Nodejs async/await not working as expected

Question

I have an express API that reads from my fauna database. I have a class setup called Database that handles all my fauna queries, and an async method called getQuotes that essentially gets a fauna document, the general structure is like this:

async getQuotes() {
    // note, `query` is an async function
    const doc = this.client.query(...).then((res) => {
        const data = ...; // to keep this short I'm not gonna show this
        console.log("faunadb handler: " + data); // just for debug
        return Promise.resolve(data);
    })
}

Then when I start the express API, I have it call the getQuotes method and log the data (just for debug).

const quotes = db.getQuotes().then((quotes) => {
    console.log("fauna consumer: " + quotes);
})

Now, when I run the app, I get the following output:

starting server on port.... ussual stuff
fauna consumer: undefined
faunadb handler: { ... }

The fauna consumer code runs before we actually get the promise from the fauna query API. I need to use .then because node for some reason doesn't allow me to use await. Does anyone know how to solve this? Thank you! (using node version v16.14.0, running on Arch Linux)

Does this answer your question? [How to return the response from an asynchronous call](https://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-asynchronous-call) — jabaa, Feb 19 '22 at 18:48
Well, the problem was (credits to @Nick Bailey) that I wasn't really returning anything. I got it messed up with how returns work in local functions in javascript. — FilipToth, Feb 19 '22 at 19:01

score 0 · Accepted Answer · answered Feb 19 '22 at 18:35

You aren't returning anything from your getQuotes() function. The return in your callback is only going to return from the callback, not from the enclosing function. You also aren't awaiting anywhere, which means your async function is not going to actual wait on the result of your query.

Anatoly · Answer 2 · 2022-02-20T14:52:18.453

0

From my perspective it's better not to mix up await and then. Either use one or another.
I'd recommend to go with await.

async getQuotes() {
    // note, `query` is an async function
    const res = await this.client.query(...)
    const doc = ...; // to keep this short I'm not gonna show this
    console.log("faunadb handler: " + doc); // just for debug
    return doc;
}

edited Feb 20 '22 at 14:52

answered Feb 19 '22 at 18:35

Anatoly

20,799
3
28
42

It will work but in such line `then` is redundant., you can simply do `const res = await f()` – Anatoly Feb 19 '22 at 18:55
In case of `then` you decrease readability by adding a new level of indentation and nested function. It's much better to have two straightforward lines (even with an extra variable). Technically you also have an extra variable `g` so there no benefit in that. – Anatoly Feb 20 '22 at 08:17

score 0 · Answer 3 · answered Feb 19 '22 at 18:37

try to use await

async getQuotes() {
    const res = await this.client.query(...);
    const data = ...;
    console.log("faunadb handler: " + data); // just for debug
    return data;
}

// must be in async function
const quotes = await db.getQuotes();
console.log("fauna consumer: " + quotes);

score 0 · Answer 4 · answered Feb 19 '22 at 18:38

The problem is that you need to return data fromgetQuotesPlease just change the code as follows (replace the placeholders):

  async getQuotes() {
    // note, `query` is an async function
    const res = await this.client.query(...)
    const data = ...; // to keep this short I'm not gonna show this
    console.log("faunadb handler: " + data); // just for debug
    return data
  }

score 0 · Answer 5 · answered Feb 19 '22 at 18:41

getQuotes function should return the result like this.

async getQuotes() {
    // note, `query` is an async function
    return this.client.query(...).then((res) => {
        const data = ...; // to keep this short I'm not gonna show this
        console.log("faunadb handler: " + data); // just for debug
        return Promise.resolve(data);
    })
}

or

async getQuotes() {
    // note, `query` is an async function
    const doc = await this.client.query(...);
    const data = ...; // to keep this short I'm not gonna show this
    console.log("faunadb handler: " + data); // just for debug
    return Promise.resolve(data);
}

Nodejs async/await not working as expected

5 Answers5