我有一个向API发出请求的请求-承诺函数。我受此 API 的速率限制,并且不断收到错误消息:
Exceeded 2 calls per second for api client. Reduce request rates to resume uninterrupted service.
我正在并行运行几个Promise.each
循环,这导致了问题,如果我只运行Promise.each
的一个实例,一切正常。在这些Promise.each
调用中,它们通过请求-承诺调用导致相同的函数 a。我想用另一个queue
函数包装这个函数,并将间隔设置为 500
毫秒,这样request
就不会在队列上彼此或并行,而是设置为该时间。 问题是我仍然需要这些承诺来获得它们的内容,即使需要相当长的时间才能得到回应。
有什么可以为我做到这一点吗?我可以将函数包装进去的东西,它会以设定的间隔而不是并行响应或一个接一个地触发函数?
更新:也许它确实需要特定于承诺,我尝试使用下划线的油门功能
var debug = require("debug")("throttle")
var _ = require("underscore")
var request = require("request-promise")
function requestSite(){
debug("request started")
function throttleRequest(){
return request({
"url": "https://www.google.com"
}).then(function(response){
debug("request finished")
})
}
return _.throttle(throttleRequest, 100)
}
requestSite()
requestSite()
requestSite()
我得到的只是这个:
$ DEBUG=* node throttle.js
throttle request started +0ms
throttle request started +2ms
throttle request started +0ms
更新
最后一个答案是错误的,这有效,但我仍然认为我可以做得更好:
// call fn at most count times per delay.
const debounce = function (fn, delay, count) {
let working = 0, queue = [];
function work() {
if ((queue.length === 0) || (working === count)) return;
working++;
Promise.delay(delay).tap(() => working--).then(work);
let {context, args, resolve} = queue.shift();
resolve(fn.apply(context, args));
}
return function debounced() {
return new Promise(resolve => {
queue.push({context: this, args: arguments, resolve});
if (working < count) work();
});
};
};
function mockRequest() {
console.log("making request");
return Promise.delay(Math.random() * 100);
}
var bounced = debounce(mockRequest, 800, 5);
for (var i = 0; i < 5; i++) bounced();
setTimeout(function(){
for (var i = 0; i < 20; i++) bounced();
},2000);
因此,您需要使请求在功能范围内受到限制 - 这很好。承诺几乎内置了排队功能。
var p = Promise.resolve(); // our queue
function makeRequest(){
p = p.then(function(){ // queue the promise, wait for the queue
return request("http://www.google.com");
});
var p2 = p; // get a local reference to the promise
// add 1000 ms delay to queue so the next caller has to wait
p = p.delay(1000);
return p2;
};
现在,makeRequest 调用将至少相隔 1000 毫秒。
jfriend 指出,您每秒需要两个请求,而不是一个 - 这同样可以通过第二个队列轻松解决:
var p = Promise.resolve(1); // our first queue
var p2 = Promise.resolve(2); // our second queue
function makeRequest(){
var turn = Promise.any([p, p2]).then(function(val){
// add 1000 ms delay to queue so the next caller has to wait
// here we wait for the request too although that's not really needed,
// check both options out and decide which works better in your case
if(val === 1){
p = p.return(turn).delay(1, 1000);
} else {
p2 = p2.return(turn).delay(1, 1000);
}
return request("http://www.google.com");
});
return turn; // return the actual promise
};
这可以推广到类似地使用数组n
承诺