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

}

The following code (in Kotlin) works from API 21 until at least current API version (API 29). The function getWifiState() returns one of 3 possible values for the WiFi network state: Disable, EnabledNotConnected and Connected that were defined in an enum class. This allows to take more granular decisions like informing the user to enable WiFi or, if already enabled, to connect to one of the available networks. But if all that is needed is a boolean indicating if the WiFi interface is connected to a network, then the other function isWifiConnected() will give you that. It uses the previous one and compares the result to Connected.

它受到了之前一些答案的启发，但试图解决Android API的发展或IP V6的可用性缓慢增加所带来的问题。 诀窍是使用:

wifiManager.connectionInfo.bssid != null 

而不是:

getIpAddress() == 0，仅对IP V4或有效 getNetworkId() == -1现在需要另一个特殊权限(位置)

参考文档:https://developer.android.com/reference/kotlin/android/net/wifi/WifiInfo.html#getbssid 如果没有连接到网络，它将返回null。即使我们没有获得真实值的权限，如果我们连接了，它仍然会返回null以外的值。

同时也要记住以下几点:

在android.os.Build之前发布。VERSION_CODES#N，该对象 应该只能从Context中获得#getApplicationContext()，和 类型中的内存泄漏，而不是来自任何其他派生上下文 调用过程。

在舱单中，不要忘记添加:

<uses-permission android:name="android.permission.ACCESS_WIFI_STATE"/>

建议代码为:

class MyViewModel(application: Application) : AndroidViewModel(application) {

   // Get application context
    private val myAppContext: Context = getApplication<Application>().applicationContext

   // Define the different possible states for the WiFi Connection
    internal enum class WifiState {
        Disabled,               // WiFi is not enabled
        EnabledNotConnected,    // WiFi is enabled but we are not connected to any WiFi network
        Connected,              // Connected to a WiFi network
    }

    // Get the current state of the WiFi network
    private fun getWifiState() : WifiState {

        val wifiManager : WifiManager = myAppContext.applicationContext.getSystemService(Context.WIFI_SERVICE) as WifiManager

        return if (wifiManager.isWifiEnabled) {
                    if (wifiManager.connectionInfo.bssid != null)
                        WifiState.Connected
                    else
                        WifiState.EnabledNotConnected
               } else {
                    WifiState.Disabled
               }
    }

    // Returns true if we are connected to a WiFi network
    private fun isWiFiConnected() : Boolean {
        return (getWifiState() == WifiState.Connected)
    }
}

我看了一些像这样的问题，然后想到了这个:

ConnectivityManager connManager = (ConnectivityManager) getSystemService(CONNECTIVITY_SERVICE);
NetworkInfo wifi = connManager.getNetworkInfo(ConnectivityManager.TYPE_WIFI);
NetworkInfo mobile = connManager .getNetworkInfo(ConnectivityManager.TYPE_MOBILE);

if (wifi.isConnected()){
    // If Wi-Fi connected
}

if (mobile.isConnected()) {
    // If Internet connected
}

我在Root Toolbox PRO中使用if进行许可证检查，它似乎工作得很好。

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>

从API级别29开始，NetworkInfo类以及相关的访问方法(如connectivitymanager# getNetworkInfo()和connectivitymanager# getActiveNetworkInfo())都已弃用。

文档现在建议人们使用ConnectivityManager。NetworkCallback API用于异步回调监控，或使用connectivitymanager# getNetworkCapabilities或connectivitymanager# getLinkProperties来同步访问网络信息

调用者应该使用ConnectivityManager。NetworkCallback API来了解连接变化，或切换到使用connectivitymanager# getNetworkCapabilities或connectivitymanager# getLinkProperties来同步获取信息。

要检查是否连接了WiFi，下面是我使用的代码:

科特林：

val connMgr = applicationContext.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager?
connMgr?: return false
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
    val network: Network = connMgr.activeNetwork ?: return false
    val capabilities = connMgr.getNetworkCapabilities(network)
    return capabilities != null && capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)
} else {
    val networkInfo = connMgr.activeNetworkInfo ?: return false
    return networkInfo.isConnected && networkInfo.type == ConnectivityManager.TYPE_WIFI
}

Java:

ConnectivityManager connMgr = (ConnectivityManager) getApplicationContext().getSystemService(Context.CONNECTIVITY_SERVICE);
if (connMgr == null) {
    return false;
}
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
    Network network = connMgr.getActiveNetwork();
    if (network == null) return false;
    NetworkCapabilities capabilities = connMgr.getNetworkCapabilities(network);
    return capabilities != null && capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI);
} else {
    NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();
    return networkInfo.isConnected() && networkInfo.getType() == ConnectivityManager.TYPE_WIFI;
}

记住还要向Manifest文件添加ACCESS_NETWORK_STATE权限。