async/await issue (returns undefined) mongoose

Question

When I call the method interval, the parameter user, which I get from the method allUserStats, is always undefined. Here is the method

const interval = async () => {
    const allUsers = await getAllUser();
    for (const u of allUsers) {
            let user;
            let followerScore;
            let subscriberScore;
            //let hostsScore;
            let viewerScore;
            await getAllUserStats(u.login_name).then(async (userStats)=>{
                user = userStats;
...

The method getAllUserStats looks like this:

const getAllUserStats = async (login_name) => {
    await userstats.findOne({login_name}).then(async(userStats)=>{
        if(userStats===null){
            let user = new userstats({
                login_name:login_name,
                followers: [],
                subscribers: [],
                hosts: [],
                viewers: []
            })
            await user.save();
            return user;
        }
        else{
            return userStats;
        }
    })
}

I think it is an async-await issue, but I dont know. Any advice how I can fix that?

Have you tried returning something from the getAllUserStats function? eg return await userstats... — badsyntax
– badsyntax, Commented Jan 15, 2021 at 13:03
Yeah, getAllUserStats doesn't return anything. You can either a) return a Promise/then-chain and await the function call or b) await the result, then actually return it. You are still trying to return from a callback function, which will simply discard the value. A working return has to be directly inside the function, it cannot be wrapped with a callback function. — user5734311
– user5734311, Commented Jan 15, 2021 at 13:06
@ChrisG thank you, now it works. Could you add your comment as an answer, so I can accept it and close the thread? — JackJackson12
– JackJackson12, Commented Jan 15, 2021 at 13:09
I've shown @ChrisG comment (B) option might look like. That's generally the best option -- if you're using async/await, go for it and use it! forget .then — TKoL
– TKoL, Commented Jan 15, 2021 at 13:12

TKoL · Accepted Answer · 2021-01-15 13:11:30Z

1

You're mixing async/await and using promise.then -- you can just write it all with async/await that should make it easier:

const getAllUserStats = async (login_name) => {
    const userStats = await userstats.findOne({login_name})
    if(userStats===null){
        let user = new userstats({
            login_name:login_name,
            followers: [],
            subscribers: [],
            hosts: [],
            viewers: []
        })
        await user.save();
        return user;
    }
    else{
        return userStats;
    }
}

answered Jan 15, 2021 at 13:11

TKoL

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1 Comment

user5734311 Over a year ago

Although not always good practice, combining await and then is usually fine (for instance awaiting a fetch call with a chained .json()). However this question is about a single issue: OP simply forgot to return something from their function. It's primarily a dupe of the most linked dupe of all times: stackoverflow.com/questions/14220321/…

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