我只允许在主菜单中有4个选项,我创建了5.我无法弄清楚如何将最后2个输出语句放在一起,这样就可以从另一个内部调用。所以他们都可以使用相同的菜单选项。
选项3和4需要以某种方式在彼此内部,以便我可以显示输入的所有联系人的姓名,然后提示用户选择联系人ID以显示该联系人的其余详细信息。
尝试将两者合并到一个if语句中,并且没有为添加的第二个联系人正确显示全名。它再次显示第一个名称,然后要求提供联系人ID以显示详细信息。如果您显示详细信息,则会返回正确的详细信息以及输入的最后一个联系人的姓名。
以下是我们分开时所拥有的:
package ooo1;
import java.util.ArrayList;
import java.util.Scanner;
public class ContactList {
public static void main(String[] args) {
ArrayList<Contact> contacts = new ArrayList<>();
Scanner input1 = new Scanner(System.in);
int type = 0;
while(type != 5){
System.out.println("Please select an option: ");
System.out.println("Add a Personal Contact: Enter 1 ");
System.out.println("Add a Business Contact: Enter 2 ");
System.out.println("Display Contacts List: Enter 3 ");
System.out.println("Display Contact Details: Enter 4 ");
System.out.println("To Quit: Enter 5 ");
type = input1.nextInt();
if(type == 5){
System.out.println("Goodbye ");
break;
}
if (type == 1 || type == 2){
Contact contact = null;
Scanner input = new Scanner(System.in);
System.out.println("Please enter ContactId : ");
String contactId = input.nextLine();
System.out.println("Please enter First Name : ");
String firstName = input.nextLine();
System.out.println("Please enter Last Name : ");
String lastName = input.nextLine();
System.out.println("Please enter Address in the following format : ");
System.out.println("Street Address, City, State, Zip Code");
String address = input.nextLine();
System.out.println("Please enter Phone Number : ");
String phoneNumber = input.nextLine();
System.out.println("Please enter Email Address : ");
String emailAddress = input.nextLine();
//Create a personal contact.
if(type == 1){
System.out.println("Please enter Birthday: ");
String dateofBirth = input.nextLine();
Contact pcontact = new PersonalContact(contactId, firstName, lastName, address, phoneNumber, emailAddress, dateofBirth);
contacts.add(pcontact);
System.out.println("Contact Added Successfully");
System.out.println();
}
//Create a business contact.
else if(type == 2){
System.out.println("Please enter Job Title: ");
String jobTitle = input.nextLine();
System.out.println("Please enter Organization: ");
String organization = input.nextLine();
Contact bcontact = new BusinessContact(contactId, firstName, lastName, address, phoneNumber, emailAddress, jobTitle, organization);
contacts.add(bcontact);
System.out.println("Contact Added Successfully");
System.out.println();
}
}
if (type == 3 || type == 4){
//Print full name of each Contact.
if(type == 3){
for (Contact namecontact: contacts)
{
System.out.println(namecontact.displayFullName());
System.out.println();
}
}
//Print contact details for selected contact.
else if(type == 4){
System.out.println("Enter a Contact ID to display Contact Details: ");
Scanner input2 = new Scanner(System.in);
String soughtID;
soughtID = input2.nextLine();
for (Contact showcontact1: contacts)
{
if (showcontact1.displayId().equals(soughtID))
System.out.println(showcontact1.displayContact());
System.out.println();
}
}
}
}
}
}
这是他们调用的父类:
package ooo1;
public abstract class Contact {
String contactId;
String firstName;
String lastName;
String address;
String phoneNumber;
String emailAddress;
public Contact(String contactId,String firstName,String lastName, String address, String phoneNumber, String emailAddress)
{
this.contactId = contactId;
this.firstName = firstName;
this.lastName = lastName;
this.address = address;
this.phoneNumber = phoneNumber;
this.emailAddress = emailAddress;
}
public void setContactId(String input){
this.contactId = input;
}
public String getContactId(){
return contactId;
}
public void setFirstName(String input){
this.firstName = input;
}
public String getFirstName(){
return firstName;
}
public void setLastName(String input){
this.lastName = input;
}
public String getLastName(){
return lastName;
}
public void setAddress(String input){
this.address = input;
}
public String getAddress(){
return address;
}
public void setPhoneNumber(String input){
this.phoneNumber = input;
}
public String getPhoneNumber(){
return phoneNumber;
}
public void setEmailAddress(String input){
this.emailAddress = input;
}
public String getEmailAddress(){
return emailAddress;
}
@Override
public String toString(){
return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName() + "\nAddress: " + this.getAddress() + "\nPhone Number: " + this.getPhoneNumber() + "\nEmail Address " + this.getEmailAddress());
}
public String displayFullName(){
return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName());
}
public String displayContact(){
return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName() + "\nAddress :" + this.getAddress() + "\nPhone Number :" + this.getPhoneNumber() + "\nEmail Address " + this.getEmailAddress());
}
public String displayId(){
return (this.getContactId());
}
}
这是我试图在主要方面改变以摆脱其他选项并且结果不佳:
if(type == 3){ for(联系方式contactcontact:contacts) { 的System.out.println(namecontact.displayFullName()); 的System.out.println(); System.out.println(“输入联系人ID以显示联系人详细信息:”); 扫描仪输入2 =新扫描仪(System.in); String seekID; seekID = input2.nextLine(); for(联系showcontact1:contacts) { if(showcontact1.displayId()。equals(seekID)) 的System.out.println(showcontact1.displayContact()); 的System.out.println();
以下是更改之前的输出方式,以及更改后的方式。
这是我在改变之前所做的事情,除非我在主菜单中有太多选项。
请选择一个选项: 添加个人联系人:输入1 添加业务联系人:输入2 显示联系人列表:输入3 显示联系人详细信息:输入4 退出:输入5 3 ContactID:2 名字:汤姆 姓氏:琼斯
请选择一个选项: 添加个人联系人:输入1 添加业务联系人:输入2 显示联系人列表:输入3 显示联系人详细信息:输入4 退出:输入5 4 输入联系人ID以显示联系人详细信息: 2 个人联系: ContactID:2 名字:汤姆 姓氏:琼斯 地址:西街234号 电话号码:123-345-2345 电子邮件地址tjones@www.com 出生日期:12-12-1893
当我把2放在一起时会发生这种情况。它需要新的肠道。简史密斯补充道。当我输入3来显示联系人。我希望它能让我回到简和汤姆,但它让我回到汤姆身边。当我要求Toms的详细信息时,我得到了Toms的详细信息,但我也得到了Janes的名字和我之前想要的名字。
请输入ContactId: 1 请输入名字: 简 请输入姓氏: 工匠 请按以下格式输入地址: 街道地址,城市,州,邮政编码 234霍华德 请输入电话号码: 123-235-2345 请输入电子邮件地址: jsmith@yahoo.com 请输入生日: 1978年12月12日 联系已成功添加
请选择一个选项:
添加个人联系人:输入1
添加业务联系人:输入2
显示联系人列表:输入3
退出:输入5
3
ContactID:12
名字:汤姆
姓氏:Hones
输入联系人ID以显示联系人详细信息: 12 个人联系: ContactID:12 名字:汤姆 姓氏:Hones 地址:234 south st 电话号码:234-232-2356 电子邮件地址thones@www.com 出生日期:12-12-45
ContactID:1 名字:简 姓氏:史密斯
答案 0 :(得分:1)
你可以这样做,
System.out.println("Please select an option (1-4), any other number to Quit");
System.out.println("Add a Personal Contact: Enter 1 ");
System.out.println("Add a Business Contact: Enter 2 ");
System.out.println("Display Contacts List: Enter 3 ");
System.out.println("Display Contact Details: Enter 4 ");
type = input1.nextInt();
if (type < 1 || type > 4) {
System.out.println("Goodbye ");
break;
}
答案 1 :(得分:1)
因此,如果我理解正确,您希望嵌套选项3和4,以便选项3显示所有联系人,然后选择显示特定联系人:
// Other menu code here...
System.out.println("Query Contacts: Enter 3");
System.out.println("To Quit: Enter 4");
// Other code here...
// Bring up sub-menu
// Feel free to extend to return to main menu etc.
if (type == 3){
while (true) {
System.out.println("List Contacts: Enter 1");
System.out.println("Display Contact Details: Enter 2");
System.out.println("To Quit: Enter 3");
// If you need, store in another int etc.
type = input1.nextInt();
//Print full name of each Contact.
if (type == 1) {
for (Contact namecontact: contacts) {
System.out.println(namecontact.displayFullName());
System.out.println();
}
}
//Print contact details for selected contact.
else if(type == 2){
System.out.println("Enter a Contact ID to display Contact Details: ");
Scanner input2 = new Scanner(System.in);
String soughtID;
soughtID = input2.nextLine();
for (Contact showcontact1: contacts) {
// Although correct, I'd recommend you add a braces for if statements
// Saving a line for a closing brace is not generally worth it
if (showcontact1.displayId().equals(soughtID))
System.out.println(showcontact1.displayContact());
System.out.println();
}
} else if (type == 3) {
break;
}
}
}