在Java中使用equals方法查找相等对象
我有一个对象列表，该对象具有名称，地址第1行，地址第2行，城市，listOfPeopleMatched属性。我正在通过覆盖equals和hashcode方法找到基于地址line1，地址line2和城市（而非名称）的相等性。现在，我想获取对象匹配的人员的姓名，并将其存储在listOfPeopleMatched中。例如：[[“ Val”，“ Ashish”]，[“ Steve”，“ Alex”]]。仅用equals方法怎么做到？
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 uniquePeople = new HashSet<>(); uniquePeople.add(person1); uniquePeople.add(person2); uniquePeople.add(person3); uniquePeople.add(person4);
System.out.println(uniquePeople);
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下一站是幸福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 + '}'; }2020-04-27 23:16:47赞同 展开评论 打赏 -
想将地址数据封装在一个对象中。这样，“Person”不等于其地址，你可以进行更精细的搜索。另外，你可以通过这种方式分离问题。
我在这里写了一个示例：
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 + '}'; }}
2019-10-16 18:35:13赞同 展开评论 打赏
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