Node.js - Working with an API limit of 5 requests per second

Question

I've got a 'script' that does thousands of requests to a specific API. This API will only allow 5 requests per second (and probably it measures differently then me). To make the requests I'm using request-promise framework, and I've superseded the normal request-promise function with this:

const request_promise  = require('request-promise')

function waitRetryPromise() {
  var count = 0 // keeps count of requests
  function rp(options) {
    const timedCall = (resolve) => setTimeout( ()=>resolve(rp(options)),1000) // recursive call
    count += 1
    if (count % 3 == 0) { // recalls after a second on every third request
      return new Promise(timedCall)
    } else {
      return request_promise(options)
    }
  }
  return rp
}

const rp = waitRetryPromise()

Once around 300 requests (give or take) are fired off in short succession, these requests start to interfere with each other. Does anyone have a better solution? I thought the recursive call to this same function would help, and It did but it didn't solve the problem. Maybe there is a pattern to queue requests, and do them a few at a time? A library perhaps?

Thanks!

Answer 1

OK, rather than recursing the call to rp etc, just make sure you delay between requests by an appropriate amount ... for 5 per second, that's 200ms

function waitRetryPromise() {
    let promise = Promise.resolve();
    return function rp(options) {
        return promise = promise
        .then(() => new Promise(resolve => setTimeout(resolve, 200)))
        .then(() => request_promise(options));
    }
}
const rp = waitRetryPromise();

Answer 2

My code will run the TimedQueue so as long as their is work to be done. The process() method resolves when all work is finished:

class Queue {
    constructor() {
        this.queue = [];
    }

    enqueue(obj) {
        return this.queue.push(obj);
    }

    dequeue() {
        return this.queue.shift();
    }

    hasWork() {
        return (this.queue.length > 0);
    }
}

function t_o(delay) {
    return new Promise(function (resolve, reject) {
        setTimeout(function () {
            resolve();
        }, delay);
    });
}

class TimedQueue extends Queue {
    constructor(delay) {
        super();
        this.delay = delay;
    }

    dequeue() {
        return t_o(this.delay).then(() => {
            return super.dequeue();
        });
    }
    
    process(cb) {
        return this.dequeue().then(data => {
            cb(data);
            if (this.hasWork())
                return this.process(cb);
        });
    }
}


var q = new TimedQueue(500);

for (var request = 0; request < 10; ++request)
    q.enqueue(request);

q.process(console.log).then(function () {
    console.log('done');
});

Answer 3

I am not sure but maybe you get some idea from below

function placeAnOrder(orderNumber) {
    console.log("customer order:", orderNumber)
    cookAndDeliverFood(function () {
        console.log("food delivered order:", orderNumber);
    });
}

//  1 sec need to cook

function cookAndDeliverFood(callback){
    setTimeout(callback, 1000);
}

//users web request
placeAnOrder(1);
placeAnOrder(2);
placeAnOrder(3);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>