How to return value in sequential forEach loop?

Question

This is how I'm using a sequential forEach loop to handle some file uploads. Now I want to get a total result for each upload: For each file I resolve with an object (success or error). At the end I want to get a returned array to display the result (File 1 successful, file 2 error and so on).

But with this code I only get undefined for the last output. What am I missing?

Maybe it would even be better to return two string arrays: one successful and one failed array with filenames.

Array.prototype.forEachSequential = async function (
    func: (item: any) => Promise<void>
): Promise<void> {
    for (let item of this) await func(item)
}
async uploadFiles(files: [File]): Promise<any> {
    const result = await files.forEachSequential(
        async (file): Promise<any> => {
        const res = await new Promise(async (resolve, reject) => {
            // do some stuff
            resolve({ success: file.filename })
            // or maybe
            resolve({ error: file.filename })
        })
        console.log(res) // returns object as expected
        return res
        }
    )
    // Get all results after running forEachSequential
    console.log('result', result) // result: undefined
})

Answer 1

Your forEachSequential doesn't really return anything. That's why you get undefined. When awaiting, you should attach a result to some variable, store it somewhere and then at the end of this function you should return it.

Now, you would like to return a tuple [string[], string[]]. To do so you have to create to arrays at the beginning of forEachSequential body, then call func in the loop and add a result to first array (let's say - success array) or second (failure array).

const successArray = [];
const failureArray = [];

for (let item of this) {
    const result = await func(item);

    if ('success' in result) {
        successArray.push(result.success);
    } else {
        failureArray.push(result.error);
    }
}

return [successArray, failureArray];