这个问题有点老了，但我用的就是这个。要求最低api级别21也考虑到废弃的Networkinfo api。

boolean isWifiConn = false;
    ConnectivityManager connMgr = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        Network network = connMgr.getActiveNetwork();
        if (network == null) return false;
        NetworkCapabilities capabilities = connMgr.getNetworkCapabilities(network);
        if(capabilities != null && capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)){
            isWifiConn = true;
            Toast.makeText(context,"Wifi connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }else{
            Toast.makeText(context,"Wifi not connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }
    } else {
        for (Network network : connMgr.getAllNetworks()) {
            NetworkInfo networkInfo = connMgr.getNetworkInfo(network);
            if (networkInfo.getType() == ConnectivityManager.TYPE_WIFI && networkInfo.isConnected()) {
                isWifiConn = true;
                Toast.makeText(context,"Wifi connected ",Toast.LENGTH_LONG).show();
                break;
            }else{
                Toast.makeText(context,"Wifi not connected ",Toast.LENGTH_LONG).show();
            }
        }
    }
    return isWifiConn;

这适用于最新版本的android:

fun getConnectionType(context: Context): ConnectivityType {
    var result = NONE
    val cm = context.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager?
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        if (cm != null) {
            val capabilities = cm.getNetworkCapabilities(cm.activeNetwork)
            if (capabilities != null) {
                when {
                    capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI) -> {
                        result = WIFI
                    }
                    capabilities.hasTransport(NetworkCapabilities.TRANSPORT_CELLULAR) -> {
                        result = MOBILE_DATA
                    }
                    capabilities.hasTransport(NetworkCapabilities.TRANSPORT_VPN) -> {
                        result = VPN
                    }
                }
            }
        }
    } else {
        if (cm != null) {
            val activeNetwork = cm.activeNetworkInfo
            if (activeNetwork != null) {
                // connected to the internet
                when (activeNetwork.type) {
                    ConnectivityManager.TYPE_WIFI -> {
                        result = WIFI
                    }
                    ConnectivityManager.TYPE_MOBILE -> {
                        result = MOBILE_DATA
                    }
                    ConnectivityManager.TYPE_VPN -> {
                        result = VPN
                    }
                }
            }
        }
    }
    return result
}

enum class ConnectivityType {
    NONE,
    MOBILE_DATA,
    WIFI,
    VPN,
}

在清单中:

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

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)
    }
}

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。

我看了一些像这样的问题，然后想到了这个:

ConnectivityManager connManager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
NetworkInfo wifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
NetworkInfo mobile = connManager .getNetworkInfo(ConnectivityManager.TYPE_MOBILE);

if (wifi.isConnected()){
    // If Wi-Fi connected
}

if (mobile.isConnected()) {
    // If Internet connected
}

我在Root Toolbox PRO中使用if进行许可证检查，它似乎工作得很好。

由于方法NetworkInfo.isConnected()现在在API-23中已弃用，下面是一个方法，它可以检测Wi-Fi适配器是否打开，并使用WifiManager连接到接入点:

private boolean checkWifiOnAndConnected() {
    WifiManager wifiMgr = (WifiManager) getSystemService(Context.WIFI_SERVICE);

    if (wifiMgr.isWifiEnabled()) { // Wi-Fi adapter is ON

        WifiInfo wifiInfo = wifiMgr.getConnectionInfo();

        if( wifiInfo.getNetworkId() == -1 ){
            return false; // Not connected to an access point
        }
        return true; // Connected to an access point
    }
    else {
        return false; // Wi-Fi adapter is OFF
    }
}