android - 将对象序列化为android中的文件

Question

现在写下我希望能够使用序列化保存和打开的此类:

public class Region
implements Serializable
{
private final int      inputNumberOfColumnsAlongXAxis;
private final int      inputNumberOfColumnsAlongYAxis;
private double         inputDataScaleReductionOnXAxis;
private double         inputDataScaleReductionOnYAxis;

private int            numberOfColumnsAlongXAxis;
private int            numberOfColumnsAlongYAxis;
private int            cellsPerColumn;                // Z-Axis dimension
private float          inhibitionRadius;
private final float    percentMinimumOverlapScore;
private final float    minimumOverlapScore;

// ----------------------------------------------------------
/**
 * Save the current Region object on the view into a file named
 * "TrainedRegion".
 */
public static byte[] serializeObject(Object region)
{
    // TODO: save with current timestamp and use open pop-up
    // String timeStamp = "" + System.currentTimeMillis() / 1000;
    // String fileName = ("Region created at timestamp(seconds): " +
    // timeStamp);
    ByteArrayOutputStream baos = new ByteArrayOutputStream();

    try {
      ObjectOutput objectOutput = new ObjectOutputStream(baos);
      objectOutput.writeObject(region);
      objectOutput.close();

      // Get the bytes of the serialized object
      byte[] bytesOfSerializedObject = baos.toByteArray();

      return bytesOfSerializedObject;
    } catch(IOException ioe) {
      Log.e("serializeObject", "error", ioe);

      return null;
    }
}


// ----------------------------------------------------------
/**
 * Returns a previously saved Region object with the given name.
 *
 * @param fileName
 * @return A previously saved Region object.
 */
public static Object deserializeObject(byte[] bytes)
{
    // TODO: later read Region object saved in file named by the time stamp during
    // saving.
    // ObjectInputStream inputStream = new ObjectInputStream(new
    // FileInputStream(fileName));

    try {
        ObjectInputStream in = new ObjectInputStream(new ByteArrayInputStream(bytes));
        Object object = in.readObject();
        in.close();

        return object;
      } catch(ClassNotFoundException cnfe) {
        Log.e("deserializeObject", "class not found error", cnfe);

        return null;
      } catch(IOException ioe) {
        Log.e("deserializeObject", "io error", ioe);

        return null;
      }
}
}

以下是我的“屏幕”类，它充当我上面的模型和我的 View (屏幕)的 Controller :

// ----------------------------------------------------------
/**
 * I want to be able to open a Region instance that was saved.
 */
public void openClicked() 
{
    // open the file created below in save method and write all the bytes into
    // the global region instance for this class. How can do this?

    this.region = (Region) this.region.deserializeObject(savedRegionInBytes);
}


// ----------------------------------------------------------
/**
 *
 */
public void saveClicked()
{
    this.savedRegionInBytes = this.region.serializeObject(this.region);

    // TODO: write savedRegionInBytes to a file. How can I do this?
}

此外，如果有更简单的方法将对象序列化为 android 中的文件，我很乐意听到它。谢谢!

Answer 1

基本上就是这样。但是你不必像你那样处理字节。我使用这两种方法将 Serializable 对象写入和读取到 Android 应用程序中的私有(private)文件。

public static boolean saveRegion(Context context, Region region) {
    try {
        FileOutputStream fos = context.openFileOutput(REGION_FILENAME, Context.MODE_PRIVATE);
        ObjectOutputStream oos = new ObjectOutputStream(fos);
        oos.writeObject(region);
        oos.close();
    } catch (IOException e) {
        e.printStackTrace();
        return false;
    }

    return true;
}

public static Region getRegion(Context context) {
    try {
        FileInputStream fis = context.openFileInput(REGION_FILENAME);
        ObjectInputStream is = new ObjectInputStream(fis);
        Object readObject = is.readObject();
        is.close();

        if(readObject != null && readObject instanceof Region) {
            return (Region) readObject;
        }
    } catch (IOException e) {
        e.printStackTrace();
    } catch (ClassNotFoundException e) {
        e.printStackTrace();
    }

    return null;
}