为什么我的自定义操作按钮不起作用

Question

因此，我试图通过自定义操作按钮推送通知，但监听器似乎无法正常工作。

importScripts("https://www.gstatic.com/firebasejs/4.12.1/firebase.js");
importScripts("https://www.gstatic.com/firebasejs/4.12.1/firebase-messaging.js");


const config = {
  messagingSenderId: "X"
};

const app = firebase.initializeApp(config);
const MESSAGING = firebase.messaging(app);

self.addEventListener('notificationclick', function(e) {
  var notification = e.notification;
  var action = e.action;
  if(action == "campaign"){
    clients.openWindow('http://www.google.com');
    notification.close();
  }

});

MESSAGING.setBackgroundMessageHandler(function(payload) {
  console.log(
    "[firebase-messaging-sw.js] Received background message ",
    payload
  );
  // Customize notification here
  var notificationTitle = payload.notification.title;
  var notificationOptions = {
    body: payload.notification.body,
    icon: payload.notification.icon,
    image: payload.notification.image,
    click_action: payload.notification.click_action,
    actions: [
      {
        action: payload.notification.actions.action,
        title: payload.notification.actions.title
      }
    ]
  };

  return self.registration.showNotification(
    notificationTitle,
    notificationOptions
  );
});

任何想法为何？单击我的自定义操作按钮后，应该按照事件侦听器的建议将其链接到google.com。