在firebase云函数中发送推送通知时出错

Question

我正在尝试使用Firebase云功能创建消息推送。以下是我的代码：

const admin = firebase
exports.sendNewTripNotification = functions.firestore
  .document('notifications/{notificationId}')
  .onWrite(async (event) => {
    console.log("event: ", event)
    const key = "cK0e5ZRAVRUAlx8hXHkNkXp7mj43"
    const ref = admin.database().ref(`users/${key}/deviceToken`);
    console.log("ref: ", ref)
    const deviceToken = admin.database().ref(`/users/${key}/deviceToken`).once('value');
    return deviceToken.then(result => {
      const token_id = result.val();
      console.log("token_id", token_id)
      const payload = {
        notification: {
          title: "Friend Request",
          body: "You just got a new friend request",
          icon: "default"
        }
      };
      console.log("payload", payload)
      return admin.messaging().sendToDevice(token_id, payload).then(Response => {
        console.log('this is the notification',Response)
      });
    });
  })

我得到以下错误：

错误：提供给sendToDevice（）的注册令牌必须是非空字符串或非空数组。在FirebaseMessagingeror。FirebaseError[作为构造函数]（/srv/node_modules/firebase admin/lib/utils/error.js:42:28）位于FirebaseMessagingError。PrefixedFirebaseError[作为构造函数]（/srv/node_modules/firebase admin/lib/utils/error.js:88:28）

我确实参考了以下两个链接，但没有得到任何帮助。我做错什么了吗？

Firebase Cloud Functions在创建实时数据库时获取数据

Firebase Cloud Functions代币

Answer 1

问题已修复。我犯了一个愚蠢的错误。我没有在登录另一台设备时覆盖设备令牌。

Answer 2

您将async/await与then（）混合使用，这是不推荐的。以下内容应该可以做到（未经测试）：

exports.sendNewTripNotification = functions.firestore
    .document('notifications/{notificationId}')
    .onWrite(async (event) => {
        console.log("event: ", event)

        const key = "cK0e5ZRAVRUAlx8hXHkNkXp7mj43"
        const ref = admin.database().ref(`users/${key}/deviceToken`);
        console.log("ref: ", ref)

        const deviceToken = await ref.once('value');
        const token_id = deviceToken.val();

        console.log("token_id", token_id)

        const payload = {
            notification: {
                title: "Friend Request",
                body: "You just got a new friend request",
                icon: "default"
            }
        };

        const messagingDevicesResponse = await admin.messaging().sendToDevice(token_id, payload);

        console.log('this is the notification', messagingDevicesResponse)

        return null;

    })