我需要在运行时检查手机中的wifi是打开还是关闭?
如果没有连接,我想显示对话框并直接转到设置/无线控件以便用户启用它。
用于手机的wifi和 Gps staus 。怎么做?哪个意图为此醒来?有什么想法吗?
答案 0 :(得分:11)
要检查设备是通过手机还是wifi连接,您可以使用以下代码:
ConnectivityManager conMan = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
//mobile
State mobile = conMan.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).getState();
//wifi
State wifi = conMan.getNetworkInfo(ConnectivityManager.TYPE_WIFI).getState();
然后像这样使用它:
if (mobile == NetworkInfo.State.CONNECTED || mobile == NetworkInfo.State.CONNECTING) {
//mobile
} else if (wifi == NetworkInfo.State.CONNECTED || wifi == NetworkInfo.State.CONNECTING) {
//wifi
}
答案 1 :(得分:6)
您可以使用WifiManager课程来获取Wi-Fi的状态。
答案 2 :(得分:0)
private boolean isNetworkAvailable() {
ConnectivityManager connManager = (ConnectivityManager)
getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo activeNetworkInfo = connManager.getActiveNetworkInfo();
return activeNetworkInfo.isConnected();
}
public void onClick(DialogInterface dialog, int id) {
// ...
if (isNetworkAvailable()) {
t3.setText("The Internet is available");
} else {
t3.setText("internet is not available");
}
}
答案 3 :(得分:0)
ConnectivityManager conMan = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
//mobile
State mobile = conMan.getNetworkInfo(ConnectivityManager.TYPE_MOBILE).getState();
//wifi
State wifi = conMan.getNetworkInfo(ConnectivityManager.TYPE_WIFI).getState();