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
类似于@Jason Knight的回答，但以Kotlin的方式:
val connManager = getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager
val mWifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI)
if (mWifi.isConnected) {
// Do whatever
}
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
}
}
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" />
val wifi = context!!.applicationContext.getSystemService(Context.WIFI_SERVICE) as WifiManager?
if (wifi!!.isWifiEnabled)
//do action here
else
//do action here
许多答案使用了废弃的代码，或者在更高API版本上可用的代码。现在我用这样的东西
ConnectivityManager connectivityManager = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
if(connectivityManager != null) {
for (Network net : connectivityManager.getAllNetworks()) {
NetworkCapabilities nc = connectivityManager.getNetworkCapabilities(net);
if (nc != null && nc.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)
&& nc.hasCapability(NetworkCapabilities.NET_CAPABILITY_INTERNET))
return true;
}
}
return false;
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();
}
