[英]Finding equal objects using equals method in Java
I have a list of objects which has name, address line1, address line2, city, listOfPeopleMatched properties.我有一个对象列表,其中包含名称、地址 line1、地址 line2、城市、listOfPeopleMatched 属性。 I am finding equality based on address line1, address line2 and city (not name) by overriding equals and hashcode methods.
我通过覆盖 equals 和 hashcode 方法找到基于地址 line1、地址 line2 和城市(不是名称)的相等性。 Now I want to get the name of people whose objects matches and store them in listOfPeopleMatched.
现在我想获取对象匹配的人的姓名并将它们存储在 listOfPeopleMatched 中。 For example: [["Val","Ashish"], ["Steve","Alex"]].
例如:[["Val","Ashish"], ["Steve","Alex"]]。 How can this be done in equals method only?
如何仅在 equals 方法中完成?
public class Person {
private String name;
private String addressLine1;
private String addressLine2;
private String city;
private List<List<String>> listOfPeopleMatched =
new ArrayList<List<String>>();
public String getName() {
return name;
}
public List<List<String>> getListOfPeopleMatched() {
return listOfPeopleMatched;
}
public void setListOfPeopleMatched(List<List<String>> listOfPeopleMatched) {
this.listOfPeopleMatched = listOfPeopleMatched;
}
public void setName(String name) {
this.name = name;
}
public String getAddressLine1() {
return addressLine1;
}
public void setAddressLine1(String addressLine1) {
this.addressLine1 = addressLine1;
}
public String getAddressLine2() {
return addressLine2;
}
public void setAddressLine2(String addressLine2) {
this.addressLine2 = addressLine2;
}
public String getCity() {
return city;
}
public void setCity(String city) {
this.city = city;
}
public Person(String name, String addressLine1,
String addressLine2, String city) {
super();
this.name = name;
this.addressLine1 = addressLine1;
this.addressLine2 = addressLine2;
this.city = city;
}
@Override
public String toString() {
return "Person [name=" + name +
", addressLine1=" +
addressLine1 + ", addressLine2=" +
addressLine2 + ", city="
+ city + "]";
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((addressLine1 == null) ?
0 : addressLine1.hashCode());
result = prime * result + ((addressLine2 == null) ?
0 : addressLine2.hashCode());
result = prime * result + ((city == null) ?
0 : city.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;
Person other = (Person) obj;
if (addressLine1 == null) {
if (other.addressLine1 != null)
return false;
} else if (!addressLine1.equals(other.addressLine1))
return false;
if (addressLine2 == null) {
if (other.addressLine2 != null)
return false;
} else if (!addressLine2.equals(other.addressLine2))
return false;
if (city == null) {
if (other.city != null)
return false;
} else if (!city.equals(other.city))
return false;
return true;
}
}
Person person1 = new Person("Val", "ABC", "Shivaji Nagar", "Pune");
Person person2 = new Person("Ashish", "ABC", "Shivaji Nagar", "Pune");
Person person3 = new Person("Steve", "MNO", "Shivaji Nagar", "Pune");
Person person4 = new Person("Alex", "MNO", "Shivaji Nagar", "Pune");
Set<Person> uniquePeople = new HashSet<>();
uniquePeople.add(person1);
uniquePeople.add(person2);
uniquePeople.add(person3);
uniquePeople.add(person4);
System.out.println(uniquePeople);
I think you want to encapsulate your Address data in an object.我认为您想将您的地址数据封装在 object 中。 That way a Person isn't equal to it's address and you can do more granular searching.
这样一个 Person 不等于它的地址,您可以进行更精细的搜索。 Plus you are separating the concerns this way.
另外,您以这种方式分离关注点。
I wrote up a working example here:我在这里写了一个工作示例:
import java.util.ArrayList;
import java.util.List;
import java.util.Objects;
import java.util.stream.Collectors;
public class Person {
public static void main(String[] args) {
// what we're searching for
Address address = new Address("123 N 3rd st", "east ohg", "this-city");
// init
List<Person> persons = new ArrayList<>();
persons.add(new Person("Jim", "123 N 56 st", "east ohg", "this-city"));
persons.add(new Person("Kyle", "145 N 67th st", "east ohg", "this-city"));
persons.add(new Person("Sam", "12 beach av", "east ohg", "this-city"));
persons.add(new Person("Tracy", "123 N 3rd st", "east ohg", "this-city"));
persons.add(new Person("Ashley", "123 N 3rd st", "east ohg", "this-city"));
// search
List<Person> people = persons.stream().filter(person -> person.address.equals(address)).collect(Collectors.toList());
people.forEach(System.out::println);
}
String name;
Address address;
public Person(String name,
String addressLine1,
String addressLine2,
String city) {
this.name = name;
this.address = new Address(addressLine1,
addressLine2,
city);
}
private static final class Address {
String addressLine1;
String addressLine2;
String city;
public Address(String addressLine1, String addressLine2, String city) {
this.addressLine1 = addressLine1;
this.addressLine2 = addressLine2;
this.city = city;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Address address = (Address) o;
return Objects.equals(addressLine1, address.addressLine1) &&
Objects.equals(addressLine2, address.addressLine2) &&
Objects.equals(city, address.city);
}
@Override
public int hashCode() {
return Objects.hash(addressLine1, addressLine2, city);
}
@Override
public String toString() {
return "Address{" +
"addressLine1='" + addressLine1 + '\'' +
", addressLine2='" + addressLine2 + '\'' +
", city='" + city + '\'' +
'}';
}
}
@Override
public String toString() {
return "Person{" +
"name='" + name + '\'' +
", address=" + address +
'}';
}
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.