What is the most elegant way to sort a list of objects in Java? [duplicate]

This question already has an answer here:

这个问题在这里已有答案:

Sorting by property in Java 8 stream 1 answer

按Java 8流1属性排序

I have the following java class:

我有以下java类:

public class Customer {
    private long id;

    public long getId() {
        return id;
    }
}

And I also have a List<Customer> customers. What is the most elegant way to sort the list by Customer.getId() using a lambda expression?

我还有一个List 客户。 Customer.getId()使用lambda表达式对列表进行排序的最优雅方法是什么?

Right now I have this code:

现在我有这个代码:

Comparator<Customer> customerComparator = (previousCustomer, nextCustomer) ->
        Long.compare(previousCustomer.getId(), nextCustomer.getId());

customers.sort(customerComparator);

But it feels clunky especially if I need to write a comparator for each different sort or for each different generic list. Could this be done with one line of code?

但是,如果我需要为每个不同的排序或每个不同的通用列表编写比较器,它会感到笨拙。这可以用一行代码完成吗?

3 个解决方案

#1

I think you may be looking for the Comparator.comparing...() method:

我想你可能正在寻找Comparator.comparing ...()方法:

customers.sort(comparingLong(Customer::getId))

#2

In Java 8 you can write on one line.

在Java 8中,您可以在一行上书写。

customers.sort((a, b) -> Long.compare(a.getId(), b.getId());

If you make the Customer implements Comparable<Customer> you can just do

如果您使Customer实现Comparable ,您就可以这样做

customers.sort(Customer::compare);

customers.sort(null);

#3

Sort the List using Lambda Expression

使用Lambda表达式对列表进行排序

customers.sort((c1, c2) -> Long.compare(c1.getId(),c2.getId()));

#1