这段代码将帮助您找到互联网是否打开。

public final boolean isInternetOn() {
        ConnectivityManager conMgr = (ConnectivityManager) this.con
                .getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo info = conMgr.getActiveNetworkInfo();
        return (info != null && info.isConnected());
}

此外，您应该提供以下权限

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

要让getActiveNetworkInfo()工作，您需要向清单中添加以下内容。

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

使用这个Kotlin扩展:

/**
 * Check whether network is available
 *
 * @param context
 * @return Whether device is connected to Network.
 */
fun Context.isNetworkAvailable(): Boolean {
    with(getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager) {
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
            //Device is running on Marshmallow or later Android OS.
            with(getNetworkCapabilities(activeNetwork)) {
                return hasTransport(NetworkCapabilities.TRANSPORT_WIFI) || hasTransport(
                    NetworkCapabilities.TRANSPORT_CELLULAR
                )
            }
        } else {
            activeNetworkInfo?.let {
                // connected to the internet
                @Suppress("DEPRECATION")
                return listOf(ConnectivityManager.TYPE_WIFI, ConnectivityManager.TYPE_MOBILE).contains(it.type)
            }
        }
    }
    return false
}

有不止一种方法

第一，最短但效率低的方法

只需要网络状态权限

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

然后这个方法，

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

        NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
        boolean isConnected = activeNetwork != null &&
                activeNetwork.isConnected();

        return isConnected;

    }

正如在回答中所看到的ConnectivityManager是一个解决方案，我只是在一个方法中添加了它，这是一个简化的方法 ConnectivityManager返回true，如果有网络访问而不是互联网访问，这意味着如果你的WiFi连接到路由器，但路由器没有互联网，它返回true，它检查连接可用性

二、高效的方式

需要网络状态和Internet权限

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

然后这门课，

 public class CheckInternetAsyncTask extends AsyncTask<Void, Integer, Boolean> {

        private Context context;

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

        @Override
        protected Boolean doInBackground(Void... params) {

            ConnectivityManager cm =
                    (ConnectivityManager)context.getSystemService(Context.CONNECTIVITY_SERVICE);

            assert cm != null;
            NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
            boolean isConnected = activeNetwork != null &&
                    activeNetwork.isConnected();


            if (isConnected) {
                try {
                    HttpURLConnection urlc = (HttpURLConnection)
                            (new URL("http://clients3.google.com/generate_204")
                                    .openConnection());
                    urlc.setRequestProperty("User-Agent", "Android");
                    urlc.setRequestProperty("Connection", "close");
                    urlc.setConnectTimeout(1500);
                    urlc.connect();
                    if (urlc.getResponseCode() == 204 &&
                            urlc.getContentLength() == 0)
                        return true;

                } catch (IOException e) {
                    Log.e("TAG", "Error checking internet connection", e);
                    return false;
                }
            } else {
                Log.d("TAG", "No network available!");
                return false;
            }


            return null;
        }

        @Override
        protected void onPostExecute(Boolean result) {
            super.onPostExecute(result);
            Log.d("TAG", "result" + result);

            if(result){
                // do ur code
            }

        }


    }

叫CheckInternetAsyncTask

new CheckInternetAsyncTask(getApplicationContext()).execute();

部分解释:-

you have to check Internet on AsyncTask, otherwise it can throw android.os.NetworkOnMainThreadException in some cases ConnectivityManager used to check the network access if true sends request (Ping) Request send to http://clients3.google.com/generate_204, This well-known URL is known to return an empty page with an HTTP status 204 this is faster and more efficient than http://www.google.com , read this. if you have website it's preferred to put you website instead of google, only if you use it within the app Timeout can be changed range (20ms -> 2000ms), 1500ms is commonly used

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();

不需要太复杂。最简单的框架方式是使用ACCESS_NETWORK_STATE权限并创建一个连接的方法

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

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

如果您有特定的主机和连接类型(wifi/移动)，也可以使用requestRouteToHost。

你还需要:

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

在你的android清单中。