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
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
}
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
}
}
使用WifiManager你可以做到:
WifiManager wifi = (WifiManager) getSystemService (Context.WIFI_SERVICE);
if (wifi.getConnectionInfo().getNetworkId() != -1) {/* connected */}
方法getNeworkId只有在没有连接到网络时才返回-1;
这适用于最新版本的android:
fun getConnectionType(context: Context): ConnectivityType {
var result = NONE
val cm = context.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager?
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
if (cm != null) {
val capabilities = cm.getNetworkCapabilities(cm.activeNetwork)
if (capabilities != null) {
when {
capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI) -> {
result = WIFI
}
capabilities.hasTransport(NetworkCapabilities.TRANSPORT_CELLULAR) -> {
result = MOBILE_DATA
}
capabilities.hasTransport(NetworkCapabilities.TRANSPORT_VPN) -> {
result = VPN
}
}
}
}
} else {
if (cm != null) {
val activeNetwork = cm.activeNetworkInfo
if (activeNetwork != null) {
// connected to the internet
when (activeNetwork.type) {
ConnectivityManager.TYPE_WIFI -> {
result = WIFI
}
ConnectivityManager.TYPE_MOBILE -> {
result = MOBILE_DATA
}
ConnectivityManager.TYPE_VPN -> {
result = VPN
}
}
}
}
}
return result
}
enum class ConnectivityType {
NONE,
MOBILE_DATA,
WIFI,
VPN,
}
在清单中:
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
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>
val wifi = context!!.applicationContext.getSystemService(Context.WIFI_SERVICE) as WifiManager?
if (wifi!!.isWifiEnabled)
//do action here
else
//do action here
