What would the code be for check开发者_如何转开发ing whether the Wi-Fi is enabled or not?
WifiManager wifiManager = (WifiManager) getSystemService(Context.WIFI_SERVICE);
if (wifiManager.isWifiEnabled()) {
// wifi is enabled
}
For details check here
The above answers work fine. But don't forget to add the right permissions in the Manifest:
<uses-permission android:name="android.permission.ACCESS_WIFI_STATE"/>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE"/>
<uses-permission android:name="android.permission.CHANGE_WIFI_STATE"/>
Hope it helps ..
The top answer is correct, but not up to date because this code may leak memory on certain devices.
Therefore the better answer would be:
WifiManager wifiManager = (WifiManager) getApplicationContext().getSystemService(Context.WIFI_SERVICE);
if (wifiManager.isWifiEnabled()) {
// wifi is enabled
}
Permissions in app=>mainfests=>AndroidManifest.xml:
<uses-permission android:name="android.permission.ACCESS_WIFI_STATE" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permission.CHANGE_WIFI_STATE" />
Reference: https://www.mysysadmintips.com/other/programming/759-the-wifi-service-must-be-looked-up-on-the-application-context
public static boolean wifiState() {
WifiManager wifiManager = (WifiManager) getSystemService(Context.WIFI_SERVICE);
return wifiManager.isWifiEnabled();
}
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