17

如何以编程方式检查 android 设备上的网络是否可用，当我们尝试连接 Wifi 和 3G 等网络时会抛出消息或 toast 消息。

4

10 回答 10

56

要检查网络（即 3G 或 WiFi）是否可用，我们可以在开始活动之前使用以下方法进行验证。

ConnectivityManager manager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);

//For 3G check
boolean is3g = manager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE)
            .isConnectedOrConnecting();
//For WiFi Check
boolean isWifi = manager.getNetworkInfo(ConnectivityManager.TYPE_WIFI)
            .isConnectedOrConnecting();

System.out.println(is3g + " net " + isWifi);

if (!is3g && !isWifi) 
{ 
Toast.makeText(getApplicationContext(),"Please make sure your Network Connection is ON ",Toast.LENGTH_LONG).show();
} 
 else 
{ 
        " Your method what you want to do "
}

希望这会对某人有所帮助。

于 2012-07-05T11:19:29.437 回答
7 
final ConnectivityManager connMgr = (ConnectivityManager)
    this.getSystemService(Context.CONNECTIVITY_SERVICE);

    final android.net.NetworkInfo wifi =
    connMgr.getNetworkInfo(ConnectivityManager.TYPE_WIFI);

    final android.net.NetworkInfo mobile =
    connMgr.getNetworkInfo(ConnectivityManager.TYPE_MOBILE);

    if( wifi.isAvailable() && wifi.getDetailedState() == DetailedState.CONNECTED){
        Toast.makeText(this, "Wifi" , Toast.LENGTH_LONG).show();
    }
    else if( mobile.isAvailable() && mobile.getDetailedState() == DetailedState.CONNECTED ){
        Toast.makeText(this, "Mobile 3G " , Toast.LENGTH_LONG).show();
    }
    else
    {   
        Toast.makeText(this, "No Network " , Toast.LENGTH_LONG).show();
    }

此代码检查您是否使用 wifi 或 3g 或什么都没有，如果 wifi 打开但未连接到网络或 3g 有信号问题，它会检测此详细信息，并使用详细状态

于 2012-12-07T18:20:32.297 回答
6

您可以使用此方法检查您的互联网连接是2G、3G 还是 4G： 

public String getNetworkClass(Context context) {
    TelephonyManager mTelephonyManager = (TelephonyManager)
            context.getSystemService(Context.TELEPHONY_SERVICE);
    int networkType = mTelephonyManager.getNetworkType();
    switch (networkType) {
        case TelephonyManager.NETWORK_TYPE_GPRS:
        case TelephonyManager.NETWORK_TYPE_EDGE:
        case TelephonyManager.NETWORK_TYPE_CDMA:
        case TelephonyManager.NETWORK_TYPE_1xRTT:
        case TelephonyManager.NETWORK_TYPE_IDEN:
            return "2G";
        case TelephonyManager.NETWORK_TYPE_UMTS:
        case TelephonyManager.NETWORK_TYPE_EVDO_0:
        case TelephonyManager.NETWORK_TYPE_EVDO_A:
        case TelephonyManager.NETWORK_TYPE_HSDPA:
        case TelephonyManager.NETWORK_TYPE_HSUPA:
        case TelephonyManager.NETWORK_TYPE_HSPA:
        case TelephonyManager.NETWORK_TYPE_EVDO_B:
        case TelephonyManager.NETWORK_TYPE_EHRPD:
        case TelephonyManager.NETWORK_TYPE_HSPAP:
            return "3G";
        case TelephonyManager.NETWORK_TYPE_LTE:
            return "4G";
        default:
            return "Unknown";
    }
}

使用以下方法，您可以检查互联网是否可用，并了解您是通过移动网络还是 WiFi访问互联网：

public String getNetworkType(Context context){
    String networkType = null;
    ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetwork = connectivityManager.getActiveNetworkInfo();
    if (activeNetwork != null) { // connected to the internet
        if (activeNetwork.getType() == ConnectivityManager.TYPE_WIFI) {
                networkType = "WiFi";
        } else if (activeNetwork.getType() == ConnectivityManager.TYPE_MOBILE) {
            networkType = "Mobile";
        }
    } else {
        // not connected to the internet
    }
    return networkType;
}
于 2016-12-15T11:25:17.673 回答
3

这对我有用

NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();   
String name networkInfo.getTypeName();
于 2013-08-22T21:58:27.827 回答
3

Rahul Baradia 的回答包括manager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE)并且已弃用。

以下是针对最新 Android SDK 的改进解决方案。

ConnectivityManager connManager = (ConnectivityManager) context.getSystemService(CONNECTIVITY_SERVICE);
        boolean is3gEnabled = false;
        if(Build.VERSION.SDK_INT >= Build.VERSION_CODES.LOLLIPOP) {
            Network[] networks = connManager.getAllNetworks();
            for(Network network: networks)
            {
                NetworkInfo info = connManager.getNetworkInfo(network);
                if(info!=null) {
                    if (info.getType() == ConnectivityManager.TYPE_MOBILE) {
                        is3gEnabled = true;
                        break;
                    }
                }
            }
        }
        else
        {
            NetworkInfo mMobile = connManager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE);
            if(mMobile!=null)
                is3gEnabled = true;
        }
于 2017-04-22T21:10:41.903 回答
1

使用以下代码作为NetworkChecker.java并在您的代码中重用它：

package das.soumyadip.util;

import android.net.ConnectivityManager;
import android.util.Log;

public class NetworkChecker {
    private final static String TAG = "NwtworkChecker";

    public static boolean isNetworkConnected(
            final ConnectivityManager connectivityManager) {
        boolean val = false;

        Log.d(TAG, "Checking for Mobile Internet Network");
        final android.net.NetworkInfo mobile = connectivityManager
                .getNetworkInfo(ConnectivityManager.TYPE_MOBILE);
        if (mobile.isAvailable() && mobile.isConnected()) {
            Log.i(TAG, "Found Mobile Internet Network");
            val = true;
        } else {
            Log.e(TAG, "Mobile Internet Network not Found");
        }

        Log.d(TAG, "Checking for WI-FI Network");
        final android.net.NetworkInfo wifi = connectivityManager
                .getNetworkInfo(ConnectivityManager.TYPE_WIFI);
        if (wifi.isAvailable() && wifi.isConnected()) {
            Log.i(TAG, "Found WI-FI Network");
            val = true;
        } else {
            Log.e(TAG, "WI-FI Network not Found");
        }

        return val;
    }
}
于 2012-07-05T11:25:27.310 回答
1 
        // TODO Auto-generated method stub
        ConnectivityManager connMgr =
                (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();

        final android.net.NetworkInfo mobile = connectivityManager
                .getNetworkInfo(ConnectivityManager.TYPE_MOBILE);
        if (mobile.isAvailable() && mobile.isConnected()) {
            Log.i(TAG, "Found Mobile Internet Network");
            val = true;
        }
        // Checks the user prefs and the network connection. Based on the result, decides
        // whether
        // to refresh the display or keep the current display.
        // If the userpref is Wi-Fi only, checks to see if the device has a Wi-Fi connection.
        if (WIFI.equals(sPref) && networkInfo != null
                && networkInfo.getType() == ConnectivityManager.TYPE_WIFI) {
            // If device has its Wi-Fi connection, sets refreshDisplay
            // to true. This causes the display to be refreshed when the user
            // returns to the app.
            refreshDisplay = true;
            Toast.makeText(context, R.string.wifi_connected, Toast.LENGTH_SHORT).show();

            // If the setting is ANY network and there is a network connection
            // (which by process of elimination would be mobile), sets refreshDisplay to true.
        }

}
        else if (ANY.equals(sPref) && networkInfo != null) {
            refreshDisplay = true;

            // Otherwise, the app can't download content--either because there is no network
            // connection (mobile or Wi-Fi), or because the pref setting is WIFI, and there
            // is no Wi-Fi connection.
            // Sets refreshDisplay to false.
        } else {
            refreshDisplay = false;
            Toast.makeText(context, R.string.lost_connection, Toast.LENGTH_SHORT).show();
        }
于 2012-09-06T06:46:36.807 回答
1

我们可以使用ConnectivityManager 类获取与网络相关的任何信息。

它还会在网络连接发生变化时通知应用程序。通过调用获取此类的实例

这个类的主要职责是：

  1. 监控网络连接（Wi-Fi、GPRS、UMTS 等）
  2. 当网络连接发生变化时发送广播意图
  3. 当与网络的连接丢失时，尝试“故障转移”到另一个网络
  4. 提供 API 允许应用程序查询可用网络的粗粒度或细粒度状态
  5. 提供允许应用程序为其数据流量请求和选择网络的 API

GetNetworkInfo 函数返回有关特定网络类型的状态信息。

此方法需要调用者持有权限

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

     /**
     * Checks the type of data connection that is currently available on the device.
     *
     * @return <code>ConnectivityManager.TYPE_*</code> as a type of internet connection on the
     *This method does not support multiple connected networks
     *             of the same type.
     * device. Returns -1 in case of error or none of
     * <code>ConnectivityManager.TYPE_*</code> is found.
     **/

--

public static int getDataConnectionType(Context ctx) {

        ConnectivityManager connectivityManager = (ConnectivityManager) ctx.getSystemService(Context.CONNECTIVITY_SERVICE);

        if (connectivityManager != null && connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE) != null) {
            if (connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).isConnected()) {
                return ConnectivityManager.TYPE_MOBILE;
            } else if (connectivityManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI).isConnected()) {
                return ConnectivityManager.TYPE_WIFI;
            } else
                return -1;
        } else
            return -1;
    }
于 2017-09-04T10:00:54.043 回答
1 
public boolean isInternetAvailable(Context context) {
    ConnectivityManager cm = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
    if (activeNetwork != null) { // connected to the internet
        if (activeNetwork.getType() == ConnectivityManager.TYPE_WIFI) {
            // connected to wifi
            return true;

        } else if (activeNetwork.getType() == ConnectivityManager.TYPE_MOBILE) {
            // connected to the mobile provider's data plan
            return true;
        }
    } else {
        // not connected to the internet
        return false;
    }
    return false;
}
于 2019-12-20T06:47:29.067 回答
0

在上面的任何代码中，我都没有看到 getNetworkInfo() 是否返回 null 的检查，根据文档，当不支持请求的网络类型时会发生这种情况。这表明在没有 3g 的设备上，应用程序将因空指针异常而崩溃。

于 2014-04-06T19:42:51.030 回答