Node.js - 使用每秒5个请求的API限制

时间:2017-10-12 04:02:48

标签: javascript node.js promise request-promise

我有一个'脚本'这会对特定API执行数千次请求。这个API每秒只允许5个请求(可能它的测量方式与我不同)。要使用request-promise框架发出请求,我已经用这个取代了正常的request-promise函数:

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()

一旦大约300个请求(发出或接受)短时间内被触发,这些请求就会开始相互干扰。有没有人有更好的解决方案?我认为对这个函数的递归调用会有所帮助,而且确实如此,但它并没有解决问题。也许有一种模式可以对请求进行排队,并且一次只能执行一些操作?或许图书馆?

谢谢!

3 个答案:

答案 0 :(得分:3)

我的代码将运行TimedQueue,只要他们的工作要完成。所有工作完成后process()方法解决:



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');
});
&#13;
&#13;
&#13;

答案 1 :(得分:3)

好的,不是递归调用rp等,而是确保你在请求之间延迟适当的数量......每秒5次,即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();

答案 2 :(得分:-1)

我不确定,但也许你从下面得到一些想法

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>