我得到了一个AsyncTask,应该检查对主机名的网络访问。但是doInBackground()永远不会超时。有人知道吗?

public class HostAvailabilityTask extends AsyncTask<String, Void, Boolean> {

    private Main main;

    public HostAvailabilityTask(Main main) {
        this.main = main;
    }

    protected Boolean doInBackground(String... params) {
        Main.Log("doInBackground() isHostAvailable():"+params[0]);

        try {
            return InetAddress.getByName(params[0]).isReachable(30); 
        } catch (UnknownHostException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return false;       
    }

    protected void onPostExecute(Boolean... result) {
        Main.Log("onPostExecute()");

        if(result[0] == false) {
            main.setContentView(R.layout.splash);
            return;
        }

        main.continueAfterHostCheck();
    }   
}

当前回答

网络连接/互联网接入

isConnectedOrConnecting()(在大多数回答中使用)检查任何网络连接 要了解这些网络是否有internet接入,请使用以下方法之一

A) Ping服务器(简单)

// ICMP 
public boolean isOnline() {
    Runtime runtime = Runtime.getRuntime();
    try {
        Process ipProcess = runtime.exec("/system/bin/ping -c 1 8.8.8.8");
        int     exitValue = ipProcess.waitFor();
        return (exitValue == 0);
    }
    catch (IOException e)          { e.printStackTrace(); }
    catch (InterruptedException e) { e.printStackTrace(); }

    return false;
}

+可以在主线程上运行

在一些旧设备上不能工作(Galays S3等),如果没有网络,它会阻塞一段时间。

B)连接到Internet上的Socket(高级)

// TCP/HTTP/DNS (depending on the port, 53=DNS, 80=HTTP, etc.)
public boolean isOnline() {
    try {
        int timeoutMs = 1500;
        Socket sock = new Socket();
        SocketAddress sockaddr = new InetSocketAddress("8.8.8.8", 53);

        sock.connect(sockaddr, timeoutMs);
        sock.close();

        return true;
    } catch (IOException e) { return false; }
}

+非常快(任何一种方式),适用于所有设备,非常可靠

-不能在UI线程上运行

这工作非常可靠,在每个设备上,非常快。它需要在一个单独的任务中运行(例如ScheduledExecutorService或AsyncTask)。

可能的问题

Is it really fast enough? Yes, very fast ;-) Is there no reliable way to check internet, other than testing something on the internet? Not as far as I know, but let me know, and I will edit my answer. What if the DNS is down? Google DNS (e.g. 8.8.8.8) is the largest public DNS in the world. As of 2018 it handled over a trillion queries a day [1]. Let 's just say, your app would probably not be the talk of the day. Which permissions are required? <uses-permission android:name="android.permission.INTERNET" /> Just internet access - surprise ^^ (Btw have you ever thought about, how some of the methods suggested here could even have a remote glue about internet access, without this permission?)

 

额外:一次性RxJava/RxAndroid示例(Kotlin)

fun hasInternetConnection(): Single<Boolean> {
  return Single.fromCallable {
    try {
      // Connect to Google DNS to check for connection
      val timeoutMs = 1500
      val socket = Socket()
      val socketAddress = InetSocketAddress("8.8.8.8", 53)
    
      socket.connect(socketAddress, timeoutMs)
      socket.close()
  
      true
    } catch (e: IOException) {
      false
    }
  }
  .subscribeOn(Schedulers.io())
  .observeOn(AndroidSchedulers.mainThread())
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    hasInternetConnection().subscribe { hasInternet -> /* do something */}

额外:一次性RxJava/RxAndroid示例(Java)

public static Single<Boolean> hasInternetConnection() {
    return Single.fromCallable(() -> {
        try {
            // Connect to Google DNS to check for connection
            int timeoutMs = 1500;
            Socket socket = new Socket();
            InetSocketAddress socketAddress = new InetSocketAddress("8.8.8.8", 53);

            socket.connect(socketAddress, timeoutMs);
            socket.close();

            return true;
        } catch (IOException e) {
            return false;
        }
    }).subscribeOn(Schedulers.io()).observeOn(AndroidSchedulers.mainThread());
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    hasInternetConnection().subscribe((hasInternet) -> {
        if(hasInternet) {

        }else {

        }
    });

额外:一次性AsyncTask示例

注意:这是如何执行请求的另一个示例。然而,由于AsyncTask已弃用,它应该被你的应用程序的线程调度,Kotlin协程,Rx,…

class InternetCheck extends AsyncTask<Void,Void,Boolean> {

    private Consumer mConsumer;
    public  interface Consumer { void accept(Boolean internet); }

    public  InternetCheck(Consumer consumer) { mConsumer = consumer; execute(); }

    @Override protected Boolean doInBackground(Void... voids) { try {
        Socket sock = new Socket();
        sock.connect(new InetSocketAddress("8.8.8.8", 53), 1500);
        sock.close();
        return true;
    } catch (IOException e) { return false; } }

    @Override protected void onPostExecute(Boolean internet) { mConsumer.accept(internet); }
}

///////////////////////////////////////////////////////////////////////////////////
// Usage

    new InternetCheck(internet -> { /* do something with boolean response */ });

其他回答

只需创建下面的类来检查internet连接:

public class ConnectionStatus {

    private Context _context;

    public ConnectionStatus(Context context) {
        this._context = context;
    }

    public boolean isConnectionAvailable() {
        ConnectivityManager connectivity = (ConnectivityManager) _context
                .getSystemService(Context.CONNECTIVITY_SERVICE);
        if (connectivity != null) {
            NetworkInfo[] info = connectivity.getAllNetworkInfo();
            if (info != null)
                for (int i = 0; i < info.length; i++)
                    if (info[i].getState() == NetworkInfo.State.CONNECTED) {
                        return true;
                    }
        }
        return false;
    }
}

该类仅包含一个返回连接状态布尔值的方法。因此,简单来说,如果该方法找到一个到Internet的有效连接,则返回值为true,否则为false,如果没有找到有效连接。

MainActivity中的下面的方法调用前面描述的方法的结果,并提示用户进行相应的操作:

public void addListenerOnWifiButton() {
        Button btnWifi = (Button)findViewById(R.id.btnWifi);

        iia = new ConnectionStatus(getApplicationContext());

        isConnected = iia.isConnectionAvailable();
        if (!isConnected) {
            btnWifi.setOnClickListener(new View.OnClickListener() {

                @Override
                public void onClick(View v) {
                    startActivity(new Intent(Settings.ACTION_WIFI_SETTINGS));
                    Toast.makeText(getBaseContext(), "Please connect to a hotspot",
                            Toast.LENGTH_SHORT).show();
                }
            });
        }
        else {
            btnWifi.setVisibility(4);
            warning.setText("This app may use your mobile data to update events and get their details.");
        }
    }

在上面的代码中,如果结果为假,(因此没有互联网连接,用户将被带到Android wi-fi面板,在那里他将被提示连接到wi-fi热点。

这个线程中的大多数答案只检查是否有可用的连接,但不检查该连接是否工作,其他答案不是设备范围,我的解决方案应该在每个设备上工作。

你可以在启动应用程序之前在你的主要活动中删除我的代码,它会快速确定是否有实际的互联网连接,如果有对话框将立即删除,应用程序将被启动,如果没有警报会弹出说应用程序需要互联网连接才能工作。

final AlertDialog alertDialog = new AlertDialog.Builder(this).create();
        alertDialog.setTitle("Checking Connection");
        alertDialog.setMessage("Checking...");
        alertDialog.show();
        new CountDownTimer(5000, 1000) {
            @Override
            public void onTick(long millisUntilFinished) {

                new Thread(new Runnable() {
                    public void run() {
                        try {
                            URL url = new URL("http://web.mit.edu/");
                            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
                            connection.setRequestMethod("GET");
                            connection.setConnectTimeout(5000);
                            isConnected = connection.getResponseCode() == HttpURLConnection.HTTP_OK;
                        } catch (Exception e) {
                            e.printStackTrace();
                        }
                    }
                }).start();

                if (isConnected == false){
                    alertDialog.setMessage("Try " +  (5 - millisUntilFinished/1000) + " of 5.");
                } else {
                    alertDialog.dismiss();
                }
            }
            @Override
            public void onFinish() {
                if (isConnected == false) {
                    alertDialog.dismiss();
                    new AlertDialog.Builder(activity)
                            .setTitle("No Internet")
                            .setMessage("Please connect to Internet first.")
                            .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                                public void onClick(DialogInterface dialog, int which) {
                                    // kill the app?
                                }
                            })
                            .setIcon(android.R.drawable.ic_dialog_alert)
                            .show();
                } else {
                    // Launch the app
                }
            }
        }.start();

如果你正在使用Firebase,你可以使用这个。

Java:

DatabaseReference connectedRef = FirebaseDatabase.getInstance().getReference(".info/connected");
connectedRef.addValueEventListener(new ValueEventListener() {
    @Override
    public void onDataChange(@NonNull DataSnapshot snapshot) {
        boolean connected = snapshot.getValue(Boolean.class);
        if (connected) {
            Log.d(TAG, "connected");
        } else {
            Log.d(TAG, "not connected");
        }
    }

    @Override
    public void onCancelled(@NonNull DatabaseError error) {
        Log.w(TAG, "Listener was cancelled");
    }
});

科特林:

val connectedRef = Firebase.database.getReference(".info/connected")
connectedRef.addValueEventListener(object : ValueEventListener {
    override fun onDataChange(snapshot: DataSnapshot) {
        val connected = snapshot.getValue(Boolean::class.java) ?: false
        if (connected) {
            Log.d(TAG, "connected")
        } else {
            Log.d(TAG, "not connected")
        }
    }

    override fun onCancelled(error: DatabaseError) {
        Log.w(TAG, "Listener was cancelled")
    }
})
public class Network {

Context context;

public Network(Context context){
    this.context = context;
}

public boolean isOnline() {
    ConnectivityManager cm =
            (ConnectivityManager)context.getSystemService(Context.CONNECTIVITY_SERVICE);

    NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
    return activeNetwork != null &&
                          activeNetwork.isConnectedOrConnecting();
}

}

最好的方法:

public static boolean isOnline() {
    try {
    InetAddress.getByName("google.com").isReachable(3);

    return true;
    } catch (UnknownHostException e){
    return false;
    } catch (IOException e){
    return false;
    }
    }