我正在使用netbeans制作一个GUI程序,它应该是一个用于管理视频商店中记录的界面。
这是界面。它是两个标签,一面允许一个人添加记录,另一面显示它们。当一个人添加记录时,它们会被添加到名为output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,程序加载.dat文件中的所有记录。我已经创建了我的代码,但是我收到了以下错误:
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)
然后我会粘贴下面的所有相关代码。
在GUI类的main方法中,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++;
}
VideoStore 构造函数类中的:
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:
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mutator和accessor方法......
@Override
public String toString(){
return videoName + "\t" + producer +
"\t" + rating + "\t" + genre +
"\t" + videoNumber + "\t" + videoPrice;
}
所以是的,我的问题是我无法弄清楚如何将文件中的记录加载到表格中。在我的代码中,我尝试使用一个循环,它将根据记录的大小迭代文件中的每个记录..但它似乎没有起作用。如果有人想查看我的完整代码或需要更多信息,请不要犹豫与我联系:)
答案 0 :(得分:1)
首先,您应该使用更多面向对象的方法。
您的视频类只包含静态属性,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。
List<Video> videosList = new ArrayList<>();
videoList.add(video);
然后您可以serialize将您的列表添加到文件中。
try(FileOutputStream outputFile = new FileOutputStream("/path/to/file");
ObjectOutputStream out = new ObjectOutputStream(outputFile)){
out.writeObject(videoList);
} catch (IOException e1) {
// Handle the exception
}
要从文件中回读列表,您需要对其进行反序列化:
try(FileInputStream inputFile = new FileInputStream("/path/to/file");
ObjectInputStream in = new ObjectInputStream(inputFile)){
videoList = (List<Video>)in.readObject();
} catch (IOException e1) {
// Handle the exception
}