java - 如何使远程 api 中的对象可序列化

Question

我目前正在进行我的第一个练习，使用服务来处理网络操作。我的第一个应用程序想法涉及使用 YahooFinanceApi https://financequotes-api.com/#singlestock-hist 获取具有一些历史记录的股票报价。我目前遇到的问题是将对象(Stock)返回到要显示的游戏 Activity 。我知道我需要使用可序列化或可打包，但是因为只有当我认为可序列化需要更少的编码时我才会使用它。然而 putExtra 似乎没有将我的 Stock 变量 stockInfo 识别为我可以传输的东西。谁能帮助我理解我做错了什么。

StockPul​​lService 文件编写如下。

import android.app.IntentService;
import android.content.Intent;
import android.util.Log;

import java.io.IOException;
import java.util.Calendar;

import yahoofinance.Stock;
import yahoofinance.YahooFinance;
import yahoofinance.histquotes.Interval;
import static android.content.ContentValues.TAG;

public class StockPullService extends IntentService {

    public static final String TICKER = "ticker";

    public StockPullService() {
        super("StockPullService");
    }

    @Override
    protected void onHandleIntent(Intent workIntent) {

        Log.d(TAG, "Service Started");
        if(workIntent != null) {
            int toMonth = (int) (-1 * Math.ceil(Math.random() * 100)); //Randomises what month to end on a

            Calendar from = Calendar.getInstance(); //Makes a calendar from point
            Calendar to = Calendar.getInstance(); //Makes a calendar to point
             from.add(Calendar.MONTH, toMonth - 5); //From a random number of months ago
             to.add(Calendar.MONTH, toMonth); //To 5 months later (To accomodate     for 20 weeks of guesses)

            try {
                Stock stockInfo = YahooFinance.get(workIntent.getStringExtra(TICKER), from, to, Interval.WEEKLY);
                 Log.d(TAG, "Service Successful");
            } catch (IOException e) {
                Log.d(TAG, "Service Failed");
                e.printStackTrace();
            }

            Intent intentResponse = new Intent();
            intentResponse.addCategory(Intent.CATEGORY_DEFAULT);
            intentResponse.putExtra("FetchedStock", stockInfo);
            Log.d(TAG, "Service Completed");
        }

    }
}

游戏 Activity 如下。

公共(public)类 GameActivity 扩展了 Activity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_game);

    Intent intent = new Intent(getApplicationContext(), StockPullService.class);
    intent.putExtra(StockPullService.TICKER, "HM-B.ST");
    startService(intent);

    Stock stock = (Stock) getIntent().getSerializableExtra("FetchedStock");
}

}

Answer 1

您可以为 Stock 创建可序列化的包装器，如下所示:

public class SerializableStock implements Serializable {

    private static final long serialVersionUID = 1L;

    private Stock stock;

    public SerializableStock(final Stock stock) {
        this.stock = stock;
    }

    public Stock getStock() {
        return this.stock;
    }

    private void writeObject(java.io.ObjectOutputStream stream)
        throws IOException {
        stream.writeObject(stock.getSymbol());
        stream.writeObject(stock.getName());
        // TODO: serialize the rest of the fields
    }

    private void readObject(java.io.ObjectInputStream stream)
        throws IOException, ClassNotFoundException {
        this.stock = new Stock((String) stream.readObject());
        this.stock.setName((String) stream.readObject();
        // TODO: deserialize and set the rest of the fields
    }
}

在 StockPul​​lService 中序列化之前将其包装:

@Override
protected void onHandleIntent(Intent workIntent) {
    ...
    intentResponse.putExtra("FetchedStock", new SerializableStock(stockInfo));
    ...
}

并解开它的 GameActivity 类:

@Override
protected void onCreate(Bundle savedInstanceState) {
    ...
    Stock stock = ((SerializableStock) getIntent().getSerializableExtra("FetchedStock")).getStock();
}