如何按字符串对象的升序对对象数组进行排序？

Question

我必须接受用户输入（int id、String title、String 文件夹、int pages）。我必须按字符串标题（字典）的升序获取输出。我已经写了代码，但是我的输出是完全不同的。

package zzz;
import java.util.Scanner;
import java.util.Arrays;
import java.util.Collections;

class Book{
    int id;
    String title;
    String folder;
    int pages;
}


public class practice {

    public static void main(String[] args)throws Exception {
        // TODO Auto-generated method stub
    Scanner sc = new Scanner(System.in);
    Book b1 = new Book();
    Book b2 = new Book();
    Book b3 = new Book();
    Book b[]= {b1,b2,b3};
    for(int i=0;i<b.length;i++) {
        b[i].id=sc.nextInt();
        sc.nextLine();
    b[i].title=sc.next();
        sc.nextLine();
        b[i].folder=sc.next();
        b[i].pages=sc.nextInt();
    }
    Book temp = null;
    for(int i=0;i<b.length;i++) {
        for(int j=0;j<b.length-1-i;j++) {
            if(b[i].title.compareTo(b[j].title)<0) {
             temp =b[j];
            b[j]=b[j+1];
            b[j+1]=temp;
        }}
    }
    for(int i=0;i<b.length;i++) {
        System.out.println(b[i].id+" "+b[i].title+" "+b[i].folder+" "+b[i].pages);
    }

    }}

Answer 1

我会删除开头的类并在main方法之后添加类：

static class Book implements Comparable<Book>{
    int id;
    String title;
    String folder;
    int pages;

    @Override
    public int compareTo(Book other) {
        //If this is backword then switch it to -> other.title.compareTo(this.title);
        return this.title.compareTo(other.title);
     }
}

同：Implements Comparable to get alphabetical sort with Strings

然后你就可以获取书籍数组并使用Arrays.sort(book_arr);

为你：

package zzz;
import java.util.Scanner;
import java.util.Arrays;
import java.util.Collections;


public class practice {

    public static void main(String[] args)throws Exception {
        // TODO Auto-generated method stub
    Scanner sc = new Scanner(System.in);
    Book b1 = new Book();
    Book b2 = new Book();
    Book b3 = new Book();
    Book b[]= {b1,b2,b3};
    for(int i=0;i<b.length;i++) {
        b[i].id=sc.nextInt();
        sc.nextLine();
        b[i].title=sc.next();
        sc.nextLine();
        b[i].folder=sc.next();
        b[i].pages=sc.nextInt();
    }

    //Sort!
    Arrays.sort(b);

    for(int i=0;i<b.length;i++) {
        System.out.println(b[i].id+" "+b[i].title+" "+b[i].folder+" "+b[i].pages);
    }

    }
    static class Book implements Comparable<Book>{
        int id;
        String title;
        String folder;
        int pages;

        @Override
         public int compareTo(Book other) {
             return this.title.compareTo(other.title);
         }
     }
}

Answer 2

您收到 NullPointerException 是因为 b[i].title=sc.next(); 已被注释掉。

在不使 Book Comparable 的情况下这样做的一种方法是对 Stream 进行排序：

Stream.of(b)
    .sorted((thisBook,anotherBook) -> thisBook.title.compareTo(anotherBook.title))
    .forEach(bk -> System.out.println(bk.id+" "+bk.title+" "+bk.folder+" "+bk.pages));
}

Answer 3

您可以在 java 8 中使用stream 简单地处理您的问题，因此您可以使用List 而不是array，然后您必须流它，init，最后根据 title 对书籍进行排序。

夏天

1- 创建 Book 对象（与您的相同）

2- 创建图书列表（而不是数组）

3- Stream bookList，初始化每个书本对象，然后根据标题排序它们

4- 打印结果

我将您的场景编码如下

public static void main(String[] args)  {

        //Create Book Objects
        Scanner sc = new Scanner(System.in);
        Book b1 = new Book();
        Book b2 = new Book();
        Book b3 = new Book();

        //Create BookList
        List<Book> bookList = new ArrayList<>();
        bookList.add(b1);
        bookList.add(b2);
        bookList.add(b3);

        //Stream bookList, Init Books, Sort them based on the title
        bookList.stream().peek(book -> {

            //Init book objects (id , folder , title , pages)
            book.id = sc.nextInt();
            sc.nextLine();
            book.folder = sc.next();
            sc.nextLine();
            book.title = sc.next();
            sc.nextLine();
            book.pages = sc.nextInt();          

        }).sorted(Comparator.comparing(book -> book.title)).collect(Collectors.toList());


        //Print results
        bookList.forEach(book -> {
            System.out.println(book.title);
        });
    }