1

我有一个对象数组列表,其中包含例如:

  • 姓名
  • 地址
  • 电话
  • 许多其他属性...

如果某些属性与数组列表中的其他对象具有相同的值,我希望删除此列表中的某些对象。我需要遍历整个列表并查看名称、地址和电话是否已存在于此列表中。我不能做一个简单的:

for (...)
if (!newlist.contains(element)) { newlist.add(element); }

因为我只需要在添加element到新列表之前检查特定属性是否相同。

谁能指导我正确的方向?

4

2 回答 2

3

将 Set 与自定义 Comparator 一起使用怎么样?让您的对象类实现 Comparable。在 compare 方法中,您可以编写测试以完全匹配您需要的对象。

于 2013-01-30T11:55:01.873 回答
2

创建一个 Key Class,让我们用下面的代码说Employee.java 。

package com.innovation;


public class Employee {
private String name;
private String address;
private String phone;

public Employee() {
    super();
}

public Employee(String name, String address, String phone) {
    super();
    this.name = name;
    this.address = address;
    this.phone = phone;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getAddress() {
    return address;
}

public void setAddress(String address) {
    this.address = address;
}

public String getPhone() {
    return phone;
}

public void setPhone(String phone) {
    this.phone = phone;
}

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + ((address == null) ? 0 : address.hashCode());
    result = prime * result + ((name == null) ? 0 : name.hashCode());
    result = prime * result + ((phone == null) ? 0 : phone.hashCode());
    return result;
}

@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    Employee other = (Employee) obj;
    if (address == null) {
        if (other.address != null)
            return false;
    } else if (!address.equals(other.address))
        return false;
    if (name == null) {
        if (other.name != null)
            return false;
    } else if (!name.equals(other.name))
        return false;
    if (phone == null) {
        if (other.phone != null)
            return false;
    } else if (!phone.equals(other.phone))
        return false;
    return true;
}

@Override
public String toString() {
    return "Employee [name=" + name + ", address=" + address + ", phone="
            + phone + "]";
}



}

现在创建一个要应用逻辑的客户端类,让我们假设一个包含主方法的类,比如Client.java

package com.innovation;

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Client {

    public static void main(String[] args) {


        Set<Employee> empSet = new HashSet<Employee>(populateList());

        for (Employee employee : empSet)
        {
            System.out.println(employee);
        }

    }   

    public static List<Employee> populateList()
    {
        List<Employee> lsts = new ArrayList<Employee>();

        lsts.add(new Employee("rais","gurgaon","123456"));
        lsts.add(new Employee("alam","Delhi","123685"));
        lsts.add(new Employee("shyam","Mumbai","1257456"));
        lsts.add(new Employee("ramesh","Ahmadabad","196356"));
        lsts.add(new Employee("rais","gurgaon","123456"));
        lsts.add(new Employee("rais","gurgaon","123456"));
        lsts.add(new Employee("rais","gurgaon","123456"));


        return lsts;

    }

}

你会看到下面的输出。可以清楚地看到列表中存在的重复条目在集合中被删除。这一切都是很好地实现equals和hashcode方法的魔力。

Employee [name=rais, address=gurgaon, phone=123456]
Employee [name=ramesh, address=Ahmadabad, phone=196356]
Employee [name=alam, address=Delhi, phone=123685]
Employee [name=shyam, address=Mumbai, phone=1257456]
于 2013-01-30T12:28:12.227 回答