我正在尝试创建一个登录注册应用程序。显然我的应用程序需要互联网连接才能启动。我跟随其他用户的答案有同样的问题,但即使没有互联网连接我的应用程序仍然启动。提前谢谢。
这是我的清单
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.claude.pickupcartbeta">
<uses-sdk
android:maxSdkVersion="24"
android:minSdkVersion="16" />
<supports-screens
android:anyDensity="true"
android:largeScreens="true"
android:normalScreens="true"
android:resizeable="true"
android:smallScreens="true"
android:xlargeScreens="true" />
<uses-permission android:name="android.permission.INTERNET" ></uses-permission>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE"></uses-permission>
<application
android:allowBackup="true"
android:icon="@drawable/logo"
android:label="@string/app_name"
android:supportsRtl="true"
android:theme="@style/AppTheme">
<activity
android:name=".LogInActivity"
android:screenOrientation="portrait">
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>
<activity
android:name=".MainScreen"
android:screenOrientation="landscape" />
<activity android:name=".Register"
android:screenOrientation="portrait"></activity>
</application>
</manifest>
活动在这里:
public class LogInActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_NO_TITLE);
getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
setContentView(R.layout.activity_log_in);
}
public void MScreen(View view) {
Intent intent = new Intent(this, MainScreen.class);
startActivity(intent);
}
}
答案 0 :(得分:0)
在您的活动的onCreate功能中添加以下内容;
ConnectivityManager cm = (Connectivity Manager) context.getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo activeNetwork = cm.getActiveNetworkInfo();
boolean isConnected = activeNetwork != null && activeNetwork.isConnectedOrConnecting();
if (!isConnected)
{
// Toast... (Show some notification.
finish();
}