Dnyaneshwar Wakchaure

Question

这是我发送的json但是我只在IOS中接收前台通知但不在后台接收通知。

remoteMessage.appData {          colorCode = XXXXX;         description = "XXXXX";     from = XXXXX;     notification =     {         body = "XXXXX";         e=1;     };     notificationName = "XXXXX";     notificationType = XXXXX;     outbid = XXXXX;     paused = XXXXX;     sound = "XXXXX.wav";     suspended = XXXXX; }

这是我用来生成上面json的php代码：

public function sendPushNotification（$ registration_ids，array $ notification，array $ message = null）{

       $url = 'https://fcm.googleapis.com/fcm/send';
    $notification['notification']['sound'] = $this->_notificationSoundFile;
    $fields = array(
        'registration_ids' => array($registration_ids),
        'notification' =>  $notification['notification'],
        'data' =>  $message['message']
    );
    $headers = array(
        'Authorization:key=' . $this->_fcmKey,
        'Content-Type: application/json'
    );


    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($fields));

    $result = curl_exec($ch);


    if ($result === false)
        throw new Exception('Curl failed ' . curl_error());

    curl_close($ch);
    if ($result) {
        return json_decode($result, true)['success'] == true ? true : false;
    }
} :

    enter code here

Expected Output :

{
    aps =     {
        alert =         {
            body = "XXXX";
            title = "";
        };        
        sound = "XXXX.wav";
    };
    "gcm.message_id" = "XXXX";
    "gcm.notification.appointmentId" = XXXX;
    "gcm.notification.carCode" = XXXX;
    "gcm.notification.deal_lost" = XXXX;
    "gcm.notification.dealerCode" = XXXX;
    "gcm.notification.notificationName" = "XXXX";
    "gcm.notification.notificationType" = XXXX;
    "gcm.notification.outbid" = XXXX;
    "gcm.notification.paused" = XXXX;
    "gcm.notification.suspended" = XXXX;
}

Help appreciated.

Answer 1

我已在目标C的app Delegate中实现了以下方法

Dnyaneshwar Wakchaure

- (void)applicationReceivedRemoteMessage:(FIRMessagingRemoteMessage *)remoteMessage {
}
- (void)userNotificationCenter:(UNUserNotificationCenter *)center willPresentNotification:(UNNotification *)notification
         withCompletionHandler:(void (^)(UNNotificationPresentationOptions))completionHandler {
}
- (void)userNotificationCenter:(UNUserNotificationCenter *)center didReceiveNotificationResponse:(UNNotificationResponse *)response
         withCompletionHandler:(void (^)())completionHandler {
}
- (void) application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo {
}
- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo fetchCompletionHandler:(void (^)(UIBackgroundFetchResult))completionHandler {
}

Answer 2

您应该检查 content_available 标记。这是Apple自动转换为aps-＆gt;警报的标准FCM有效负载（用于后台通知处理）。

{
   "notification" : {
      "title": "XXX",
      "body" : "xxx",
      "title": "xxx",
      "content_available": 1
   },
   "data" : {
       //contain the payload
   }
}

仅供参考：FCM guidelines to send notification

Answer 3

在notification = {alert：＆＃34;＆＃34;}

中

警报应该在那里。还有一件事就是在appdelegate类中实现了backgroundfetch处理程序方法。

func application（_ application：UIApplication，didReceiveRemoteNotification userInfo：[AnyHashable：Any]，fetchCompletionHandler completionHandler：@escaping（UIBackgroundFetchResult） - ＆gt; Void）{

} 希望它能奏效。

我没有在IOS的后台接收fcm推送通知

3 个答案:

Dnyaneshwar Wakchaure