java - 如何将记录从文件加载到 jTable 中？

Question

我正在使用 netbeans 制作一个 GUI 程序，它应该是用于管理视频商店中的记录的界面。

这是界面。它有两个选项卡，一侧允许用户添加记录，而另一侧选项卡则显示它们。当用户添加记录时，它们将被添加到名为“输出”的 .dat 文件中。我想使用 .dat 文件作为视频记录的永久存储区域，基本上我想要发生的是，当加载 GUI 类时，程序会加载 .dat 文件中的所有记录。我已经创建了代码，但收到以下错误:

run:
java.io.EOFException
    at java.io.RandomAccessFile.readChar(RandomAccessFile.java:773)
    at videostore.BinaryFile.getString(BinaryFile.java:82)
    at videostore.BinaryFile.load(BinaryFile.java:116)
    at videostore.VideoStore.main(VideoStore.java:409)
Exception in thread "main" java.lang.NullPointerException
    at videostore.VideoStore.main(VideoStore.java:420)
/Users/(my Name)/Library/Caches/NetBeans/8.1/executor-snippets/run.xml:53: Java returned: 1
BUILD FAILED (total time: 2 seconds)

然后我将粘贴下面所有相关代码。

在GUI类的main方法中，命名为VideoStore.java:

file = new BinaryFile("/Users/hanaezz/Desktop/output.dat");
        int length = file.length();
        int ID = 1;

        for (int xx = 0; xx < length; xx += file.getRecordLength()) {
            Video load = file.load(xx);
            String tempName = load.getVideoName();
            String tempProd = load.getProducer();
            String tempRat = load.getRating();
            String tempGenre = load.getGenre();
            short tempNum = load.getVidNum();
            float tempPrice = load.getvideoPrice(); 

            Object[] row = {ID, tempName, tempProd, tempGenre, tempRat, tempNum, tempPrice};


            model.addRow(row);

            ID++;
        }

在VideoStore构造函数类中:

public VideoStore() {
        initComponents();
        model = (DefaultTableModel) displayVideos.getModel();
    }

在BinaryFile类中:

private static final int RecordLength = 112;
public static Video load(int place){
        String name = "", prod="", rat="", genre="";

        float price = 1;
        short number = 1;
        try {
            raf.seek(place);
            name = getString(20);
            prod = getString(15);
            rat = getString(20);
            genre = getString(10);
            price = Float.parseFloat(getString(4));
            number = Short.parseShort(getString(4));

            writeString(20, name);
            writeString(15, prod);
            writeString(10, genre);
            writeString(4, VideoStore.vPrice.getText());
            writeString(4, VideoStore.vNumber.getText());
            writeString(4, rat);
        } catch (Exception e) {
            e.printStackTrace();
        }
        Video r = new Video(name, prod, genre, rat, number, price);
        return r;
    }

    public static int getRecordLength() throws IOException{
        return RecordLength;
    }

    public static int length() throws IOException {
        return (int)raf.length();
        }

最后，我的视频类(class):

    private static String videoName;
    private static String producer;
    private static String rating;
    private static String genre;
    private static short videoNumber;
    private static float videoPrice;

public Video(String a, String b, String c, String d, short e, float f){
    videoName = a;
    producer = b;
    rating = c;
    genre = d;
    videoNumber = e;
    videoPrice = f;
}

...然后是类中每个变量的修改器和访问器方法...

@Override
public String toString(){
    return videoName + "\t" + producer +
            "\t" + rating + "\t" + genre +
            "\t" + videoNumber + "\t" + videoPrice;
}

所以，是的，我的问题是我不知道如何将记录从文件加载到表中。在我的代码中，我尝试使用一个循环，该循环将根据记录的大小迭代文件中的每个记录。但是它似乎不起作用。如果有人想查看我的完整代码或需要更多信息，请随时与我联系:)

Answer 1

首先，您应该使用更加面向对象的方法。

您的视频类只包含静态属性，它应该如下所示:

public class Video implements Serializable{

    private String name;
    private String producer; //consider using an object for this
    private String rating; //consider using a numeric type for this
    private String genre; //consider using an object for this
    private int number;
    private double price;


    //getters and setters

}

检查Object-Oriented Programming Concepts .

要添加新视频，您可以从图形界面获取用户输入，并使用它来创建视频对象。

Video video = new Video();

video.setName(nameTextField.getText());
//same for the other attributes

然后您可以将所有视频保存在 List 中.

List<Video> videosList = new ArrayList<>();

videoList.add(video);

然后你就可以serialize将您的列表保存到文件中。

try(FileOutputStream outputFile = new FileOutputStream("/path/to/file");
    ObjectOutputStream out = new ObjectOutputStream(outputFile)){

      out.writeObject(videoList);

} catch (IOException e1) {

      // Handle the exception

}

要从文件中读回列表，您需要反序列化它:

try(FileInputStream inputFile = new FileInputStream("/path/to/file");
    ObjectInputStream in = new ObjectInputStream(inputFile)){

      videoList = (List<Video>)in.readObject();

} catch (IOException e1) {

      // Handle the exception

}