2

当我的应用程序启动时，它会连接到数据库并以 XML 格式下载一些数据。这是布局：

<Customers>
  <Customer> 
    <name>...</name>
    ...
  </Customer>
   <Customer> 
    <name>...</name>
    ...
  </Customer>
  ...
</Customer>

对于每个客户，我创建一个Customer存储在 ArrayList 中的类的对象。用户可以编辑和添加每个客户的一些数据，他们可以将其上传回服务器。

问题是我不知道在本地存储数据的最佳方式是什么，所以如果我的应用程序关闭或用户不再有互联网，他们将有一个备份。我现在将所有Customer对象转换回 XML，然后将其存储在 SD 卡上，但我认为这不是一个好的解决方案。每个客户有大约 40 个字符串、10 个整数和一些我必须存储的布尔值。有没有更好的方法在本地存储数据？

我发现一些代码可以存储一个 ArrayList 但它不适用于自定义类。

编辑：我使用本教程来解决我的问题。

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutput;
import java.io.ObjectOutputStream;
import java.util.ArrayList;

import android.os.Environment;

public class SerializeData {
    public static byte[] serializeObject(Object o) {
        ByteArrayOutputStream bos = new ByteArrayOutputStream();

        try {
            ObjectOutput out = new ObjectOutputStream(bos);
            out.writeObject(o);
            out.close();

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

            return buf;
        } catch (IOException ioe) {
            CreateLog.addToLog(ioe.toString());
            return null;
        }
    }

    public static Object deserializeObject(byte[] b) {
        try {
            ObjectInputStream in = new ObjectInputStream(
                    new ByteArrayInputStream(b));
            Object object = in.readObject();
            in.close();

            return object;
        } catch (ClassNotFoundException cnfe) {
            CreateLog.addToLog(cnfe.toString());

            return null;
        } catch (IOException ioe) {
            CreateLog.addToLog(ioe.toString());

            return null;
        }
    }

    public static void saveData(){
        byte[] arrayData = SerializeData
                .serializeObject(MyTasks.allCustomers);

        BufferedOutputStream bos;
        try {
            bos = new BufferedOutputStream(new FileOutputStream(
                    Environment.getExternalStorageDirectory()
                            + "/myprogram/myarray.txt"));
            bos.write(arrayData);
            bos.flush();
            bos.close();
        } catch (FileNotFoundException e) {
            CreateLog.addToLog(e.toString());
        } catch (IOException e) {
            CreateLog.addToLog(e.toString());
        }
    }

    public static ArrayList<Customer> getData(){
        File afile = new File(Environment.getExternalStorageDirectory()
                + "/myprogram/myarray.txt");
        int size = (int) afile.length();
        byte[] bytes = new byte[size];
        try {
            BufferedInputStream buf = new BufferedInputStream(new FileInputStream(afile));
            buf.read(bytes, 0, bytes.length);
            buf.close();
        } catch (FileNotFoundException e) {
            CreateLog.addToLog(e.toString());
        } catch (IOException e) {
            CreateLog.addToLog(e.toString());
        }

        return (ArrayList<Customer>) SerializeData.deserializeObject(bytes); 

    }
}
4

1 回答 1

1

As I understand, the Customer structure will be read and written only locally and also by the same or highly related program. For such situation, I think, Java serialization is appropriate as it adds very little code.

Make the Customer to implement Serializable (put immediately serialVersionUID so you could control versions later). ArrayList is already Serializable. Obtain the file name as explained here and use ObjectInput/Output stream to serialize. You can always migrate to XML/JSON later if the need arises. A simple "hello world" example for serialization can be found here.

于 2013-01-13T15:34:52.587 回答