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
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
其他回答
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
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
}
这是一个更简单的解决方案。参见Stack Overflow 在Android上检查是否启用Wi-Fi。
注:不要忘记将代码添加到manifest.xml文件中以允许权限。如下图所示。
<uses-permission android:name="android.permission.ACCESS_WIFI_STATE" >
</uses-permission>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" >
</uses-permission>
<uses-permission android:name="android.permission.CHANGE_WIFI_STATE" >
</uses-permission>
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
