返回结果数组的JavaScript函数

Question

[嗨，我正在使用puppeteer和cheerio在我的网站上运行抓取工具。 Scrapper运行良好，可以正常运行所有操作，但是当我尝试返回这些数组时，它仅返回一个对象而不是返回所有对象，这是我的代码：

    async function mainScraper(url) {
        try {
            const page = await browser.newPage();
            await page.goto(url);
            let html = await page.evaluate(() => document.body.innerHTML);
            const $ = await cheerio.load(html);
            const link = $('.business-name').map((i, elem) =>
                "http://mysiteURL.com" + $(elem).attr("href")).get()
            return link
        } catch (e) {
            console.error(e)
        }
    }

    async function scrapeLinks(url, page) {
        try {
            let results = [];
            await page.goto(url)
            const html = await page.evaluate(() => document.body.innerHTML);
            const $ = cheerio.load(html);
            const title = $(".title").text();
            const websiteURL = $(".link").attr("href");
            let email = $(".email").attr("href");
            const metadata = {
                email: email,
                websiteURL: websiteURL,
                title: title,
            }
            results.push(metadata);
            console.log(results)
            return results
        }
        catch (e) {
            console.log(e)
        }
    }
    async function clientScraper(city, searchTerm, index) {
        browser = await puppeteer.launch({ headless: false });
        const linkPage = await browser.newPage();
        for (i = 1; i <= index; i = index + 1) {
            const list = await mainScraper(`http://mysiteURL.com/${city}/${searchTerm}&page=${index}`)
            for (let i = 0; i < list.length; i++) {
// here when I return it gives only single value and looks like scrapper only scrape once and then stops
                return await scrapeLinks(list[i], linkPage);
            }
        }
    }

Answer 1

这可能是因为您在两个循环中都使用了变量i，并且也可能存在闭包问题。尝试更新两个for循环：

for (let j = 1; j <= index; j = index + 1) {
  const list = await mainScraper(`http://mysiteURL.com/${city}/${searchTerm}&page=${index}`)
  for (let i = 0; i < list.length; i++) {
    (function(x) {
      let result = await scrapeLinks(list[x], linkPage);
      console.log(result) // Debug to see if all results are returned or not
    })(i)
  }
}