Reputation: 3875
How can I open my application when an user enters a zone that has wi-fi? Is this possible? Suppose my application is onPause() state (means My Device's homescreen). now when device connected with wifi. it will automatically open my application.
Upvotes: 5
Views: 6380
Reputation: 873
Maybe doable with the Tasker app from Play Store (not free though). Or you can create a Service (http://developer.android.com/guide/components/services.html) that will have code outlined in the other answers and then launch your app (Activity) when wifi is available.
Upvotes: 0
Reputation: 6778
What I can imagine is an overwriting of the onPause()
Method like this:
@Override
public void onPause() {
String conn_context = Context.WIFI_SERVICE;
final WifiManager wifi = (WifiManager) getSystemService(conn_context);
if (wifi.isWifiEnabled())
{
super.onResume();
}
else
{
super.onPause();
}
}
But you must also figure a way to handle the real onPause event.
Upvotes: 0
Reputation: 439
Try add broadcast receiver and listen network changes, when wi-fi connected start your activity. Something like this solution
public class ConnectivityReceiver extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
ConnectivityManager connMgr = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo wifi = connMgr.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
NetworkInfo mobile = connMgr.getNetworkInfo(ConnectivityManager.TYPE_MOBILE);
if (((null != wifi)&&(wifi.isAvailable())) || ((null != mobile)&&(mobile.isAvailable()))){
Intent uplIntent = new Intent(context, YourActivity.class);
uplIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
context.startActivity(uplIntent);
}
}
}
And add to manifest
<receiver android:name=".receiver.ConnectivityReceiver">
<intent-filter>
<action android:name="android.net.conn.CONNECTIVITY_CHANGE" />
</intent-filter>
</receiver>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE"/>
Upvotes: 5