ConnectivityManager manager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
boolean is3g = manager.getNetworkInfo(
                  ConnectivityManager.TYPE_MOBILE).isConnectedOrConnecting();
boolean isWifi = manager.getNetworkInfo(
                    ConnectivityManager.TYPE_WIFI).isConnectedOrConnecting();

Log.v("", is3g + " ConnectivityManager Test " + isWifi);
if (!is3g && !isWifi) {
    Toast.makeText(
        getApplicationContext(),
        "Please make sure, your network connection is ON ",
        Toast.LENGTH_LONG).show();
}
else {
    // Put your function() to go further;
}

试试这个方法。

public boolean isInternetConnected() {
    ConnectivityManager conMgr = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    boolean ret = true;
    if (conMgr != null) {
        NetworkInfo i = conMgr.getActiveNetworkInfo();

        if (i != null) {
            if (!i.isConnected()) {
                ret = false;
            }

            if (!i.isAvailable()) {
                ret = false;
            }
        }

        if (i == null)
            ret = false;
    } else
        ret = false;
    return ret;
}

这种方法将有助于找到互联网连接是否可用。

下面是我在我的应用程序中使用的实用方法:

public static boolean isDeviceOnWifi(final Context context) {
        ConnectivityManager connManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
        NetworkInfo mWifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
        return mWifi != null && mWifi.isConnectedOrConnecting();
}

在新版Android中

private void getWifiInfo(Context context) {
    ConnectivityManager connManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    Network[] networks = connManager.getAllNetworks();

    if(networks == null || networks.length == 0)
        return;

    for( int i = 0; i < networks.length; i++) {
        Network ntk = networks[i];
        NetworkInfo ntkInfo = connManager.getNetworkInfo(ntk);
        if (ntkInfo.getType() == ConnectivityManager.TYPE_WIFI && ntkInfo.isConnected() ) {
            final WifiManager wifiManager = (WifiManager) context.getSystemService(Context.WIFI_SERVICE);
            final WifiInfo connectionInfo = wifiManager.getConnectionInfo();
            if (connectionInfo != null) {
                // add some code here
            }
        }

    }
}

还要加上前提

The following code (in Kotlin) works from API 21 until at least current API version (API 29). The function getWifiState() returns one of 3 possible values for the WiFi network state: Disable, EnabledNotConnected and Connected that were defined in an enum class. This allows to take more granular decisions like informing the user to enable WiFi or, if already enabled, to connect to one of the available networks. But if all that is needed is a boolean indicating if the WiFi interface is connected to a network, then the other function isWifiConnected() will give you that. It uses the previous one and compares the result to Connected.

它受到了之前一些答案的启发，但试图解决Android API的发展或IP V6的可用性缓慢增加所带来的问题。 诀窍是使用:

wifiManager.connectionInfo.bssid != null 

而不是:

getIpAddress() == 0，仅对IP V4或有效 getNetworkId() == -1现在需要另一个特殊权限(位置)

参考文档:https://developer.android.com/reference/kotlin/android/net/wifi/WifiInfo.html#getbssid 如果没有连接到网络，它将返回null。即使我们没有获得真实值的权限，如果我们连接了，它仍然会返回null以外的值。

同时也要记住以下几点:

在android.os.Build之前发布。VERSION_CODES#N，该对象 应该只能从Context中获得#getApplicationContext()，和 类型中的内存泄漏，而不是来自任何其他派生上下文 调用过程。

在舱单中，不要忘记添加:

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

建议代码为:

class MyViewModel(application: Application) : AndroidViewModel(application) {

   // Get application context
    private val myAppContext: Context = getApplication<Application>().applicationContext

   // Define the different possible states for the WiFi Connection
    internal enum class WifiState {
        Disabled,               // WiFi is not enabled
        EnabledNotConnected,    // WiFi is enabled but we are not connected to any WiFi network
        Connected,              // Connected to a WiFi network
    }

    // Get the current state of the WiFi network
    private fun getWifiState() : WifiState {

        val wifiManager : WifiManager = myAppContext.applicationContext.getSystemService(Context.WIFI_SERVICE) as WifiManager

        return if (wifiManager.isWifiEnabled) {
                    if (wifiManager.connectionInfo.bssid != null)
                        WifiState.Connected
                    else
                        WifiState.EnabledNotConnected
               } else {
                    WifiState.Disabled
               }
    }

    // Returns true if we are connected to a WiFi network
    private fun isWiFiConnected() : Boolean {
        return (getWifiState() == WifiState.Connected)
    }
}

Try

wifiManager.getConnectionInfo().getIpAddress()

这将返回0，直到设备有一个可用的连接(在我的机器上，三星SM-T280, Android 5.1.1)。