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

检查这段代码…这对我很有用:)

public static void isNetworkAvailable(final Handler handler, final int timeout) {
    // ask fo message '0' (not connected) or '1' (connected) on 'handler'
    // the answer must be send before before within the 'timeout' (in milliseconds)

    new Thread() {
        private boolean responded = false;   
        @Override
        public void run() { 
            // set 'responded' to TRUE if is able to connect with google mobile (responds fast) 
            new Thread() {      
                @Override
                public void run() {
                    HttpGet requestForTest = new HttpGet("http://m.google.com");
                    try {
                        new DefaultHttpClient().execute(requestForTest); // can last...
                        responded = true;
                    } 
                    catch (Exception e) {
                    }
                } 
            }.start();

            try {
                int waited = 0;
                while(!responded && (waited < timeout)) {
                    sleep(100);
                    if(!responded ) { 
                        waited += 100;
                    }
                }
            } 
            catch(InterruptedException e) {} // do nothing 
            finally { 
                if (!responded) { handler.sendEmptyMessage(0); } 
                else { handler.sendEmptyMessage(1); }
            }
        }
    }.start();
}

然后,我定义处理程序:

Handler h = new Handler() {
    @Override
    public void handleMessage(Message msg) {

        if (msg.what != 1) { // code if not connected

        } else { // code if connected

        }   
    }
};

...并启动测试:

isNetworkAvailable(h,2000); // get the answser within 2000 ms

其他回答

public boolean isOnline() {
    boolean var = false;
    ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    if ( cm.getActiveNetworkInfo() != null ) {
        var = true;
    }
    return var;
} 

我是这样做的。 我想会更短更有可读性。

干杯!

Saiyan

不要忘记将这些权限添加到您的manifest:

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

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();

Jetpack组成/芬兰湾的科特林

根据Levite的回答,我们可以在Jetpack Compose中使用这个组合:

val DNS_SERVERS = listOf("8.8.8.8", "1.1.1.1", "4.2.2.4")
const val INTERNET_CHECK_DELAY = 3000L
@Composable
fun InternetAwareComposable(
    dnsServers: List<String> = DNS_SERVERS,
    delay: Long = INTERNET_CHECK_DELAY,
    successContent: (@Composable () -> Unit)? = null,
    errorContent: (@Composable () -> Unit)? = null,
    onlineChanged: ((Boolean) -> Unit)? = null
) {
    suspend fun dnsAccessible(
        dnsServer: String
    ) = try {
        withContext(Dispatchers.IO) {
            Runtime.getRuntime().exec("/system/bin/ping -c 1 $dnsServer").waitFor()
        } == 0
    } catch (e: Exception) {
        false
    }

    var isOnline by remember { mutableStateOf(false) }
    LaunchedEffect(Unit) {
        while (true) {
            isOnline = dnsServers.any { dnsAccessible(it) }
            onlineChanged?.invoke(isOnline)
            delay(delay)
        }
    }
    if (isOnline) successContent?.invoke()
    else errorContent?.invoke()
}

这对我很有用。试试吧。

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    try {
        URL url = new URL("http://stackoverflow.com/posts/11642475/edit" );
        //URL url = new URL("http://www.nofoundwebsite.com/" );
        executeReq(url);
        Toast.makeText(getApplicationContext(), "Webpage is available!", Toast.LENGTH_SHORT).show();
    }
    catch(Exception e) {
        Toast.makeText(getApplicationContext(), "oops! webpage is not available!", Toast.LENGTH_SHORT).show();
    }
}

private void executeReq(URL urlObject) throws IOException
{
    HttpURLConnection conn = null;
    conn = (HttpURLConnection) urlObject.openConnection();
    conn.setReadTimeout(30000);//milliseconds
    conn.setConnectTimeout(3500);//milliseconds
    conn.setRequestMethod("GET");
    conn.setDoInput(true);

    // Start connect
    conn.connect();
    InputStream response =conn.getInputStream();
    Log.d("Response:", response.toString());
}}