Question

我一直在尝试将我的一个Java类保存到xml文件中，以便在重新运行我的应用程序时加载它，但是我得到了一个奇怪的结果。这是我要保存的课程：

public //public definition of a Song object
class Song implements Comparable, Serializable {
    //fields
    private StringProperty name;
    public StringProperty artist;
    public StringProperty album;
    public IntegerProperty year;

    public boolean emptySong;
    //for serialization
    private static final long serialVersionUID = 1738L;

    //minimum constructor
    public Song(){
        this.emptySong = true;
    }
    public Song(String name, String artist){
        this.name = new SimpleStringProperty(name);
        this.artist = new SimpleStringProperty(artist);
    }

    //setter methods
    public void setName(String name){
        this.name = new SimpleStringProperty(name);
    }

    public void setArtist(String artist){
        this.artist = new SimpleStringProperty(artist);
    }

    public void setAlbum(String album){
        this.album = new SimpleStringProperty(album);
    }

    public void setYear(int i){
        this.year = new SimpleIntegerProperty(i);
    }

    //getter methods
    public String getName(){
        return name.get();
    }
    public StringProperty nameProperty(){
        return name;
    }

    public StringProperty getArtist(){
        return artist;
    }
    public String getArtistString(){
        return artist.get();
    }
    public StringProperty getAlbum(){
        return album;
    }
    public String getAlbumString(){
        if(album.get() != null){
            return album.get();
        }else{
            return null;
        }
    }

    public IntegerProperty getYear(){
        return year;
    }
    public int getYearInt(){
        return year.get();
    }

    @Override
    public String toString() {
        return name.toString();
    }

    @Override
    public int compareTo(Object o) {
        if (!(o instanceof Song)){
            return -1;
        }
        String curr = name + "" + artist;
        String comp = ((Song) o).getName().toString() + ((Song) o).getArtist().toString();
        return curr.compareTo(comp);
    }
}

这是我的保存方法：

public Songs songs;
public void saveSongDataToFile() {
    try {
        File file = new File("data/data.xml");
        JAXBContext context = JAXBContext
                .newInstance(SaveData.class);
        Marshaller m = context.createMarshaller();
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

        // Wrapping our person data.
        SaveData wrapper = new SaveData();
        wrapper.setSongs(songs.songlist);

        // Marshalling and saving XML to the file.
        m.marshal(wrapper, file);

        // Save the file path to the registry.
        //setSongFilePath(file);
    } catch (Exception e) { // catches ANY exception
        Alert alert = new Alert(AlertType.ERROR);
        alert.setTitle("Error");
        alert.setHeaderText("Could not save data");
        alert.setContentText("Could not save data to file:\n");
        e.printStackTrace();
        alert.showAndWait();
    }
}

但是我的输出文件正在保存没有艺术家，专辑或年份信息的数据。所有元素都与此类似：

<song>
    <artist/>
    <album/>
    <year/>
    <emptySong>false</emptySong>
    <name>Song 1</name>
</song>

我不确定为什么会这样。

编辑：这是saveData类：

@XmlRootElement(name="songs")
public class SaveData {
    private List<Song> songs;

    @XmlElement(name="song")
    public List<Song> getSongs(){
        return songs;
    }

    public void setSongs(List<Song> songs){
        this.songs = songs;
    }

}

歌曲只是在saveSongDataToFile（）方法上面的内存中声明。

Answer 1

我发现问题所在，JAXB不知道如何处理StringProperty，所以它需要字符串，默认情况下，它将访问getter和setter属性。我把线放了

@XmlElement(name="artist")

在getArtistString（）方法之上，这使得艺术家标签工作，所以我对其余部分做了同样的事情，现在它完美地工作了。

Java XML不保存所有元素

1 个答案: