假设你有一些对象，它们有几个字段可以比较:
public class Person {
private String firstName;
private String lastName;
private String age;
/* Constructors */
/* Methods */
}
所以在这个例子中，当你问if:
a.compareTo(b) > 0
你可能会问a的姓是不是在b的姓之前，或者a的年龄是不是比b大，等等……
在不增加不必要的混乱或开销的情况下，在这些类型的对象之间进行多重比较的最干净的方法是什么?
comparable接口只允许通过一个字段进行比较 在我看来，添加大量的比较方法(如compareByFirstName()， compareByAge()等)是混乱的。
那么最好的解决办法是什么呢?
当前回答
如果实现Comparable接口，则需要选择一个简单的属性进行排序。这就是所谓的自然排序。把它看作默认值。通常在没有提供特定比较器时使用。通常这是名称，但您的用例可能调用不同的东西。您可以自由地使用任何数量的其他比较器，您可以提供给各种集合api来覆盖自然排序。
还要注意，通常如果a.c omareto (b) == 0，则a.c omareto (b) == true。如果没有也没关系，但是有副作用要注意。在Comparable接口上查看优秀的javadocs，您将找到许多关于这方面的有用信息。
其他回答
从Steve的回答开始，三元操作符可以使用:
public int compareTo(Person other) {
int f = firstName.compareTo(other.firstName);
int l = lastName.compareTo(other.lastName);
return f != 0 ? f : l != 0 ? l : Integer.compare(age, other.age);
}
我认为如果你的比较算法是“聪明的”，那就更令人困惑了。我会选择你建议的众多比较方法。
对我来说唯一的例外就是平等。对于单元测试，重写. equals(在.net中)对我来说很有用，以便确定两个对象之间的几个字段是否相等(而不是引用是否相等)。
//Following is the example in jdk 1.8
package com;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
class User {
private String firstName;
private String lastName;
private Integer age;
public Integer getAge() {
return age;
}
public User setAge(Integer age) {
this.age = age;
return this;
}
public String getFirstName() {
return firstName;
}
public User setFirstName(String firstName) {
this.firstName = firstName;
return this;
}
public String getLastName() {
return lastName;
}
public User setLastName(String lastName) {
this.lastName = lastName;
return this;
}
}
public class MultiFieldsComparision {
public static void main(String[] args) {
List<User> users = new ArrayList<User>();
User u1 = new User().setFirstName("Pawan").setLastName("Singh").setAge(38);
User u2 = new User().setFirstName("Pawan").setLastName("Payal").setAge(37);
User u3 = new User().setFirstName("Anuj").setLastName("Kumar").setAge(60);
User u4 = new User().setFirstName("Anuj").setLastName("Kumar").setAge(43);
User u5 = new User().setFirstName("Pawan").setLastName("Chamoli").setAge(44);
User u6 = new User().setFirstName("Pawan").setLastName("Singh").setAge(5);
users.add(u1);
users.add(u2);
users.add(u3);
users.add(u4);
users.add(u5);
users.add(u6);
System.out.println("****** Before Sorting ******");
users.forEach(user -> {
System.out.println(user.getFirstName() + " , " + user.getLastName() + " , " + user.getAge());
});
System.out.println("****** Aftre Sorting ******");
users.sort(
Comparator.comparing(User::getFirstName).thenComparing(User::getLastName).thenComparing(User::getAge));
users.forEach(user -> {
System.out.println(user.getFirstName() + " , " + user.getLastName() + " , " + user.getAge());
});
}
}
Java 8通过lambda方式我们可以通过方法引用进行比较。 学生POJO
public class Student {
int id;
String firstName;
String lastName;
String subject;
public Student(int id, String firstName, String lastName, String subject) {
this.id = id;
this.firstName = firstName;
this.lastName = lastName;
this.subject = subject;
}
enter code here
现在我们可以根据
1. id - > FirstName - > LastName - > 2。主题- > id - > FirstName - > LastName
我们将在数组Stream中使用Comparator
public class TestComprator {
public static void main(String[] args) {
Student s1= new Student(108, "James", "Testo", "Physics");
Student s2= new Student(101, "Fundu", "Barito", "Chem");
Student s3= new Student(105, "Sindhu", "Sharan", "Math");
Student s4= new Student(98, "Rechel", "Stephen", "Physics");
System.out.printf("----------id->FirstName->LastName->Subject-------------");
Arrays.asList(s1,s2,s3,s4).stream()
.sorted(Comparator.comparing(Student::getId)
.thenComparing(Student::getFirstName)
.thenComparing(Student::getLastName)
.thenComparing(Student::getSubject))
.forEach(System.out::println);
System.out.printf("----Subject->id->FirstName->LastName ------\n");
Arrays.asList(s1,s2,s3,s4).stream()
.sorted(Comparator. comparing(Student::getSubject)
.thenComparing(Student::getId)
.thenComparing(Student::getFirstName)
.thenComparing(Student::getLastName)
)
.forEach(System.out::println);
}
}
输出:
`----------id->FirstName->LastName->Subject-------------
Student{id=98, firstName='Rechel', lastName='Stephen', subject='Physics'}
Student{id=101, firstName='Fundu', lastName='Barito', subject='Chem'}
Student{id=105, firstName='Sindhu', lastName='Sharan', subject='Math'}
Student{id=108, firstName='James', lastName='Testo', subject='Physics'}
----Subject->id->FirstName->LastName ------
Student{id=101, firstName='Fundu', lastName='Barito', subject='Chem'}
Student{id=105, firstName='Sindhu', lastName='Sharan', subject='Math'}
Student{id=98, firstName='Rechel', lastName='Stephen', subject='Physics'}
Student{id=108, firstName='James', lastName='Testo', subject='Physics'}
在java中使用hashcode方法比较两个对象很容易。
public class Sample{
String a=null;
String b=null;
public Sample(){
a="s";
b="a";
}
public Sample(String a,String b){
this.a=a;
this.b=b;
}
public static void main(String args[]){
Sample f=new Sample("b","12");
Sample s=new Sample("b","12");
//will return true
System.out.println((s.a.hashCode()+s.b.hashCode())==(f.a.hashCode()+f.b.hashCode()));
//will return false
Sample f=new Sample("b","12");
Sample s=new Sample("b","13");
System.out.println((s.a.hashCode()+s.b.hashCode())==(f.a.hashCode()+f.b.hashCode()));
}
