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();
}

}

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。

我已经应用了@Levit提供的解决方案，并创建了不会调用额外Http请求的函数。

它将解决无法解析主机的错误

public static boolean isInternetAvailable(Context context) {
    ConnectivityManager cm = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
    if (activeNetwork == null) return false;

    switch (activeNetwork.getType()) {
        case ConnectivityManager.TYPE_WIFI:
            if ((activeNetwork.getState() == NetworkInfo.State.CONNECTED ||
                    activeNetwork.getState() == NetworkInfo.State.CONNECTING) &&
                    isInternet())
                return true;
            break;
        case ConnectivityManager.TYPE_MOBILE:
            if ((activeNetwork.getState() == NetworkInfo.State.CONNECTED ||
                    activeNetwork.getState() == NetworkInfo.State.CONNECTING) &&
                    isInternet())
                return true;
            break;
        default:
            return false;
    }
    return false;
}

private static boolean isInternet() {

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

    return false;
}

现在叫它，

if (!isInternetAvailable(getActivity())) {
     //Show message
} else {
     //Perfoem the api request
}

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();

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());
}}

代码:

    fun isInternetConnection(): Boolean {
    var returnVal = false
    thread {
        returnVal = try {
            khttp.get("https://www.google.com/")
            true
        }catch (e:Exception){
            false
        }
    }.join()
    return returnVal
}

Gradle:

implementation 'io.karn:khttp-android:0.1.0'

我使用khttp是因为它很容易使用。

在上面的代码中，如果它成功连接到google.com，它返回true否则为false。很简单。我不明白为什么每个人都要写这么长的代码，即使是为了这么简单的事情。