我读过关于使用比较器排序数组列表的内容，但在所有的例子中，人们都使用了compareTo，根据一些研究，它是字符串的一种方法。
我想根据自定义对象的一个属性(Date对象)对其数组列表进行排序 (getStartDay())。通常我通过item1.getStartDate().before(item2.getStartDate())比较它们，所以我想知道我是否可以写一些像这样的东西:
public class CustomComparator {
public boolean compare(Object object1, Object object2) {
return object1.getStartDate().before(object2.getStartDate());
}
}
public class RandomName {
...
Collections.sort(Database.arrayList, new CustomComparator);
...
}
当前回答
我已经尝试了很多不同的解决方案，可在互联网上，但解决方案，为我是可在下面的链接。
https://www.java67.com/2017/07/how-to-sort-arraylist-of-objects-using.html
其他回答
java8 lambda表达式
Collections.sort(studList, (Student s1, Student s2) ->{
return s1.getFirstName().compareToIgnoreCase(s2.getFirstName());
});
OR
Comparator<Student> c = (s1, s2) -> s1.firstName.compareTo(s2.firstName);
studList.sort(c)
在Java 8中，你可以为你的比较器使用方法引用:
import static java.util.Comparator.comparing;
Collections.sort(list, comparing(MyObject::getStartDate));
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Date;
public class test {
public static class Person {
public String name;
public int id;
public Date hireDate;
public Person(String iname, int iid, Date ihireDate) {
name = iname;
id = iid;
hireDate = ihireDate;
}
public String toString() {
return name + " " + id + " " + hireDate.toString();
}
// Comparator
public static class CompId implements Comparator<Person> {
@Override
public int compare(Person arg0, Person arg1) {
return arg0.id - arg1.id;
}
}
public static class CompDate implements Comparator<Person> {
private int mod = 1;
public CompDate(boolean desc) {
if (desc) mod =-1;
}
@Override
public int compare(Person arg0, Person arg1) {
return mod*arg0.hireDate.compareTo(arg1.hireDate);
}
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
SimpleDateFormat df = new SimpleDateFormat("mm-dd-yyyy");
ArrayList<Person> people;
people = new ArrayList<Person>();
try {
people.add(new Person("Joe", 92422, df.parse("12-12-2010")));
people.add(new Person("Joef", 24122, df.parse("1-12-2010")));
people.add(new Person("Joee", 24922, df.parse("12-2-2010")));
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Collections.sort(people, new Person.CompId());
System.out.println("BY ID");
for (Person p : people) {
System.out.println(p.toString());
}
Collections.sort(people, new Person.CompDate(false));
System.out.println("BY Date asc");
for (Person p : people) {
System.out.println(p.toString());
}
Collections.sort(people, new Person.CompDate(true));
System.out.println("BY Date desc");
for (Person p : people) {
System.out.println(p.toString());
}
}
}
具有自然排序顺序的类(例如类Number)应该实现Comparable接口，而没有自然排序顺序的类(例如类Chair)应该提供Comparator(或匿名Comparator类)。
两个例子:
public class Number implements Comparable<Number> {
private int value;
public Number(int value) { this.value = value; }
public int compareTo(Number anotherInstance) {
return this.value - anotherInstance.value;
}
}
public class Chair {
private int weight;
private int height;
public Chair(int weight, int height) {
this.weight = weight;
this.height = height;
}
/* Omitting getters and setters */
}
class ChairWeightComparator implements Comparator<Chair> {
public int compare(Chair chair1, Chair chair2) {
return chair1.getWeight() - chair2.getWeight();
}
}
class ChairHeightComparator implements Comparator<Chair> {
public int compare(Chair chair1, Chair chair2) {
return chair1.getHeight() - chair2.getHeight();
}
}
用法:
List<Number> numbers = new ArrayList<Number>();
...
Collections.sort(numbers);
List<Chair> chairs = new ArrayList<Chair>();
// Sort by weight:
Collections.sort(chairs, new ChairWeightComparator());
// Sort by height:
Collections.sort(chairs, new ChairHeightComparator());
// You can also create anonymous comparators;
// Sort by color:
Collections.sort(chairs, new Comparator<Chair>() {
public int compare(Chair chair1, Chair chair2) {
...
}
});
因为技术每天都在出现，所以答案会随着时间的推移而改变。我看了一下LambdaJ，看起来很有趣。
您可以尝试使用LambdaJ解决这些任务。你可以在这里找到它:http://code.google.com/p/lambdaj/
这里有一个例子:
这种迭代
List<Person> sortedByAgePersons = new ArrayList<Person>(persons);
Collections.sort(sortedByAgePersons, new Comparator<Person>() {
public int compare(Person p1, Person p2) {
return Integer.valueOf(p1.getAge()).compareTo(p2.getAge());
}
});
用排序
List<Person> sortedByAgePersons = sort(persons, on(Person.class).getAge());
当然，拥有这种美感会影响性能(平均2次)，但你能找到更可读的代码吗?
