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javascript - Javascript：Promise 实现中的错误

问题描述

我正在尝试执行异步功能，然后在Promise的帮助下控制台记录结果。恐怕我还没有完全掌握这个概念。

getlinks performs async action.

async function getLinks(param, data) {
  return new Promise((resolve, reject) => {
    let psub;
    var name;
    let g;

    psub = checkForPsub(param);
    var ultUrls = [];

    _.each(data, o => {
      title = sanitizeString(o.title);
      if (psub == true) {
        name = title + " u -- " + o.author;
      } else {
        name = title;
      }

      switch (o.domain) {
        case "i.imgur.com":
          {
            // downloadImgur(o,name)
          }
          break;
        case "imgur.com":
          {
            id = o.url.substring(o.url.lastIndexOf("/") + 1);
            if (
              o.url.includes("https://imgur.com/a/") ||
              o.url.includes("https://imgur.com/gallery/") ||
              o.url.includes("http://imgur.com/a/") ||
              o.url.includes("http://imgur.com/gallery/")
            ) {
              let urls = [];
              let file_name;
              axios
                .get(
                  "https://api.imgur.com/3/album/" + id,

                  { headers: { Authorization: "Client-ID 295ebd07bdc0ae8" } }
                )
                .then(res => {
                  let images = res.data.data.images;

                  _.each(images, function(v) {
                    var ext = v.link.split(".").pop();
                    if (ext == "gifv") {
                      ext = "mp4";
                    }
                    if (psub == true) {
                      file_name =
                        title + "--" + v.id + " " + "u--" + auth + "." + ext;
                    } else {
                      file_name = title + "--" + v.id + "." + ext;
                    }

                    let p = { url: v.link, file_name: file_name };
                    ultUrls.push(p);
                  });
                })
                .catch(err => {
                  console.log(err);
                });
            }
          }
          break;
        case "i.redd.it":
          {
          }
          break;
        default:
          console.log("other", o.domain);
      }
    }); //end each

    return resolve(ultUrls);
  });
}

我想等到 getlinks 完成执行任务，然后控制台记录结果。

 getLinks(sub,result).then(res =>  console.log({res}))

但是即使在 getlink 完成之前，它也会将结果记录为空。

标签： javascriptasynchronouspromise

解决方案

最简单的答案是您承诺在异步代码 ( ) 完成之前return resolve(utlUrls)解析 ( ) 。axios.get(...).then(...)

这是重现您的问题的最小示例：

let timeout = ms => new Promise(resolve => setTimeout(() => resolve(ms), ms));

async function getLinks(urls) {
  return new Promise((resolve, reject) => {
    let ultUrls = [];
    urls.forEach(url =>
        timeout(500).then(res => ultUrls.push(res)))
    return resolve(ultUrls);
  });
}

getLinks([1, 2, 3]).then(a => console.log(a));

它不起作用，因为我们在填充它之前返回了 ultUrls。我们不等待超时完成。

要解决此问题，只需使用Promise.all. 另外删除一些不必要的承诺包装，我们得到：

let timeout = ms => new Promise(resolve => setTimeout(() => resolve(ms), ms));

function getLinks(urls) {
  let ultUrls = [];
  let promises = urls.map(url =>
      timeout(500).then(res => ultUrls.push(res)))
  return Promise.all(promises).then(a => ultUrls);
}

getLinks([1, 2, 3]).then(a => console.log(a));

此外，如果您想使用该async/await语法，尽管在您有多个并行请求的情况下它不会给您带来太多好处，您可以将其编写为：

let timeout = ms => new Promise(resolve => setTimeout(() => resolve(ms), ms));

async function getLinks(urls) {
  let ultUrls = [];
  let promises = urls.map(url =>
      timeout(500).then(res => ultUrls.push(res)))
  await Promise.all(promises);
  return ultUrls;
}

getLinks([1, 2, 3]).then(a => console.log(a));

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