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
你可以打开WIFI，如果它没有被激活如下 1. 查看@Jason Knight回复的WIFI状态 2. 如果没有激活，请激活它 不要忘记在manifest文件中添加WIFI权限
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
您的Java类应该是这样的
public class TestApp extends Activity {
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
//check WIFI activation
ConnectivityManager connManager = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo mWifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
if (mWifi.isConnected() == false) {
showWIFIDisabledAlertToUser();
}
else {
Toast.makeText(this, "WIFI is Enabled in your devide", Toast.LENGTH_SHORT).show();
}
}
private void showWIFIDisabledAlertToUser(){
AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(this);
alertDialogBuilder.setMessage("WIFI is disabled in your device. Would you like to enable it?")
.setCancelable(false)
.setPositiveButton("Goto Settings Page To Enable WIFI",
new DialogInterface.OnClickListener(){
public void onClick(DialogInterface dialog, int id){
Intent callGPSSettingIntent = new Intent(
Settings.ACTION_WIFI_SETTINGS);
startActivity(callGPSSettingIntent);
}
});
alertDialogBuilder.setNegativeButton("Cancel",
new DialogInterface.OnClickListener(){
public void onClick(DialogInterface dialog, int id){
dialog.cancel();
}
});
AlertDialog alert = alertDialogBuilder.create();
alert.show();
}
}
其他回答
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
}
}
val wifi = context!!.applicationContext.getSystemService(Context.WIFI_SERVICE) as WifiManager?
if (wifi!!.isWifiEnabled)
//do action here
else
//do action 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
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
}
ConnectivityManager manager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
boolean is3g = manager.getNetworkInfo(
ConnectivityManager.TYPE_MOBILE).isConnectedOrConnecting();
boolean isWifi = manager.getNetworkInfo(
ConnectivityManager.TYPE_WIFI).isConnectedOrConnecting();
Log.v("", is3g + " ConnectivityManager Test " + isWifi);
if (!is3g && !isWifi) {
Toast.makeText(
getApplicationContext(),
"Please make sure, your network connection is ON ",
Toast.LENGTH_LONG).show();
}
else {
// Put your function() to go further;
}
