我不希望我的用户尝试下载任何东西,除非他们连接了Wi-Fi。然而,我似乎只能判断是否启用了Wi-Fi,但他们仍然可能有3G连接。

android.net.wifi.WifiManager m = (WifiManager) getSystemService(WIFI_SERVICE);
android.net.wifi.SupplicantState s = m.getConnectionInfo().getSupplicantState();
NetworkInfo.DetailedState state = WifiInfo.getDetailedStateOf(s);
if (state != NetworkInfo.DetailedState.CONNECTED) {
    return false;
}

然而,这种状态并不是我所期望的。即使Wi-Fi是连接的,我得到OBTAINING_IPADDR作为状态。


当前回答

这适用于Q之前和之后的Android设备

fun isWifiConnected(context: Context):Boolean {
    val cm = context.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager
    return if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
             val capabilities = cm.getNetworkCapabilities(cm.activeNetwork)
             capabilities?.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)==true
        } else {
            val activeNetwork: NetworkInfo? = cm.activeNetworkInfo
            activeNetwork?.typeName?.contains("wifi",ignoreCase = true)?:false
        }
}

其他回答

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
    }
}

使用WifiManager你可以做到:

WifiManager wifi = (WifiManager) getSystemService (Context.WIFI_SERVICE);
if (wifi.getConnectionInfo().getNetworkId() != -1) {/* connected */}

方法getNeworkId只有在没有连接到网络时才返回-1;

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;
}

我在我的应用程序中使用这个来检查活动网络是否为Wi-Fi:

ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo ni = cm.getActiveNetworkInfo();
if (ni != null && ni.getType() == ConnectivityManager.TYPE_WIFI)
{

    // Do your work here

}

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。