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Android Firebase-当用户单击通知时,活动重新启动_Android_Firebase - Fatal编程技术网

Android Firebase-当用户单击通知时,活动重新启动

Android Firebase-当用户单击通知时,活动重新启动,android,firebase,Android,Firebase,我正在编写一个使用firebase实现通知的应用程序。在我的MainActivity中,我有一个带有url的WebView,但问题是当用户单击通知时,我想在WebView中使用不同的url打开MainActivity。我读了很多书,并在intent中添加了一个bundle(单击通知时打开MainActivity),该bundle包含所需的url。但当我点击通知时,MainActivity会重新启动,我的意思是,它不会转到onNewIntent,而是运行onCreate。我就是这样实现的: pri

我正在编写一个使用firebase实现通知的应用程序。在我的MainActivity中,我有一个带有url的WebView,但问题是当用户单击通知时,我想在WebView中使用不同的url打开MainActivity。我读了很多书,并在intent中添加了一个bundle(单击通知时打开MainActivity),该bundle包含所需的url。但当我点击通知时,MainActivity会重新启动,我的意思是,它不会转到onNewIntent,而是运行onCreate。我就是这样实现的:

private void sendNotification(String messageTitle, String messageBody, String url){

    Intent intent = new Intent(this, MainActivity.class);
    intent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);

    //This adds the url to the intent
    if(!url.equals("")){
        Bundle bundle = new Bundle();
        bundle.putString("url", url);
        intent.putExtras(bundle);
    }

    PendingIntent pendingIntent = PendingIntent.getActivity(this, 0,intent, PendingIntent.FLAG_ONE_SHOT);

    String channelId = getString(R.string.default_notification_channel_id);

    NotificationCompat.Builder notificationBuilder = new NotificationCompat
            .Builder(this, channelId)
            .setContentTitle(messageTitle)
            .setContentText(messageBody)
            .setPriority(NotificationCompat.PRIORITY_DEFAULT)
            .setAutoCancel(true)
            .setSmallIcon(R.drawable.ic_launcher_background)
            .setContentIntent(pendingIntent);

    NotificationManager notificationManager = (NotificationManager) getSystemService(Context.NOTIFICATION_SERVICE);

    if(Build.VERSION.SDK_INT >= Build.VERSION_CODES.O){
        NotificationChannel channel = new NotificationChannel(channelId,
                "Notification channel",
                NotificationManager.IMPORTANCE_DEFAULT);
        notificationManager.createNotificationChannel(channel);
    }

    notificationManager.notify(0, notificationBuilder.build());

}
我有这个:

Bundle bundle = intent.getExtras();
        if (bundle != null) {
            String url = bundle.getString("url");
            mWebView.loadUrl(url);
        }
但当单击通知时,活动将重新启动,因此它不会运行OnEvent,并且日志会显示以下错误:

02-08 12:51:12.140 19056-19056/com.example.android.app E/ActivityThread: Activity com.example.android.app.MainActivity has leaked IntentReceiver com.example.android.app.MainActivity$1@d7818db that was originally registered here. Are you missing a call to unregisterReceiver()?
android.app.IntentReceiverLeaked: Activity com.example.android.app.MainActivity has leaked IntentReceiver com.example.android.app.MainActivity$1@d7818db that was originally registered here. Are you missing a call to unregisterReceiver()?
    at android.app.LoadedApk$ReceiverDispatcher.<init>(LoadedApk.java:999)
    at android.app.LoadedApk.getReceiverDispatcher(LoadedApk.java:795)
    at android.app.ContextImpl.registerReceiverInternal(ContextImpl.java:1329)
    at android.app.ContextImpl.registerReceiver(ContextImpl.java:1309)
    at android.app.ContextImpl.registerReceiver(ContextImpl.java:1303)
    at android.content.ContextWrapper.registerReceiver(ContextWrapper.java:554)
    at com.example.android.app.MainActivity.onCreate(MainActivity.java:264)
    at android.app.Activity.performCreate(Activity.java:6367)
    at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1110)
    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2404)
    at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2511)
    at android.app.ActivityThread.access$900(ActivityThread.java:165)
    at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1375)
    at android.os.Handler.dispatchMessage(Handler.java:102)
    at android.os.Looper.loop(Looper.java:150)
    at android.app.ActivityThread.main(ActivityThread.java:5621)
    at java.lang.reflect.Method.invoke(Native Method)
    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:794)
    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:684)
但是在这种情况下,当用户单击通知时,它什么也不做

有人知道如何修复它,以便在用户单击时加载url吗? 提前向您表示感谢

来自it声明:

如果它已将其启动模式声明为“多个”(默认),并且您没有按照相同的意图设置
FLAG\u ACTIVITY\u SINGLE\u TOP
,则它将完成并重新创建;对于所有其他启动模式,或者如果设置了
FLAG\u ACTIVITY\u SINGLE\u TOP
,则此意图将传递到当前实例的onNewIntent()


因此,在创建新意图时设置
标志\u ACTIVITY\u SINGLE\u TOP
标志应该解决并运行onNewIntent()方法,而不是重新创建应用程序。

创建这样的方法

private PendingIntent retrievePlaybackAction(final String action) {
    Intent intent = new Intent(action);
    return PendingIntent.getBroadcast(this, 0, intent, PendingIntent.FLAG_UPDATE_CURRENT);
}
现在像这样添加您的意图

builder.setContentIntent(retrievePlaybackAction("OPEN_MAIN")); //change action text as you want
创建一个简单的类
NotificationReceiver
,我想您已经创建了这个类,现在创建这个类的对象,从中发送通知

private NotificationReceiver notificationReceiver;
onCreate()
中注册您的接收者

notificationReceiver = new NotificationReceiver();

IntentFilter intentFilterNextClick = new IntentFilter("OPEN_MAIN");

registerReceiver(notificationReceiver, intentFilterNextClick); 
//can create exception, better to surround with try catch
onDestroy()
中注销您的接收器

unregisterReceiver(notificationReceiver); 
//can create exception, better to surround with try catch
@Override
public void onReceive(Context context, Intent intent) {
    Log.e(TAG, "onReceive: received " + intent.getAction());

    String action = intent.getAction();

    //no need to create switch you can also use if
    switch (action) {
        case "OPEN_MAIN":
            openMain();
            break;
    }

}

//here is openMain();
private void openMain(Context context) {
    Intent openMainIntent = new Intent(context, MainActivity.class);
    openMainIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK); //this flag is important
    context.startActivity(openMainIntent);
}
要打开或执行某些操作,请将此添加到接收器中

unregisterReceiver(notificationReceiver); 
//can create exception, better to surround with try catch
@Override
public void onReceive(Context context, Intent intent) {
    Log.e(TAG, "onReceive: received " + intent.getAction());

    String action = intent.getAction();

    //no need to create switch you can also use if
    switch (action) {
        case "OPEN_MAIN":
            openMain();
            break;
    }

}

//here is openMain();
private void openMain(Context context) {
    Intent openMainIntent = new Intent(context, MainActivity.class);
    openMainIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK); //this flag is important
    context.startActivity(openMainIntent);
}
如果应用程序最小化或关闭,这也会起作用

希望这会有帮助


如果您需要更多帮助,请询问

很高兴它只是这么小的东西:)祝你在应用程序中的冒险事业好运!