這是用于保留重復(fù)ID的代碼。

public void addContact(Person p) { for(int i = 0; i < ArrayOfContacts.size(); i++) {Person contact = ArrayOfContacts.get(i);if(contact.getID() == p.getID()) { System.out.println('Sorry this contact already exists.'); return; // the id exists, so we exit the method. } } // Otherwise... you’ve checked all the elements, and have not found a duplicate ArrayOfContacts.add(p);}

如果您想更改此代碼以保留重復(fù)的名稱，請執(zhí)行以下操作

public void addContact(Person p) { String pName = p.getFname() + p.getLname(); for(int i = 0; i < ArrayOfContacts.size(); i++) {Person contact = ArrayOfContacts.get(i);String contactName = contact.getFname() + contact.getLname(); if(contactName.equals(pName)) { System.out.println('Sorry this contact already exists.'); return; // the name exists, so we exit the method. } } // Otherwise... you’ve checked all the elements, and have not found a duplicate ArrayOfContacts.add(p);}解決方法

switch(menuChoice) {case 1: System.out.println('Enter your contact’s first name:n'); String fname = scnr.next(); System.out.println('Enter your contact’s last name:n'); String lname = scnr.next(); Necronomicon.addContact(new Person(fname,lname)); break;// main truncated here for readability

import java.util.ArrayList;public class AddressBook { ArrayList<Person> ArrayOfContacts= new ArrayList<Person>(); public void addContact(Person p) { ArrayOfContacts.add(p); /* for(int i = 0; i < ArrayOfContacts.size(); i++) { if(ArrayOfContacts.get(i).getID() != p.getID()) ArrayOfContacts.add(p); elseSystem.out.println('Sorry this contact already exists.'); } */ }}

public class Person { private String fName = null; private String lName = null; private static int ID = 1000; public Person(String fName,String lName) { // Constructor I’m using to try and increment the ID each time a Person object is created starting at 1001. this.fName = fName; this.lName = lName; ID = ID + 1; }}

我正在嘗試創(chuàng)建一個通訊錄，其中每個聯(lián)系人都有一個名字，姓氏和唯一的ID。

我的問題是如何防止用戶輸入具有相同名字和姓氏的重復(fù)聯(lián)系人？我應(yīng)該在addContact方法中還是在main中實現(xiàn)某種檢查？怎么樣？