS'客户表'B.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAMEIS'客户表'C.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAME'客户表'D.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAME'客户表'

2.You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns:CUST_ID NUMBER(4) NOT NULLCUST_NAME VARCHAR2(100) NOT NULLCUST_ADDRESS VARCHAR2(150)CUST_PHONE VARCHAR2(20)Which SELECT statement accomplishes this task?（）A. SELECT* FROM customers;B. SELECT name, address FROM customers;C. SELECT id, name, address, phone FROM customers;D. SELECT cust_name, cust_address FROM customers;E. SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

3.阅读以下说明和 Java程序，填补代码中的空缺，将解答填入答题纸的对应栏内。【说明】以下Java代码实现一个简单客户关系管理系统（CRM）中通过工厂（CustomerFactory ）对象来创建客户（Customer）对象的功能。客户分为创建成功的客户（RealCustomer）和空客户（NullCustomer）。空客户对象是当不满足特定条件时创建或获取的对象。类间关系如图 5-1 所示。图5-1 类图【Java代码】 Abstract class Customer﹛ Protected String name; （ 1 ）boolean isNil(); （ 2 ）String getName(); ﹜ Class RealCustomer （ 3 ）Customer{ Public RealCustomer(String name){ this.name=name; } Public String getName(){ return name ; } Public boolean is Nil() { return false; } ﹜ Class NullCustomer（ 4 ）Customer﹛ Public String getName()﹛ return "Not Available in Customer Database"; ﹜ Public boolean isNil() ﹛ return true; ﹜ ﹜ class Customerfactory { public String[] names = {"Rob"，"Joe"，"Julie"}; public Customer getCustomer(String name) { for (int i = 0; i < names.length;i++) { if (names[i].（ 5 ）)﹛ return new RealCustomer(name); ﹜ ﹜ return （ 6 ）; ﹜ ﹜ Public class CrM﹛ Public viod get Customer()﹛ Customerfactory（ 7 ）; Customer customer1-cf.getCustomer("Rob"); Customer customer2=cf.getCustomer("Bob"); Customer customer3= cf.getCustomer("Julie"); Customer customer4= cf.getCustomer("Laura"); System.out.println("customers”) System.out.println(customer1.getName()); System.out.println(customer2getName()); System.out.println(customer3.getName()); System.out.println(customer4.getName()); ﹜ Public static viod main (String[]arge)﹛ CRM crm =new CRM(); Crm.getCustomer(); ﹜ ﹜ /*程序输出为： Customers rob Not Available in Customer Database Julie Not Available in Customer Database */

4.假设需要给某个客户表Customer的Customer_name列添加注释信息:客户姓名,可以使用如下()方式A.COMMENTONTABLECUSTOMERIS'客户姓名'B.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAMEIS'客户姓名'C.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAME'客户姓名'D.COMMENTONCOLUMNCUSTOMER.CUSTOMER_NAME'客户姓名'

更多“下面的XML片断中结构完整的是（）。A、＜customer name=”＜xml＞.con”＞＜address＞123 MainStreet＞＜/address＞＜/customer＞B、＜customer＞＜name＞Joe’s XML Works＜/name＞＜address＞New York＜/costomer＞C、＜ customer type=extemal＞＜name＞Partners Unlimited＜/name＞＜/customer＞D、＜customer name=”John Doe””相关问题

第1题：

阅读下列说明和C++代码，填补代码中的空缺，将解答填入答题纸的对应栏内。【说明】以下C++代码实现一个简单客户关系管理系统（CrM）中通过工厂（Customerfactory）对象来创建客户（Customer）对象的功能。客户分为创建成功的客户（realCustomer）和空客户（NullCustomer）。空客户对象是当不满足特定条件时创建或获取的对象。类间关系如图6-1所示。图6-1
【C++代码】 include<iostream> include<string> using namespace std; class Customer｛ protected: string name; public: （ 1 ） boll isNil()=0; （ 2 ） string getName()=0; ﹜; class RealCustomer ( 3 )｛ Public: realCustomer(string name)｛this->name=name;﹜ bool isNil()｛ return false; ﹜ string getName()｛ return name; ﹜ ﹜; class NullCustomer ( 4 ) ｛ public: bool isNil()｛ return true; ﹜ string getName()｛ return 〝Not Available in Customer Database〞; ﹜ ﹜; class Customerfactory｛ public: string names［3］=｛〝rob〞, 〝Joe〞,〝Julie〞﹜; public: Customer*getCustomer(string name)｛ for （int i=0;i<3;i++)｛ if （names[i].( 5 ) )｛ return new realCustomer(name); ﹜ ﹜ return ( 6 ); ﹜ ﹜; class CRM｛ public: void getCustomer()｛ Customerfactory*( 7 ); Customer*customer1=cf->getCustomer(〝Rob〞); Customer*customer2=cf->getCustomer(〝Bob〞); Customer*customer3=cf->getCustomer(〝Julie〞); Customer*customer4=cf->getCustomer(〝Laura〞); cout<<〝Customers〞<<endl; cout<<Customer1->getName()<<endl; delete customer1; cout<<Customer2->getName()<<endl; delete customer2; cout<<Customer3->getName()<<endl; delete customer3; cout<<Customer4->getName()<<endl; delete customer4; delete cf; ﹜ ﹜; int main()｛ CRM*crs=new CRM(); crs->getCustomer(); delete crs; return 0; ﹜ /*程序输出为： Customers rob Not Available in Customer Database Julie Not Available in Customer Database */

正确答案：1)virtual
2)virtual
3):public Customer
4):public Customer
5)compare(name)==0
6)new Null Customer()
7)cf=New CustomerFactory();
第2题：

BookStore数据库中有客户表Customer（CustomerCode，Name，Sex，Hometown，Email，Telephone，LoginDate，Password）查询Customer表中所有客户的信息，要求显示CustomerCode、Name、Hometown和Telephone字段信息。SQL语句：SELECT（）FROM Customer
- A、*
- B、ALL
- C、CustomerCode、Name、Hometown、Telephone
- D、CustomerCode，Name，Hometown，Telephone
正确答案:D
第3题：

BookStore数据库中有客户表Customer（CustomerCode，Name，Sex，Hometown，Email，Telephone，LoginDate，Password）查询Customer表中所有客户的信息，SQL语句：SELECT（）FROMCustomer
- A、*
- B、CustomerCode，Name，Hometown，Sex，Email，Telephone，LoginTime，PassWord
- C、CustomerCode，Hometown，Telephone，Name，Sex，Email，LoginTime，PassWord
- D、CustomerCode~PassWord
正确答案:A,B,C
第4题：

评估以下CREATETABLE语句的执行结果： CREATET ABLE customers （customer_id NUMBER，customer_name VARCHAR2（25）， address VARCHAR 2（25）， city VARCHAR 2（25）， region VARCHAR 2（25）， postal_code VARCHAR 2（11）， CONSTRAINT customer_id_un UNIQUE（customer_id）， CONSTRAINTcustomer_name_nnNOTNULL（customer_name））；为什么执行时此语句会失败（）
- A、NUMBER数据类型要求精度值
- B、UNIQUE约束条件必须在列级定义
- C、CREATETABLE语句不定义PRIMARYKEY
- D、不能在表级定义NOTNULL约束条件
正确答案:D
第5题：

Which statement adds a constraint that ensures the CUSTOMER_NAME column of the CUSTOMERS table holds a value?（）
- A、ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
- B、ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
- C、ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
- D、ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
- E、ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
- F、ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;
正确答案:C
第6题：

You are developing a Windows Forms application that contains a DataGridView control. The DataGridView capture the customer's name, address, and phone number. You have been asked to provide data validation in a DataGridView to prevent users from leaving the.. You need to ensure that users cannot tab out of the name field without entering data What should you do ?（）
- A、 Validate the name field in the CellValidating event. Set e.Cancel = true if the name field is
- B、 Validate the name field in the CellErrorTextChanged event. Get e.RowIndex if the name field ..
- C、 Validate the name field in the CellEnter event. Set the focus on the name field if the name ...
- D、 Validate the name field in the CancelRowEdit event. Set the focus on the name field if the name ...
正确答案:A
第7题：

单选题
BookStore数据库中有客户表Customer（CustomerCode，Name，Sex，Hometown，Email，Telephone，LoginDate，Password）查询Customer表中所有客户的信息，要求显示CustomerCode、Name、Hometown和Telephone字段信息。SQL语句：SELECT（）FROM Customer
A
*
B
ALL
C
CustomerCode、Name、Hometown、Telephone
D
CustomerCode，Name，Hometown，Telephone

正确答案： C
解析：暂无解析
第8题：

多选题
The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which two statements find the number of customers?（）
A
SELECT TOTAL(*) FROM customer;
B
SELECT COUNT(*) FROM customer;
C
SELECT TOTAL(customer_id) FROM customer;
D
SELECT COUNT(customer_id) FROM customer;
E
SELECT COUNT(customers) FROM customer;
F
SELECT TOTAL(customer_name) FROM customer;

正确答案： A,F
解析：暂无解析
第9题：

单选题
You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: CUST_ID NUMBER(4) NOT NULL CUST_NAME VARCHAR2(100) NOT NULL CUST_ADDRESS VARCHAR2(150) CUST_PHONE VARCHAR2(20) Which SELECT statement accomplishes this task?（）
A
SELECT* FROM customers;
B
SELECT name, address FROM customers;
C
SELECT id, name, address, phone FROM customers;
D
SELECT cust_name, cust_address FROM customers;
E
SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

正确答案： E
解析：暂无解析
第10题：

单选题
The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) Which statement finds the rows in the CUSTOMERS table that do not have a postal code? （）
A
SELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;
B
SELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';
C
SELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;
D
SELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;
E
SELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;

正确答案： A
解析：暂无解析
第11题：

单选题
You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: Which SELECT statement accomplishes this task?（）
A
SELECT*   FROM customers;
B
SELECT name, address   FROM customers;
C
SELECT id, name, address, phone   FROM customers;
D
SELECT cust_name, cust_address   FROM customers;
E
SELECT cust_id, cust_name, cust_address, cust_phone   FROM customers;

正确答案： D
解析：暂无解析
第12题：

( 难度：中等)在Hive中，以下哪个SQL语句是错误的
A.hive＞ SELECT ID, NAME, o.AMOUNT, o.DATE
＞ FROM CUSTOMERS c
＞ LEFT OUTER JOIN ORDERS o
＞ ON (ID = o.CUSTOMER_ID);
B.hive＞ SELECT ID, NAME, o.AMOUNT, o.DATE
＞ FROM CUSTOMERS c
＞ RIGHT OUTER JOIN ORDERS o
＞ ON (ID = o.CUSTOMER_ID);
C.hive＞ SELECT ID, NAME, o.AMOUNT, o.DATE
＞ FROM CUSTOMERS c
＞ FULL OUTER JOIN ORDERS o
＞ ON (ID = o.CUSTOMER_ID);
D.hive＞ SELECT ID, NAME, o.AMOUNT, o.DATE
＞ FROM CUSTOMERS c
＞ FULL OUTER JOIN ORDERS o
＞ ON (ID = o.CUSTOMER_ID OR Name = o.Name);

答案:D
第13题：

试题六（共 15 分）阅读下列说明和 C++代码，填补代码中的空缺，将解答填入答题纸的对应栏内。【说明】以下 C++代码实现一个简单客户关系管理系统（CrM）中通过工厂（Customerfactory）对象来创建客户（Customer）对象的功能。客户分为创建成功的客户（realCustomer）和空客户（NullCustomer）。空客户对象是当不满足特定条件时创建或获取的对象。类间关系如图6-1 所示。

【C++代码】#include#includeusing namespace std； class Customer｛protected:string name；public:（1） boll isNil（）=0；（2） string getName（）=0；﹜； class realCustomer （3）｛public:realCustomer（string name）｛this->name=name；﹜bool isNil（）｛ return false；﹜string getName（）｛ return name；﹜﹜； class NullCustomer （4）｛public:bool isNil（）｛ return true；﹜string getName（）｛ return 〝Not Available in Customer Database〞； ﹜﹜；class Customerfactory｛public:string names［3］=｛〝rob〞, 〝Joe〞,〝Julie〞﹜；public:Customer*getCustomer（string name）｛for （int i=0；i<3；i++）｛if （names［i］.（5））｛return new realCustomer（name）；﹜﹜return （6）；﹜﹜； class CrM｛public:void getCustomer（）｛Customerfactory*（7）；Customer*customer1=cf->getCustomer（〝rob〞）；Customer*customer2=cf->getCustomer（〝Bob〞）；Customer*customer3=cf->getCustomer（〝Julie〞）；Customer*customer4=cf->getCustomer（〝Laura〞）； cout<<〝Customers〞<getName（） <getName（） <getName（） <getName（） <getCustomer（）；delete crs；return 0；﹜ /*程序输出为：CustomersrobNot Available in Customer DatabaseJulieNot Available in Customer Database*/

答案：
解析：
1)virtual2)virtual3):public Customer4):public Customer5)compare(name)==06)new Null Customer()7)cf=New CustomerFactory();
【解析】

本题考察使用C++代码实现实际问题。在C++中，动态绑定是通过虚函数来实现的。此题中用到了虚函数，所以要在成员函数原型缺钱加一个关键字virtual。类RealCustomer和类NullCustomer是类Customer的派生类，因此3、4空都填public Customer。进行对比数据库中的人名compare(name)==0第6空与前面语句是相反的，一个是返回new RealCustomer(name)，那么此处应填：new Null Customer()第7空，用工厂创建对象，cf=New CustomerFactory();
第14题：

BookStore数据库中有客户表Customer（CustomerCode，Name，Sex，Hometown，Email，Telephone，LoginDate，Password）删除Customer表中姓“刘”和姓“杨”的记录。SQL语句：DELETE FROMCustomer WHERE（）
- A、LEFT（Name，1）=‘刘’ORLEFT（Name，1）=‘杨’
- B、NameLIKE‘刘%‘ORNameLIKE杨%‘
- C、NameLIKE‘[刘，杨]%‘
- D、NameLIKE‘[^刘，^杨]%
正确答案:A,B,C
第15题：

The accounts receivable department requested an accounts status to allow users to enter either a customer name, customer number, customer phone number and then display the customer out standing balance. You want to restrict the data retrieved based on values entered by the users. Which where clause would you write in the query to select statement to accomplish this task?（）
- A、&p_where_clause
- B、:p_where_clause
- C、WHERE CUST_ID=P_CUST_ID or cust_name=:p_cust_name or cust_phone=:p_cust_phone
- D、Where cust_ID=&p_custid cust_name=&p_cust_name or cust_phone=&p_cust_phone
正确答案:A
第16题：

The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which two statements find the number of customers? （）
- A、SELECT TOTAL(*) FROM customer;
- B、SELECT COUNT(*) FROM customer;
- C、SELECT TOTAL(customer_id) FROM customer;
- D、SELECT COUNT(customer_id) FROM customer;
- E、SELECT COUNT(customers) FROM customer;
- F、SELECT TOTAL(customer_name) FROM customer;
正确答案:B,D
第17题：

The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) Which statement finds the rows in the CUSTOMERS table that do not have a postal code? （）
- A、SELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;
- B、SELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';
- C、SELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;
- D、SELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;
- E、SELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;
正确答案:C
第18题：

多选题
BookStore数据库中有客户表Customer（CustomerCode，Name，Sex，Hometown，Email，Telephone，LoginDate，Password）查询Customer表中所有客户的信息，SQL语句：SELECT（）FROMCustomer
A
*
B
CustomerCode，Name，Hometown，Sex，Email，Telephone，LoginTime，PassWord
C
CustomerCode，Hometown，Telephone，Name，Sex，Email，LoginTime，PassWord
D
CustomerCode~PassWord

正确答案： C,A
解析：暂无解析
第19题：

单选题
The accounts receivable department requested an accounts status to allow users to enter either a customer name, customer number, customer phone number and then display the customer out standing balance. You want to restrict the data retrieved based on values entered by the users. Which where clause would you write in the query to select statement to accomplish this task?（）
A
&p_where_clause
B
:p_where_clause
C
WHERE CUST_ID=P_CUST_ID or cust_name=:p_cust_name or cust_phone=:p_cust_phone
D
Where cust_ID=&p_custid cust_name=&p_cust_name or cust_phone=&p_cust_phone

正确答案： B
解析：暂无解析
第20题：

单选题
You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: CUST_ID NUMBER(4) NOT NULL CUST_NAME VARCHAR2(100) NOT NULL CUST_ADDRESS VARCHAR2(150) CUST_PHONE VARCHAR2(20) Which SELECT statement accomplishes this task?（）
A
SELECT* FROM customers;
B
SELECT name, address FROM customers;
C
SELECT id, name, address, phone FROM customers;
D
SELECT cust_name, cust_address FROM customers;
E
SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

正确答案： C
解析：暂无解析
第21题：

单选题
You are developing a Windows Forms application that contains a DataGridView control. The DataGridView capture the customer's name, address, and phone number. You have been asked to provide data validation in a DataGridView to prevent users from leaving the.. You need to ensure that users cannot tab out of the name field without entering data What should you do ?（）
A
Validate the name field in the CellValidating event. Set e.Cancel = true if the name field is
B
Validate the name field in the CellErrorTextChanged event. Get e.RowIndex if the name field ..
C
Validate the name field in the CellEnter event. Set the focus on the name field if the name ...
D
Validate the name field in the CancelRowEdit event. Set the focus on the name field if the name ...

正确答案： C
解析：暂无解析
第22题：

单选题
Which statement adds a constraint that ensures the CUSTOMER_NAME column of the CUSTOMERS table holds a value?（）
A
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
B
ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
C
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
D
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
E
ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
F
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;

正确答案： F
解析：暂无解析
第23题：

单选题
评估以下CREATETABLE语句的执行结果： CREATET ABLE customers （customer_id NUMBER，customer_name VARCHAR2（25）， address VARCHAR 2（25）， city VARCHAR 2（25）， region VARCHAR 2（25）， postal_code VARCHAR 2（11）， CONSTRAINT customer_id_un UNIQUE（customer_id）， CONSTRAINTcustomer_name_nnNOTNULL（customer_name））；为什么执行时此语句会失败（）
A
NUMBER数据类型要求精度值
B
UNIQUE约束条件必须在列级定义
C
CREATETABLE语句不定义PRIMARYKEY
D
不能在表级定义NOTNULL约束条件

正确答案： D
解析：暂无解析

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