android中的Message和String之间的转换？

Question

我正在尝试监控android上的网络连接.这不能在主线程上完成,因此数据收集发生在我生成的单独线程上 - 我正在学习使用Handler类每秒向UI报告.这是我的代码的相关片段......

public class MainActivity extends Activity {
    private TextView textView;
    private int linkSpeed;
    Handler handler = new Handler() {
        @Override
        public void handleMessage(Message msg) {
            Message message = msg;
            textView.setText(message);
            setContentView(textView);

        }
    };

    /** Called when the activity is created. */

    @Override
    public void onCreate(Bundle savedInstanceState) {       
        super.onCreate(savedInstanceState);
        textView = new TextView(this);
        textView.setTextSize(25);

        Thread thread = new Thread() {
            public void run() {
                for (;;) {
                    try {
                        WifiManager wifiManager = (WifiManager) getSystemService(Context.WIFI_SERVICE);
                        while (true)
                        {
                            WifiInfo wifiInfo = wifiManager.getConnectionInfo();
                            if (wifiInfo != null) {
                                linkSpeed = wifiInfo.getLinkSpeed(); //measured using WifiInfo.LINK_SPEED_UNITS
                            }
                            else {
                                linkSpeed = -1;
                            }

                    }
                        String message = "linkSpeed = " + linkSpeed;
                        handler.handleMessage(message);
                        Thread.sleep(1000);
                    }
                    catch (Exception e)
                    {
                        e.printStackTrace();
                    }
                }
            }
        };

        thread.start();
    }

问题是,我从我生成的新线程获得的消息是"String"类型,但我似乎无法重载Handler类中的handleMessage方法来获取字符串而不是消息.我不认为在字符串和消息之间存在直接转换,因为Message类中返回String的唯一方法是toString()方法,但它返回描述,而不是消息包含的内容.我也很困惑我如何将一个字符串转换为一个消息 - 我觉得我正在做一个非常迂回的方法.任何帮助将不胜感激!

Answer 1

您不必超载或尝试在String和之间进行任何覆盖Message.你应该做的是将String放入Message类型的对象中并将其发送给Handler.然后handleMessage()从Message 中提取String.

像这样的东西:

// ....
String message = "linkSpeed = " + linkSpeed;
Message msg = Message.obtain(); // Creates an new Message instance
msg.obj = message; // Put the string into Message, into "obj" field.
msg.setTarget(handler); // Set the Handler
msg.sendToTarget(); //Send the message
//....

并在handleMessage():

@Override
public void handleMessage(Message msg) {
    String message = (String) msg.obj; //Extract the string from the Message
    textView.setText(message);
    //....
}

但除此之外,程序还有一个问题:您将无法将数据发送到处理程序,因为代码的这部分是无法访问的:

while (true) {
    WifiInfo s = wifiManager.getConnectionInfo();
    //..
}
String message = "linkSpeed = " + linkSpeed; // This line never won't be reached.

此外,不要忘记Thread在某个时间停止,否则即使应用程序关闭后它仍将继续运行.

Answer 2

您可以使用Bundle将对象(如String)附加到Message对象.

我创建了这个例子:

import android.app.Activity;
import android.os.Bundle;
import android.os.Handler;
import android.os.Message;

public class MainActivity extends Activity {

    Handler handler = new Handler()
    {
        @Override
        public void handleMessage(Message msg) {
            Bundle bundle = msg.getData();

            String text = bundle.getString("key");

            // text will contain the string "your string message"
        }
    };

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

        Thread thread = new Thread()
        {
            @Override
            public void run() {
                Message message = handler.obtainMessage();

                Bundle bundle = new Bundle();
                bundle.putString("key", "your string message");

                message.setData(bundle);

                handler.sendMessage(message);
            }
        };

        thread.start();
    }
}