我有一个正在运行的应用程序,它将字符串数据从应用程序的输入文本字段发送到我的 PC 上的 TCP 服务器。 我添加了 BroadcastReceiver 类,以便在收到消息时将消息发送到计算机。 应用程序接收消息(Toast 正在工作),但我无法让它将收到的消息发送到 PC。
没有错误。应用程序只是跳过调用的方法。
如何在onReceive中调用方法?
感谢您的帮助!
public class SmsReceiver extends BroadcastReceiver {
String phone;
String message;
@Override
public void onReceive(Context context, Intent intent) {
Bundle intentExtras = intent.getExtras();
if (intentExtras != null) {
// Get Messages
Object[] sms = (Object[]) intentExtras.get("pdus");
for (int i = 0; i < sms.length; ++i) {
// Parse Each Message
SmsMessage smsMessage = SmsMessage.createFromPdu((byte[]) sms[i]);
phone = smsMessage.getOriginatingAddress();
message = smsMessage.getMessageBody().toString();
System.out.println("Message is: "+message);
Log.i("sth","message "+message);
//What i have tried
sendMessage(message);
//creating instance
MessageSender messageSender = new MessageSender();
messageSender.execute(message);
//calling metod from MainActivity
((MainActivity)context.getApplicationContext()).send2(message);
MainActivity.send2(message);
//toast is working fine
Toast.makeText(context,"Alert:"+ phone + ": " + message, Toast.LENGTH_SHORT).show();
}
}
}
public void sendMessage(String s)
{
MessageSender messageSender = new MessageSender();
messageSender.execute(s);
}
主要 Activity
public class MainActivity extends AppCompatActivity {
EditText e1;
@Override
protected void onCreate(final Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
e1 = (EditText)findViewById(R.id.editText);
//This works
MessageSender messageSender = new MessageSender();
messageSender.execute("App launched!");
}
//On Button click. Working good.
public void send(View v)
{
MessageSender messageSender = new MessageSender();
messageSender.execute(e1.getText().toString());
}
//For test purposes
public static void send2(String s)
{
MessageSender messageSender = new MessageSender();
messageSender.execute(s);
}
消息发送者
public class MessageSender extends AsyncTask<String,Void,Void>
{
Socket s;
DataOutputStream dos;
PrintWriter pw;
@Override
public Void doInBackground(String... voids) {
String message = voids[0];
try {
s = new Socket("192.168.1.69", 7800);
pw = new PrintWriter(s.getOutputStream());
pw.write(message);
pw.flush();
pw.close();
s.close();
}catch(IOException e)
{
e.printStackTrace();
}
return null;
}
}
添加了服务器代码:
public class MyServerFrame extends javax.swing.JFrame
{
static Socket s;
static ServerSocket ss;
static InputStreamReader isr;
static BufferedReader br;
static String message;public MyServerFrame()
{
}
public static void main(String args[])
{
try
{
ss = new ServerSocket(7800);
while(true)
{
s=ss.accept();
isr = new InputStreamReader(s.getInputStream());
br = new BufferedReader(isr);
StringBuilder everything = new StringBuilder();
String line;
while( (line = reader.readLine()) != null)
{
everything.append(line);
}
message = everything.toString();//multiple to onelinestring
System.out.println(message);
}
}catch (IOException e)
{
e.printStackTrace();
}
}
}
最佳答案
问题是,由于您在 onReceive 方法中使用异步任务执行网络请求,因此 onReceive 方法在 AsyncTask 执行之前返回,因为异步任务是异步的。因此,运行 onRequest 方法的进程将变为低优先级,因为 onReceive 方法已返回,并且操作系统将在异步任务实际执行之前终止它。
这里有两个解决方案:
首先确保您的 Android list 中声明了接收器,然后执行以下操作之一。
不要调用异步任务来执行网络请求,而是创建一个服务并使用该服务来运行网络代码。
您可以在 onReceive 方法中调用
goAsync()
方法来告诉系统给接收者更多的时间来执行其异步任务。其代码如下所示:@Override public void onReceive(final Context context, final Intent intent) { //create a pending intend that you will pass to the Async task so you can tell the system when the Async Task finished so that it can recycle. final PendingResult pendingResult = goAsync(); AsyncTask<String, Integer, String> asyncTask = new AsyncTask<String, Integer, String>() { @Override protected String doInBackground(String... params) { //put the network calling code in here // Must call finish() so the BroadcastReceiver can be recycled. pendingResult.finish(); return data; } }; asyncTask.execute(); }
我认为第二种方法会更简单,所以我建议使用它,因为它不需要您创建一个全新的服务。
关于java - BroadcastReceiver onReceive 方法内的异步任务未运行?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49517983/