ios - 当应用程序在后台运行或关闭时，Flutter Firebase Messaging 在 IOS 上不起作用

Question

我已经配置了带有 flutter 通知的firebase云消息传递在前台工作。但在后台运行或应用程序被杀死时不起作用。

完成以下步骤。

生成app ID并push Key
https://developer.apple.com/account/resources/certificates/

将 info.plist 添加到 ios/Runner 文件夹

开启推送功能

已将 APNS 证书上传到 Firebase 控制台

在 AppDelegate.swift

中添加了行

还尝试按照 Flutter Firebase Cloud Messaging - Notification when app in background 上的建议删除以下行但它仍然无法正常工作。

if (@available(iOS 10.0, *)) {
[UNUserNotificationCenter currentNotificationCenter].delegate = (id) self;
}

Flutter Doctor

[✓] Flutter (Channel beta, v1.12.13+hotfix.6, on Mac OS X 10.14.5 18F132, locale en-IN)

[✗] Android toolchain - develop for Android devices

    ✗ Unable to locate Android SDK.

      Install Android Studio from: https://developer.android.com/studio/index.html

      On first launch it will assist you in installing the Android SDK components.

      (or visit https://flutter.dev/setup/#android-setup for detailed instructions).

      If the Android SDK has been installed to a custom location, set ANDROID_HOME to that location.

      You may also want to add it to your PATH environment variable.



[✓] Xcode - develop for iOS and macOS (Xcode 11.3)

[✓] Chrome - develop for the web

[!] Android Studio (not installed)

[✓] Connected device (3 available)

我注意到一件事，当第一次安装应用程序时，它并没有要求我检查我是否允许应用程序发送推送通知。

登录页面后，我的代码在页面上有以下行。

_firebaseMessaging.requestNotificationPermissions(
        const IosNotificationSettings(sound: true, badge: true, alert: true));
    _firebaseMessaging.onIosSettingsRegistered
        .listen((IosNotificationSettings settings) {
      print("Settings registered: $settings");
    });

还交叉检查是否勾选了所有以下设置。

Answer 1

经过很长时间寻找修复程序，这为我解决了问题:

在您的 info.plist 中:

 FirebaseAppDelegateProxyEnabled: false

将此添加到您的 AppDelegate:

override func application(_ application: UIApplication, didRegisterForRemoteNotificationsWithDeviceToken deviceToken: Data) {

   Messaging.messaging().apnsToken = deviceToken
   super.application(application, didRegisterForRemoteNotificationsWithDeviceToken: deviceToken)
 }