我有一个工作应用程序,它将应用程序输入文本字段中的字符串数据发送到我PC上的TCP服务器。 我已经添加了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),但我无法将收到的消息发送到PC。
没有错误。应用程序只是跳过名为。
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);
}
MainActivity
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);
}
messageSender,和
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();
}
}
}
答案 0 :(得分:0)
问题在于,由于您在onReceive方法中使用异步任务执行网络请求,因此在AsyncTask执行之前,onReceive方法会返回,因为异步任务是异步的。因此,运行onRequest方法的进程变为低优先级,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将其终止。
以下是两个解决方案:
首先确保在Android清单中声明接收器,然后执行以下操作之一。
不是调用异步任务来执行网络请求,而是创建服务并使用该服务运行网络代码。
您可以在onReceive方法中调用goAsync()
方法,告诉系统给接收者更多时间来执行其异步任务。这个代码就像这样:
@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();
}
我认为第二种方式会更简单,所以我建议使用它,因为它不需要你创建一个全新的服务。
答案 1 :(得分:0)
感谢SteelToe带领我这个简单的解决方案:
未在清单文件中声明接收方。