Node.js function async return not awaiting

Question

I think I am having a problem understanding the await/promises in Node.js, I thought I was implementing it correctly but it does not look right...

I have this function to get a list of files from my Google Drive:

const listFiles = async () => {
    const filesList = await googleDrive.listFiles();
    filesList.forEach((file)=>{
        console.log(`File is ${file.name}`);
    });
    return filesList;
  }

This function works fine, but now I tried to call it like this in my main.js:

const listFiles = async () => {
    const filesList = await googleDrive.listFiles();
    filesList.forEach((file)=>{
        console.log(`File is ${file.name} with id`);
    });
    return filesList;
  }

const getFiles =() =>{
    const files = listFiles();
    console.log(files);
};


getFiles();

So my problem here is that, from getFiles()I always get Promise { <pending> } as a console.log...but in my listFiles(), I can see the files being printed correctly after the await....I do not really get it, after the await, the filesList should be ready and resolved.

What am I doing wrong here?

hgb123 · Accepted Answer · 2020-12-05 16:19:09Z

2

async function returns a promise, so you still have to await it

Return value

A Promise which will be resolved with the value returned by the async function, or rejected with an exception thrown from, or uncaught within, the async function.

const listFiles = async () => {
    const filesList = await googleDrive.listFiles();
    filesList.forEach((file)=>{
        console.log(`File is ${file.name} with id`);
    });
    return filesList;
  }

const getFiles = async () =>{
    const files = await listFiles();
    console.log(files);
};


getFiles();

answered Dec 5, 2020 at 16:19

hgb123

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3 Comments

codeKiller Over a year ago

ok thanks, now I guess getFiles also will return a promise, and again when i call it i will have to await, and over and over again, correct?

hgb123 Over a year ago

@codeKiller yeah, if in this case there is no task need to be executed sequentially after getFiles(), you could still call it this way

Rupjyoti Over a year ago

Use .then & .catch when calling getFiles() . This is what I follow , all async await code , but the last one .then & .catch . If you don't catch the error , you will get "unhandled promise rejection" error when ever you have any error inside getFiles function.

A McBride · Accepted Answer · 2020-12-05 16:21:52Z

1

listFiles is correctly listed as being async, but in this case, getFiles should be async too - it needs to await the results of listFiles.

Your code should then look as follows:

const listFiles = async () => {
    const filesList = await googleDrive.listFiles();
    filesList.forEach((file)=>{
        console.log(`File is ${file.name} with id`);
    });
    return filesList;
  }

const getFiles = async () =>{
    const files = await listFiles();
    console.log(files);
};


getFiles();

answered Dec 5, 2020 at 16:21

A McBride

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