无法从Node.js中的异步函数返回值,但控制台会打印值

时间:2020-06-07 20:44:02

标签: node.js callback async.js request-promise

我是node.js的新手,我阅读了一些有关此问题的文章,但仍然不了解如何解决此问题。正确的结果已打印出来,但我希望将其返回或存储在变量中。这可能吗?

    const values = async () =>
{
  const URL = 'http://www.espn.com/sports/tennis/rankings';
  const response = await request(URL);
  const $ = cheerio.load(response);
  let title = $ ('#my-players-table > div.mod-container.mod-table > div.mod-content > table > tbody').text();
  var rank = 1;
  var count = title.length;
  var words = title;
  var player = "";
  var players = [];
  for ( var i = 70; i < count; i++)
  {
    if(!characterNumber(words[i]) && words[i] != "\n")
    {
      player += words[i];
    }
    else
    {
      if(player != "" && player != '\n')
      {
        players.push(player);
        player = "";
      }
    }
  }
return players;
};
let playersRankings;
const getRankings = async () => {
 playersRankings = await values();
 console.log(playersRankings);
}
getRankings();

0 个答案:

没有答案
相关问题