标签：java methods nullpointerexception class

我在行playlist1.firstSong = song中收到nullPointerException错误;下面(第9行).有任何想法吗？

播放列表类：

public class Playlist { 
Scanner console = new Scanner(System.in); 
private Playlist playlist1=null, playlist2=null; 

private Song firstSong;
private Song secondSong;
private Song thirdSong;

public void setSong(Song song) { 
    if (song != null) {
        if (playlist1.firstSong == null) {
            playlist1.firstSong = song;
            System.out.println("The song has been added to the playlist.");
        }

        else if (playlist1.secondSong == null) {
            playlist1.secondSong = song;
            System.out.println("The song has been added to the playlist.");
        }

        else if (playlist1.thirdSong == null) {
            playlist1.thirdSong = song;
            System.out.println("The song has been added to the playlist.");
        }
        else {
            System.out.println("This playlist is currently full with 3 songs. Please delete a song before attempting to add a new one.");
        }
    }
   }

addSongToPlaylist方法：

private void addSongToPlaylist() { 
  if (songCount <=3) { 

    System.out.println("Please enter the number of the song you'd like to be added to the playlist."); 
    System.out.println("");

    database.Display();

    int songNumber; 
    songNumber = console.nextInt(); 

    switch (songNumber) { 
        case 1:
            playlist.setSong(database.getSong(1)); 
            break;
        case 2:
            playlist.setSong(database.getSong(2)); 
            break;
        case 3:
            playlist.setSong(database.getSong(3)); 
            break;
        case 4:
            playlist.setSong(database.getSong(4)); 
            break;
        default:
            System.out.println("Please enter a valid song number."); 
            break;
    }
    songCount++; 
  }

getSong方法：

public Song getSong(int songNumber) { 
    if (songNumber == 1){ 
        return song1; 
    }
    else if (songNumber == 2){ 
        return song2; 
    }
    else if (songNumber == 3){ 
        return song3; 
    }
    else if (songNumber == 4){ /
        return song4; 
    }

    else {
        return song1; 
}
}

任何帮助将非常感谢,谢谢！

解决方法:

嗯……你在播放列表课上感到困惑.您的播放列表实际上存储了3首歌曲就是这样,您不需要使用播放列表1&播放列表2.
尝试使用此代替：

public class Playlist { 
private Song firstSong;
private Song secondSong;
private Song thirdSong;

public void setSong(Song song) { 
    if (song != null) {
        if (this.firstSong == null) {
            this.firstSong = song;
            System.out.println("The song has been added to the playlist.");
        }

        else if (this.secondSong == null) {
            this.secondSong = song;
            System.out.println("The song has been added to the playlist.");
        }

        else if (this.thirdSong == null) {
            this.thirdSong = song;
            System.out.println("The song has been added to the playlist.");
        }
        else {
            System.out.println("This playlist is currently full with 3 songs. Please delete a song before attempting to add a new one.");
        }
    }
   }

认为在创建播放列表实例时,您可以在那里存储3首歌曲,就是这样.如果您想要更多播放列表,可以创建更多播放列表.

但是,如果你想存储超过3首歌曲怎么办？为此,您可以通过使用数组(或任何类型的存储,实际上)来重构您的代码.让我们尝试使用Vector：

public class Playlist { 

private Vector<Song> songList;

public void setSong(Song song) { 
    if (song != null) {
             songList.add(song);
        }
   }
public Song getSong(int nb) {
    if (nb > 0 && nb < songList.size()) //We don't want to check the song #-1 or a song that would be out of bonds
         return songList.elementAT(nb);
}
 } 

而且你有一些更清洁的东西. (上面的代码肯定有一些拼写错误,我不能在这里检查它们,但它就是一个例子.)

作为评论的答案：
如果你想使用2个播放列表,那很好,只需在你的主页中使用：

   Playlist firstPlaylist;
   Playlist secondPlaylist;

标签：java,methods,nullpointerexception,class
来源： https://codeday.me/bug/20190717/1489537.html

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