【问题标题】:How to sort a list of maps on key in Java?如何在Java中对键上的地图列表进行排序?
【发布时间】:2020-06-25 10:56:26
【问题描述】:

我尝试按键对 HashMap 的链接列表进行排序


想要的输出:

[{Addres=Astreet, Name=Aron}, {Addres=Bstreet, Name=Bernie},{Addres=Cstreet, Name=Carol}]

  //Making Contacts

        HashMap contact1 = new HashMap();
        HashMap contact2 = new HashMap();
        HashMap contact3 = new HashMap();


        contact1.put("Name", "Bernie");
        contact1.put("Addres", "Bstreet");


        contact2.put("Name", "Aron");
        contact2.put("Addres", "Astreet");


        contact3.put("Name", "Carol");
        contact3.put("Addres", "Cstreet");


   //Making ContactBook

        LinkedList contactBook = new LinkedList();

        contactBook.add(contact1);
        contactBook.add(contact2);
        contactBook.add(contact3);


        System.out.println(contactBook);

【问题讨论】:

标签: java sorting linked-list maps


【解决方案1】:

您要做的是创建一个 POJO 来表示这些条目并将它们存储在一个列表中。然后,您可以根据姓名或地址等条件对该列表进行排序。

        class Person {

            private final String name;

            private final String address;

            Person(String name, String address) {
                this.name = name;
                this.address = address;
            }

            public String getName() {
                return name;
            }

            public String getAddress() {
                return address;
            }

            @Override
            public String toString() {
                return String.format("name=%s, address=%s", name, address);
            }
        }

        List<Person> people = new ArrayList<>(Arrays.asList(
             new Person("Bernie", "Bstreet"),
             new Person("Aron", "Astreet"),
             new Person("Carol", "Cstreet")
        ));

        List<Person> sortedByName = people.stream()
                .sorted(Comparator.comparing(Person::getName))
                .collect(Collectors.toList());

        List<Person> sortedByAddress = people.stream()
                .sorted(Comparator.comparing(Person::getAddress))
                .collect(Collectors.toList());

【讨论】:

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