Why does await is not working in my function?

Question

I am trying to push in an array all the ids of objects that match my condition.

I am using a recursive function and I want to return a value when my recursive function has finished.

Here is my code :

const getFamilies = async function (family, famillesReliees) {
    await strapi.query('famille-de-materiel').findOne({ id: family })
        .then(res => {
            if (res.cats_enfant.length > 0) {
                res.cats_enfant.forEach(async enfant => {
                    famillesReliees = await getFamilies(enfant.id, famillesReliees)
                })
            } else {
                famillesReliees.push(res.id)
            }
        })
        return famillesReliees
}

async search() {
    let value = await getFamilies(2, [])
    return value
}

I don't understand why the "value" return before the end of the recursive function

`.then(console.log('getFamilies then'))` -> `.then(() => console.log('getFamilies then'))` — VLAZ, Feb 01 '21 at 18:13
You're *immediately* calling `console.log` and assing its return value as the `then` function. — Quentin, Feb 01 '21 at 18:15
IN additon to the above, you're never awaiting the recursive call to `getFamilies` — Jamiec, Feb 01 '21 at 18:15
The inner `getFamilies(enfant.id)` needs to be returned in a `Promise.all` for the outer `.then` to wait for it to finish — CertainPerformance, Feb 01 '21 at 18:16
Also, what is `famillesReliees`? If it's a global array you're sooner or later going to run into the problems associated with trying to return a value from an async function — Jamiec, Feb 01 '21 at 18:17

score 1 · Answer 1 · answered Feb 01 '21 at 18:15

This isn't doing what you think:

getFamilies(2).then(console.log('getFamilies then'))

This executes console.log immediately and passes its result (which is undefined) as the function to be executed after the getFamilies operation.

Wrap the console.log operation in a function to be executed later:

getFamilies(2).then(() => console.log('getFamilies then'))

Structurally, this is just like your use of .then() in the getFamilies function above. The only differences are:

There are no parameters/arguments for the function, so just use () instead of a variable name. (Your callback function above has a parameter called res.)
There's only one line in the function, so the curly braces {} aren't needed. (Technically this also returns the result of that line, but in this case the returned result is undefined and nothing uses it so no harm done.)

Why does await is not working in my function?

1 Answers1