Promise.all in [removed] How to get resolve value for all promises?

Question

I wrote the following node.js file:

var csv = require(\'csv-parser\');
var fs = require(\'fs\')
var Promise = require(\'bluebird\');
var filename = \"device         


        

Answer 1

First question

Promise.all takes an array of promises

Change:

Promise.all(read_csv_file("devices.csv"), read_csv_file("bugs.csv"))

to (add [] around arguments)

Promise.all([read_csv_file("devices.csv"), read_csv_file("bugs.csv")])
// ---------^-------------------------------------------------------^

Second question

The Promise.all resolves with an array of results for each of the promises you passed into it.

This means you can extract the results into variables like:

Promise.all([read_csv_file("devices.csv"), read_csv_file("bugs.csv")])
  .then(function(results) {
    var first = results[0];  // contents of the first csv file
    var second = results[1]; // contents of the second csv file
  });

You can use ES6+ destructuring to further simplify the code:

Promise.all([read_csv_file("devices.csv"), read_csv_file("bugs.csv")])
  .then(function([first, second]) {

  });