Question

我具有以下功能

const getQuotes = symbol => {
  let quotes = {};
  new DeltaRestClient(api_key, api_secret).then(client => {
    const linkOptions = createIdentifiers(symbol, false);
    Object.entries(linkOptions).forEach(entry => {
      client.apis.Products.getTicker({ symbol: entry[1] }).then(response => {
        const ticker = JSON.parse(response.data.toString());
        quotes[entry[0]] = parseFloat(ticker.result.close);
      });
    });
  });
  return quotes;
};

我叫

const start = async () => {
  const quotes = await getQuotes("ABCD");
  console.log(quotes);
};

但是由于异步性，getQuotes在解析所有.then并返回空对象之前返回。

如何更改此值，以便仅在所有.then都解析后才返回值？

Answer 1

您还必须等待正在调用的函数：

const getQuotes = async symbol => {
  let quotes = {};
  const client = await new DeltaRestClient(api_key, api_secret);
  const linkOptions = createIdentifiers(symbol, false);

  for (const entry of Object.entries(linkOptions)) {
      const response = await client.apis.Products.getTicker({ symbol: entry[1] });
      const ticker = JSON.parse(response.data.toString());
      quotes[entry[0]] = parseFloat(ticker.result.close);
  }
  return quotes;
};

并相应地调用它：

const start = async () => {
  const quotes = await getQuotes("ABCD");
  console.log(quotes);
};

通常将async / await和Promise混合使用。然后/.catch导致容易被误解的可疑代码。

如果愿意，可以通过分解entry元素来提高可读性：

  for (const [key, symbol] of Object.entries(linkOptions)) {
      const response = await client.apis.Products.getTicker({ symbol, });
      const ticker = JSON.parse(response.data.toString());
      quotes[key] = parseFloat(ticker.result.close);
  }

如何等待所有.then完成，然后返回值

1 个答案: