如何对对象的 ArrayList 进行排序？

Question

我有和ArrayList 有两个字符串和一个Integer，但我只需要按字符串Name 和字符串LastName 对其进行排序，我尝试过使用Collections.sort 这样的方法

    Collections.sort(MyArrayList, (Contact v1, Contact v2) -> v1.getName().compareTo(v2.getname()));

但是这仅按名称对我的ArrayList 进行排序，并且不包括姓氏，我如何将姓氏添加到排序中？

Answer 1

使用Comparator.comparing(Contact::getName).thenComparing(Contact::getLastName) 作为您的比较器：

Collections.sort(MyArrayList, Comparator.comparing(Contact::getName).thenComparing(Contact::getLastName));

Answer 2

您只需要更改比较函数以包含其他类似的字段

    Collections.sort(contacts, (Contact c1, Contact c2) -> {
        int firstNameComparisonResult = c1.getFirstName().compareTo(c2.getFirstName());
        if (firstNameComparisonResult != 0) {
            return firstNameComparisonResult;
        } else {
            return c1.getLastName().compareTo(c2.getLastName());
        }
    });

警告：假设没有空值。

如果你想看看我使用的完整代码：

class Contact{
    final String firstName;
    final String lastName;
    final int age;

    public String getFirstName() {
        return firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public int getAge() {
        return age;
    }

    public Contact(String firstName, String lastName, int age) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.age = age;
    }

    @Override
    public String toString() {
        return "Contact{" +
                "firstName='" + firstName + '\'' +
                ", lastName='" + lastName + '\'' +
                ", age=" + age +
                '}';
    }
}

@Test
public void sortTest(){
    List<Contact> contacts = new ArrayList<>();
    contacts.add(new Contact("A","B",37));
    contacts.add(new Contact("A","C",34));
    contacts.add(new Contact("B","A",35));

    Collections.sort(contacts, (Contact c1, Contact c2) -> {
        int firstNameComparisonResult = c1.getFirstName().compareTo(c2.getFirstName());
        if (firstNameComparisonResult != 0) {
            return firstNameComparisonResult;
        } else {
            return c1.getLastName().compareTo(c2.getLastName());
        }
    });

    System.out.println(contacts);
 //[Contact{firstName='A', lastName='B', age=37}, Contact{firstName='A', lastName='C', age=34}, Contact{firstName='B', lastName='A', age=35}]

}

Answer 3

嗯...将您的闭包转换为一个完整的匿名类，并使用该功能来增加您的排序复杂性...

类似

 Collections.sort(MyArrayList, new Comparator<Contact>() {
    public int compare(Contact v1, Contact v2) {
        int returned =  v1.getvName().compareTo(v2.getname())
        if(returned==0) {
            // alas, Math.signum returns a float value in this cas
            returned = (int) Math.signum(v1.getYourInt(), v2.getYourInt())
        }
        return returned;
    }
 });

警告此代码尚未以任何方式编译。