每個 then() 都應該返回一個值或拋出 Firebase 雲函數

Question

我正在使用 javascript 為 firebase 編寫一個雲函數，但我被卡住了，我不知道錯誤的確切含義並且無法解決它..錯誤狀態：27:65 錯誤每個 then() 應該返回一個值或拋出承諾/總是回報

'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

改變這個：

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

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

    });

對此：

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

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

    });

then回調只需要返回一個值。

但是，eslint 可能會在您的代碼中抱怨嵌套then() ，這也是一種反模式。 你的代碼應該更像這樣的結構：

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');
});

請注意，每個然后相互鏈接，而不是相互嵌套。

Answer 2

改變這個：

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

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

  });

進入這個：

    return admin.messaging().sendToDevice(token_id, payload).then(response=>{
      console.log('This was the notification Feature');
      return true;
    },err=>
    {
      throw err;
    });

由於錯誤使用時，說then你需要返回一個值。

Answer 3

這是jslinting告訴你，每一個.then必須包括返回值。 換句話說，避免promise 反模式。

您可能會發現async函數更容易理解。 請注意，您需要運行 Node 8 運行時以獲得異步支持...