android 检测手机联网,给出震动提示

Posted on 2011-07-25 14:12  codingX  阅读(1132)  评论(0)    收藏  举报

判断手机联网,给出震动提示的代码:

      public static boolean isNetworkAvailable(Activity mActivity) {
Context context
= mActivity.getApplicationContext();
ConnectivityManager connectivity
= (ConnectivityManager) context
.getSystemService(Context.CONNECTIVITY_SERVICE);
if (connectivity == null) {
return false;
}
else {
NetworkInfo[] info
= connectivity.getAllNetworkInfo();
if (info != null) {
for (int i = 0; i < info.length; i++) {
if (info[i].getState() == NetworkInfo.State.CONNECTED) {
return true;
}
}
}
}
return false;
}

/**
* 检测网络是否存在
*/
public static void HttpTest(final Activity mActivity) {
if (!isNetworkAvailable(mActivity)) {
showToast(mActivity);
}
}
public static void showToast(Activity mActivity) {
TextView tv
= new TextView(mActivity);
Vibrator mVibrator01
= (Vibrator) mActivity.getApplication()
.getSystemService(Service.VIBRATOR_SERVICE);
mVibrator01.vibrate(
new long[] { 100, 10, 100, 1000 }, -1);
tv.setText(
"没有连接网络");
Toast toast
= new Toast(mActivity);
toast.setView(tv);
toast.setDuration(Toast.LENGTH_LONG);
toast.show();
}

其中还要加上手机震动和检测网络状态的权限:

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

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