Невозможно зарегистрировать токен в базе данных MySQL, используя облачную передачу сообщений Firebase

Я немного работал с push-уведомлениями, которые мне удавалось до сих пор, и с Apache Firebase Cloud Messaging, теперь я пытался сохранить токен в базе данных, следовал учебному пособию, и токен, похоже, не регистрируется. ,

MainActivity.java
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MyFirebaseInstanceIDService.java
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MyFirebaseMessagingService расширяет службу сообщений Firebase {

  private static final String TAG = "MyFirebaseMsgService";

@Override
public void onMessageReceived(RemoteMessage remoteMessage) {
//Displaying data in log
//It is optional
Log.d(TAG, "From: " + remoteMessage.getFrom());
Log.d(TAG, "Notification Message Body: " + remoteMessage.getNotification().getBody());

//Calling method to generate notification
sendNotification(remoteMessage.getNotification().getBody());
}

//This method is only generating push notification
//It is same as we did in earlier posts
private void sendNotification(String messageBody) {
Intent intent = new Intent(this, MainActivity.class);
intent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
PendingIntent pendingIntent = PendingIntent.getActivity(this, 0, intent,
PendingIntent.FLAG_ONE_SHOT);

Uri defaultSoundUri= RingtoneManager.getDefaultUri(RingtoneManager.TYPE_NOTIFICATION);
NotificationCompat.Builder notificationBuilder = new NotificationCompat.Builder(this)
.setSmallIcon(R.mipmap.ic_launcher)
.setContentTitle("Firebase Push Notification")
.setContentText(messageBody)
.setAutoCancel(true)
.setSound(defaultSoundUri)
.setContentIntent(pendingIntent);

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

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

}

AndroidManifest.xml

<?xml version="1.0" encoding="utf-8"?>

<!-- Adding Internet Permission -->

<uses-permission android:name="android.permission.INTERNET"/>
<application
android:allowBackup="true"android:icon="@mipmap/ic_launcher"android:label="@string/app_name"android:supportsRtl="true"android:theme="@style/AppTheme">
<activity android:name=".MainActivity">
<intent-filter>
<action android:name="android.intent.action.MAIN" />

<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>

<!--
Defining Services
-->
<service
android:name=".MyFirebaseMessagingService">
<intent-filter>
<action android:name="com.google.firebase.MESSAGING_EVENT"/>
</intent-filter>
</service>

<service
android:name=".MyFirebaseInstanceIDService">
<intent-filter>
<action android:name="com.google.firebase.INSTANCE_ID_EVENT"/>
</intent-filter>
</service>
</application>

1

Решение

Попробуйте это с библиотекой залпа & asyntask:

//Creating a string request
StringRequest req = new StringRequest(Request.Method.POST, Constants.REGISTER_URL,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
//dismissing the progress dialog
progressDialog.dismiss();

//if the server returned the string success
if (response.trim().equalsIgnoreCase("success")) {
//Displaying a success toast
Toast.makeText(MainActivity.this, "Registered successfully", Toast.LENGTH_SHORT).show();} else {
Toast.makeText(MainActivity.this, "Choose a different email", Toast.LENGTH_SHORT).show();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {

}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
//adding parameters to post request as we need to send firebase id and email
params.put("firebaseid", uniqueId);
params.put("email", email);
return params;
}

и зарегистрируйся

<?php

if($_SERVER['REQUEST_METHOD']=='POST'){

//getting the request values
$firebaseid = $_POST['firebaseid'];
$email  = $_POST['email'];

//connecting to database
$con = mysqli_connect('localhost','root','','pushnotification');

//creating an sql query
$sql = "INSERT INTO register (firebaseid, email) VALUES ('$firebaseid','$email')";

//executing the query to the database
if(mysqli_query($con,$sql)){
echo 'success';
}else{
echo 'failure';
}
}
0

Другие решения

Других решений пока нет …

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