每个then（）应该返回一个值或抛出Firebase云函数

Question

我正在使用javascript为firebase写云函数，但是我被困住了，我不知道错误的确切含义并且无法解决。.错误状态：27:65错误每个then（）应该返回一个值或抛出承诺/总是回报

'use strict'

const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp();

exports.sendNotification = functions.database.ref('/notifications/{user_id}/{notification_id}').onWrite((change, context) => {

    const user_id = context.params.user_id;
    const notification_id = context.params.notification_id;
    console.log('We have a notification from : ', user_id);

    if (!change.after.val()) {
        return console.log('A Notification has been deleted from the database : ', notification_id);
    }
    const deviceToken = admin.database().ref(`/ServiceProvider/${user_id}/device_token`).once('value');
    return deviceToken.then(result => {
        const token_id = result.val();
        const payload = {
            notification: {
              title : "New Friend Request",
              body: "You Have Received A new Friend Request",
              icon: "default"
            }
        };

        return admin.messaging().sendToDevice(token_id, payload).then(response => {

            console.log('This was the notification Feature');

        });

    });

});

Answer 1

更改此：

    return admin.messaging().sendToDevice(token_id, payload).then(response => {

        console.log('This was the notification Feature');

    });

对此：

    return admin.messaging().sendToDevice(token_id, payload).then(response => {

        console.log('This was the notification Feature');
        return null;   // add this line

    });

该then回调只需要返回一个值。

但是，eslint可能随后抱怨嵌套then()在您的代码中，这也是一种反模式。您的代码实际上应该更像这样构造：

const deviceToken = admin.database().ref(`/ServiceProvider/${user_id}/device_token`).once('value');
return deviceToken.then(result => {
    // redacted stuff...
    return admin.messaging().sendToDevice(token_id, payload);
}).then(() => {
    console.log('This was the notification Feature');
});

请注意，它们彼此之间是相互链接的，而不是彼此嵌套。