async函数+ await + setTimeout的组合 [英] Combination of async function + await + setTimeout

问题描述

我正在尝试使用新的异步功能，我希望解决我的问题将来会帮助其他人。这是我的代码正常工作：

I am trying to use the new async features and I hope solving my problem will help others in the future. This is my code which is working:

  async function asyncGenerator() {
    // other code
    while (goOn) {
      // other code
      var fileList = await listFiles(nextPageToken);
      var parents = await requestParents(fileList);
      // other code
    }
    // other code
  }

  function listFiles(token) {
    return gapi.client.drive.files.list({
      'maxResults': sizeResults,
      'pageToken': token,
      'q': query
    });
  }

问题是，我的while循环运行得太快而且脚本也发送了每秒向google API发出的许多请求。因此，我想建立一个延迟请求的睡眠功能。因此我也可以使用此函数来延迟其他请求。如果还有另一种方法可以延迟请求，请告诉我。

The problem is, that my while loop runs too fast and the script sends too many requests per second to the google API. Therefore I would like to build a sleep function which delays the request. Thus I could also use this function to delay other requests. If there is another way to delay the request, please let me know.

无论如何，这是我的新代码无效。请求的响应返回到setTimeout中的匿名异步函数，但我只是不知道如何将响应返回给sleep函数resp。到初始的asyncGenerator函数。

Anyway, this is my new code which does not work. The response of the request is returned to the anonymous async function within the setTimeout, but I just do not know how I can return the response to the sleep function resp. to the initial asyncGenerator function.

  async function asyncGenerator() {
    // other code
    while (goOn) {
      // other code
      var fileList = await sleep(listFiles, nextPageToken);
      var parents = await requestParents(fileList);
      // other code
    }
    // other code
  }

  function listFiles(token) {
    return gapi.client.drive.files.list({
      'maxResults': sizeResults,
      'pageToken': token,
      'q': query
    });
  }

  async function sleep(fn, par) {
    return await setTimeout(async function() {
      await fn(par);
    }, 3000, fn, par);
  }

我已经尝试了一些选项：将响应存储在全局变量中并返回它来自sleep函数，匿名函数内的回调等等。

I have already tried some options: storing the response in a global variable and return it from the sleep function, callback within the anonymous function, etc.

推荐答案

你的睡眠函数不起作用，因为 setTimeout 没有（还是？）返回一个可能 await ed的承诺。你需要手动宣传它：

Your sleep function does not work because setTimeout does not (yet?) return a promise that could be awaited. You will need to promisify it manually:

function timeout(ms) {
    return new Promise(resolve => setTimeout(resolve, ms));
}
async function sleep(fn, ...args) {
    await timeout(3000);
    return fn(...args);
}

顺便说一句，为了减慢循环你可能不想使用 sleep 接受回调并按此方式推迟的函数。我宁愿建议做类似的事情

Btw, to slow down your loop you probably don't want to use a sleep function that takes a callback and defers it like this. I'd rather recommend to do something like

while (goOn) {
  // other code
  var [parents] = await Promise.all([
      listFiles(nextPageToken).then(requestParents),
      timeout(5000)
  ]);
  // other code
}

可以计算父母至少需要5秒钟。

which lets the computation of parents take at least 5 seconds.

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