Java通讯录。如何防止我的代码中出现重复的联系人?

时间:2016-02-27 21:58:51

标签: java object arraylist

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);

    else 
       System.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;
   }
}

我正在尝试创建一个地址簿,每个联系人都有一个名字,姓氏和唯一ID。

我的问题是如何阻止用户输入具有相同名字和姓氏的重复联系人?我应该在addContact方法中还是在main中实现某种检查?怎么样?

4 个答案:

答案 0 :(得分:1)

您可以简单地使用HashSet并避免任何类型的循环来测试它。 HashSet负责此功能。

import java.util.Set;
import java.util.HashSet;

public class AddressBook {

   Set<Person> listOfContacts = new HashSet<>();

   public void addContact(Person p) {
      if (!listOfContacts.add(p))
         System.out.println("Sorry this contact already exists.");    
   }
}            

要增加ID属性,您应该有2个属性,1个静态和另一个属性,并在构造函数中递增它。看:

public class Person {
   private final int ID;
   private static int id = 1000;
   private String fName;
   private String lName;

   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.ID= ++id;
      this.fName = fName;
      this.lName = lName;   
   } 

要使HashSet不接受重复的对象,您应该设置不应该在类中复制哪些属性(在您的情况下为Person)。举个例子:

   @Override
   public int hashCode() {
      int hash = 7;
      hash = 61 * hash + Objects.hashCode(this.fName);
      hash = 61 * hash + Objects.hashCode(this.lName);
      return hash;
   }

   @Override
   public boolean equals(Object obj) {
      if (obj == null || getClass() != obj.getClass())
         return false;

      final Person other = (Person) obj;
      if (!Objects.equals(this.fName, other.fName))
         return false;

      return Objects.equals(this.lName, other.lName);
   }
}

顺便说一句,您可以使用IDE(Eclipse,NetBeans等)生成equalshashCode方法

修改

由于您无法使用HashSet,我将展示一个带有ArrayList的版本。顺便说一下,你必须使用HashCodeequals,因为我说要让它运作良好

import java.util.List;
import java.util.ArrayList;

public class AddressBook {

   List<Person> arrayOfContacts = new ArrayList<>();

   public void addContact(Person p) {
      if (listOfContacts.contains(p))
         System.out.println("Sorry this contact already exists.");

      else
         arrayOfContacts.add(p);    
   }
}

答案 1 :(得分:0)

使用HashSet。对于每个用户的名称,将其添加到HashSet。

例如:

HashSet<String> namesUsed = new HashSet<String>();

if (namesUsed.contains(userName)) {
    //do what you want here, if this is entered it means there is a duplicate
} else {
    namesUsed.add(userName); //add it to the list of used names
}

答案 2 :(得分:0)

以下是保留重复ID的代码。

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);

}

如果您想更改此代码以防止重复名称,请执行此类操作

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);

}

答案 3 :(得分:0)

你应该在你的Person类中覆盖equals

public boolean equals(Object obj) {
    if(fName.equals(obj.getfName()) && lName.equals(obj.getlName)) {
        return true;
    }
    return false;
}

然后打电话:

if(!(person1.equals(person2))) {
//not a duplicate
}

当然,用您想要的任何对象替换变量/对象。 您还应该为姓氏和名字添加getter和setter。 希望这有帮助!