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;
其他回答
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。
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
}
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
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。
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;
}
