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

这对我来说很管用:

    ConnectivityManager conMan = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);

    // Mobile
    State mobile = conMan.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).getState();

    // Wi-Fi
    State wifi = conMan.getNetworkInfo(ConnectivityManager.TYPE_WIFI).getState();

    // And then use it like this:

    if (mobile == NetworkInfo.State.CONNECTED || mobile == NetworkInfo.State.CONNECTING)
    {
        Toast.makeText(Wifi_Gprs.this,"Mobile is Enabled :) ....",Toast.LENGTH_LONG).show();
    }
    else if (wifi == NetworkInfo.State.CONNECTED || wifi == NetworkInfo.State.CONNECTING)
    {
        Toast.makeText(Wifi_Gprs.this,"Wifi is Enabled  :) ....",Toast.LENGTH_LONG).show();
    }
    else
    {
        Toast.makeText(Wifi_Gprs.this,"No Wifi or Gprs Enabled :( ....",Toast.LENGTH_LONG).show();
    }

并添加以下权限:

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

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。

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" />

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
    }
}

虽然Jason的回答是正确的，但现在getNetWorkInfo (int)是一个不推荐使用的方法。所以，下一个函数将是一个很好的选择:

public static boolean isWifiAvailable (Context context)
{
    boolean br = false;
    ConnectivityManager cm = null;
    NetworkInfo ni = null;

    cm = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    ni = cm.getActiveNetworkInfo();
    br = ((null != ni) && (ni.isConnected()) && (ni.getType() == ConnectivityManager.TYPE_WIFI));

    return br;
}