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

这个问题有点老了,但我用的就是这个。要求最低api级别21也考虑到废弃的Networkinfo api。

boolean isWifiConn = false;
    ConnectivityManager connMgr = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
        Network network = connMgr.getActiveNetwork();
        if (network == null) return false;
        NetworkCapabilities capabilities = connMgr.getNetworkCapabilities(network);
        if(capabilities != null && capabilities.hasTransport(NetworkCapabilities.TRANSPORT_WIFI)){
            isWifiConn = true;
            Toast.makeText(context,"Wifi connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }else{
            Toast.makeText(context,"Wifi not connected Api >= "+Build.VERSION_CODES.M,Toast.LENGTH_LONG).show();
        }
    } else {
        for (Network network : connMgr.getAllNetworks()) {
            NetworkInfo networkInfo = connMgr.getNetworkInfo(network);
            if (networkInfo.getType() == ConnectivityManager.TYPE_WIFI && networkInfo.isConnected()) {
                isWifiConn = true;
                Toast.makeText(context,"Wifi connected ",Toast.LENGTH_LONG).show();
                break;
            }else{
                Toast.makeText(context,"Wifi not connected ",Toast.LENGTH_LONG).show();
            }
        }
    }
    return isWifiConn;

其他回答

虽然Jason的回答是正确的,但现在getNetWorkInfo (int)是一个不推荐使用的方法。所以,下一个函数将是一个很好的选择:

public static boolean isWifiAvailable (Context context)
{
    boolean br = false;
    ConnectivityManager cm = null;
    NetworkInfo ni = null;

    cm = (ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
    ni = cm.getActiveNetworkInfo();
    br = ((null != ni) && (ni.isConnected()) && (ni.getType() == ConnectivityManager.TYPE_WIFI));

    return br;
}

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();
}

}

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;
}

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
}

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